geometry harry lewis

G EO M ET R Y

GE OME T R YA C O N TEM PO R A R Y COURSE

I M

PROGRAM IN

ODER N ATHEMATICS

M A TH E M A TIC S — A MODERN APPROACH, Book 1, Second Edition Peters, Schaaf

M A TH E M A TIC S — A MODERN APPROACH, Book 2, Second Edition Peters, Schaaf

A L G E B R A -A MODERN APPROACH, Book 1, Second Edition Peters, Schaaf

ALGEBRA AND TRIGONOMETRY— A MODERN APPROACH, Book 2 Peters, SchaafGEOM ETRY— A CONTEMPORARY COURSE, Second Edition Lewis

D r. H arry Lewis is principal o f Arts High School, Newark, New Jersey, He •was formerly the chairman uf the Mathematics Department of East Side High School of Newark, having taught mathematics for many years in the Newark Public School System. He is the coauthor of textbooks on business mathematics an d has taught at the New York University School o f Education.

V a n N o s t r a n d R e g io n a l O f f ic e s : New York, Chicago, San Francisco

D . V a n N o s t r a n d C o m p a n y , L t d . , London

D . V a n N o s t r a n d C o m p an y ( C a n a d a ) , L t d . , Toronto

D . V a n N o s t r a n d C o m p a n y A u s t r a l i a P ty . L td . , Melbourne

C opyright © 1964, 1968, by D . V A N N O ST R A N D C O M PA N Y, IN C.

Published simultaneously in Canada by D . V a n N o s t r a n d C o m p an y (Canada), L td .

N o reproduction in any form o f this' book, in whole or i n part (except fo r brief guolation in critical articles or review s), may be made without written authorization f r o m the publisher. '

Library of Congress Catalog Card No. 6 7-31126

016S1bl50PRINTED IN THE UNITED STATES OF AMERICA

Preface

G eom etry— A C ontem porary Course reflects the influence of a num ber of sources. Not the least of these is the weight of the experim entation conducted in the teaching of geom etry during the past twenty- five years by both the author and his colleagues. The data collected over this period strongly indicated the need for bo th the arrangem ent and the developm ent of elementary m athem atical concepts as presented in this text. F u rth e r evidence that this presentation appears to be desirable is the fact that the most frequent com m ent of teachers using the first edition is th a t a m uch larger percent of students understand the na ture of proof than have ever done so in the past.

E ven a cursory examination of this text reveals that the au thor has leaned heavily on the geometry program prepared by the School M athem atics Study Group. The symbolism— the distinction betw een line, line segm ent, and ray—the emphasis on the concept of “betw eenness" as re la ted to points or to rays—the use o f congruence ra ther than equality— and on and on, all indicate a very ap paren t a ttem pt to adhere to certain b road and desirable aspects of the SMSG proposals. Similarly, the b lending of synthetic, coordinate and three-space geometries gives am ple evidence th a t the report of the Commission on M athem atics o f the College E n trance Exam ination Board has left its im pact on the writer.

D istinctive Features

1. T he introductory concepts on definitions and postulates are totally separa ted from one another.

( a ) T he first chapter is devoted to the need for definitions in any discussion, the m ethods of form ulating a definition, and the ou tgrow th of undefined terms as a consequence of the structure of the connotative definition.(b ) The second chapter dwells only on the definitions of a re la tively few geometric term s and their application to geom etric figures. Distinction is m ade beUveen reasoning from a definition as against reasoning from the reverse of the definition.( c ) The third chapter is wholly devoted to justifying the need for postulates in any discussion and in particu lar to the field o f geom etry. Great pains are taken to apply the operational postulates to geom etric situations rather than algebraic ones. It has been shown

viPREFACE

th a t m any students, although d isin terested in algebra, seem “to come to life” when exposed to the n a tu re of proof in geometry. In view of this, examples involving excessive algebraic m anipulation have been om itted tvom the early chapters of the book.

2. T h e indirect p roof has been presented b o th slowly and carefully by m aking use of Aristotle’s second and third laws of logic. Experience ind icates th a t the student has difficulty in expressing him self w hen attem pting to develop an indirect proof in the “Tw o-C olum n” form, hence the “P aragraph" p roof is introduced, and this form at is req u ired for all problem s calling for an indirect proof. To establish a feeling for this proof, a section containing m any problem s ranging from those th a t are alm ost obvious to others th a t w ill challenge the brightest studen ts are included. From this point on in th e text there appear problem s requ iring the indirect proof in almost every set o f exercises.

3. T he narrative problem does not appear in the text un til the technique of expressing a proof has been firmly established. Only then are the conditional an d categorical statem ents introduced, and, again, a large num ber of problem s are available to the stu d en t so th a t he. can achieve some skill in handling this segment of the course.

4. T h e chap ters on coordinate geom etry are bo th extensive and thorough. T hey were not included m erely to pay lip service to th e Com m ission’s R eport. T he concept of locus as a "set o f poin ts” in coordinate geom etry he lps pave th e way for an understand ing of locus in synthetic geometry. In addition to the usual loci problem s involving equalities, there are m any th a t are devoted to inequalities. In keeping w ith the current trend, se t notation is used to describe these loci.

5. I t w as felt that introducing elem ents of th ree dim ensional geometry a t th e very outset w ould only add to the b u rd en of learning too many

. unfam iliar term s too early in the work. I t is only after the properties o f perpendicu larity in a p lane have been established th a t w e find any reference to space geom etry. By this time the studen t should feel secure in his understand ing of th e subject matter and be ready to extend some of the notions h e has learned to a ‘‘broader” space.

6. T h e book is designed so that the teacher w ho prefers to restrict the course to topics related only to synthetic plane geom etry can do so w ith o u t fear th a t she m ay be assigning problems in e ither three dimensional o r coord inate geom etry.

7. E ach chap ter contains a test or review of th a t chapter. In addition th e re is often a short section pertaining to an interesting and unusual p ro b lem th a t frequen tly has some historical significance.

8. In addition to a very careful grading of th e problem s within each s e t of exercises, the narrative problems are kept distinct from those 'wherein, the diagram , G iven Data, and Conclusion are given to the stud e n t, and these, in turn, are separated from the num erical problems.

PREFACEvii

Every effort has been m ade to gear the level o f the w riting to students in the tenth and eleventh grades. In fact, our experience has shown that it is possible"for students to read and com prehend the explanations w ith

out the aid of a teacher.

S econd Editionin this, the second edition, the definition of the in terio r of an angle

was revised to overcome certain inconsistencies. T his was achieved by establishing m ore rigorous definitions of “betw eenness" as related to points and as related to rays. By these changes, a d istu rb ing loophole was closed in the proof of the theorem on perpendicu larity of lines— Theorem 15. Largely aifected was the content o) C hap ter 3 of the earlier edition.—

and little else.A num ber of teachers who had used the first edition suggested that

m aterial be included in the curren t revision covering the circum ference and area o f a circle, the volum e and surface a rea of a sphere, th e volume and surface area of a cylinder, and, of course, the volum e an d surface area of a cone. In view of this, these topics w ere ad d ed to those of C hapters 17 and 18. T he postulates used in the developm ent of th e theorem s pe rtain ing to these concepts are qu ite different from those frequen tly found a t this level. T hey are, however, far more in keeping w ith th e in te rp reta tion of a “lim it" as used in advanced courses in m athem atics than the vague symbolism norm ally em ployed w hen this topic is developed in

secondary school m athem atics classes.I w ould like to express m y gratitude to M r. Sidney Flam m , Mr.

Angelo Rosamilia, and Mrs. L aura Schefter— all o f E ast Side H igh School, Newark, New Jersey— who offered valuable criticism ;after hav ing taught this m aterial in m im eographed form. And, b y all means, I am thankful for and deeply touched by the m any m oving and sincere letters th a t I have received from teachers and students across the nation. This is the

heady w ine on which an au thor nourishes.H a r r y L e w is

January, 1968

Contents

1 Definitions and Their Place in a Proof 1NEED FOR DEFINITIONS 2 W H O DETERMINES THE DEFINITIONS OF WORDS? 3 CONSTRUCTING A DEFINITION 4 NEED FOR UNDEFINED TERMS 6 THE LANGUAGE OF GEOMETRY 9 TEST 20

2 Definitions of Geometric Term s 22THE MEASURE OF A LINE SEGMENT 23 THE MEASURE OF AN ANGLE 27 DRAW ING A CONCLUSION BASED O N THE REVERSE OF A DEFINITION 40 DRAW ING CONCLUSIONS O N THE BASIS OF DEFINITIONS AND THE REVERSE OF DEFINITIONS 43 TEST 46

3 Assumptions and T heir Place in a Proof 50H O W DO THE BLIND DRAW CONCLUSIONS? 51 POSTULATES IN GEOMETRY 54 THE SUM AND DIFFERENCE OF TW O LINE SEGMENTS 5 7 THE SUM AND DIFFERENCE OF TW O ANGLES 59 THE ADDITION POSTULATE 62 THE SUBTRACTION POSTULATE 65 THE MULTIPLICATION AND DIVISION POSTULATES 68 THE POSTULATES OF EQUALITY' 72APPLICATIONS OF THE POSTULATES OF EQUALITY 73 APPLICATIONS OF THE POSTULATES OF GEOMETRY 7 7 TCST 80

4 The “ Simple” Theorems 85THEOREM O N RIGHT ANGLES 89 THEOREM O N STRAIGHT ANGLES 95THEOREMS O N SUPPLEMENTARY AND COMPLEMENTARY ANGLES 95 VERTICAL ANGLES 101 TEST 107

5 C o n g ru e n c e o f T r ia n g le s 110

CORRESPONDENCE 112 CORRESPONDENCE RELATED TO POLYGONS 114 CONGRUENT POLYGONS 118 v POSTULATES FOR PROVING TRIANGLES CONGRUENT 120 APPLICATIONS OF THE POSTULATES O N CONGRUENCE TO FORMAL PROOFS 126 PROVING LINE SEGMENTS OR ANGLES CONGRUENT THROUGH CONGRUENT TRIANGLES 131 . FURTHER CONCLUSIONS THAT CAN BE DRAWN O N THE BASIS OF CONGRUENT TRIANGLES 136 OVERLAPPING TRIANGLES 139 . THE ISOSCELES TRIANGLE 141 THE S.S.S, THEOREM 153 THE HYPOTENUSE-LEG METHOD OF CONGRUENCE 160 PROBLEMS INVOLVING CONGRUENCE

Ix

OF MORE THAN ONE PAIR OF TRIANGLES 164 TEST ' AND REVIEW168 TRY THIS FOR FUN 171

6 P erpend icu lar i ty 172

MEANING OF DISTANCE AND ITS RELATION TO PERPENDICULAR LINES 179 CONDITIONAL AND CATEGORICAL STATEMENTS 191 TEST A ND of.

CONTENTS

VIEW 196

7 Perpendicularity in Space G eom etry 198

V THE MEANING O F DETERMINE 200 FURTHER CONDITIONS UNDER' WHICH A PLANE IS DETERMINED 203 METHODS OF DETERMINING A

PLANE 2 0 5 PERPENDICULARITY BETWEEN A LINE AND A P U N E 2 0 7 TEST AND REVIEW 214

8 T h e Indirect Proof and Parallelism 218

NONINTERSECTING LINES AND THE INDIRECT PROOF 224 PARALLELISM — SECTION I 233 PARALLELISM— SECTION II 2 3 9 PARALLELISM—SECTION III 245 UNIQUENESS AND EXISTENCE 2 5 5 THE PARALLELOGRAM— PART I 2 5 8 THE PARALLELOGRAM— P^RT II 2 6 7 TEST AND REVIEW 274 TRY THIS FOR FUN 277

9 Parallelism in Space 279

DIHEDRAL ANGIES 2 8 9 TEST AND REVIEW 2 9 6 TRY THIS FORFUN 298

.0 ,.The Angles of a Polygon 300

THE ANGLES OF A POLYGON 309 A BRIEF JOURNEY INTO N O N - EUCLIDEAN GEOMETRY 313 TEST AND REVIEW 319

1 Sim ilar Triangles 323

RATIOS AND PROPORTION 3 2 9 THEOREMS BASIC TO THE PROOFS OF SIMILARITY 333 SIMILAR TRIANGLES 342 PROVING RATIOS EQUAL AND PRODUCTS EQUAL 3 5 2 THE RIGHT TRIANGLE 360 THE THEOREM OF PYTHAGORAS 3 6 5 TEST AND REVIEW 371 TRY THIS FOR FUN 374

2 Coordinate G eom etry—An Introduction 375

PLOTTING POINTS 3 7 6 DISTANCE BETWEEN TW O POINTS AND DIVIDIN G A LINE SEGMENT INTO ANY GIVEN RATIO 382 PARALLELISM ANDPERPENDICULARITY 3 9 3 TEST AND REVIEW 405 TRY THIS FOR FUN 4 0 7

CONTENTS

13 Coordinate G eom etry—The G raph 409THE STRAIGHT. LINE 414 INTERSECTION OF T W O SETS 419 ANALYTIC PROOFS OF PROBLEMS FROM SYNTHETIC GEOMETRY 423 THEGRAPHS OF INEQUALITIES 425 LOCUS OF POINTS 434 THECIRCLE 440 TEST AND REVIEW 447 TRY THIS FOR FUN 449

14 The Circle 450CHORDS EQUIDISTANT FROM THE CENTER OF A CIRCLE 4 6 0 TANGENTSAND SECANTS 466 THE SPHERE 477 ■ THE RELATION BETWEENANGLES AND ARCS 4 8 2 APPLICATIONS OF THE THEOREMS O N ANGLfj? ' MEASUREMENT 491 CHORDS, TANGENT SEGMENTS, AND SECANT SEGMENTS 498 TEST AND REVIEW 508 TRY THIS FOR FUN 513

15 Locus—Synthetic G eom etry 514THEOREM, CONVERSE, INVERSE, AND CONTRAPOSITIVE 517 LOCUS THEOREMS 524 COMPOUND LOCI IN SYNTHETIC GEOMETRY 5 3 3 STRAIGHTEDGE AND COMPASS CONSTRUCTIONS 539 MORE ABOUT CONSTRUCTION WITH STRAIGHTEDGE AND COMPASS 5 4 6 TEST AND REVIEW 5 5 7 TRY THESE FOR FUN 559

16 Inequalities 565TEST AND REVIEW 5 7 7 TRY THIS FOR FUN 578

' 17 Areas of Polygons and Circles 580AREA OF THE PARALLELOGRAM, THE TRIANGLE, AND THE TRAPEZOID 5 8 4 AREAS OF SIMILAR TRIANGLES 594 AREAS OF REGULAR POLYGONS 602CIRCUMFERENCE OF A CIRCLE 6 0 6 AREA OF A CIRCLE 6 1 2 TESTAND REVIEW 618 TRY THIS FOR FUN 621

18 Volumes 622, VOLUME OF A PRISM 6 2 6 VOLUME OF A PYRAMID 631 SUR

FACE AREA AND VOLUME OF A CYLINDER AND A .C O N E 636 VOLUME AND SURFACE AREA OF A SPHERE 6 4 2 TEST AND REVIEW 6 4 7 TRY

THIS FOR FUN 6 4 8

L IST O F SYMBOLS

AB Arc AB0 CircleO CirclesA TriangleA TrianglesO ParallelogramAB Segment AB

AB Line AB—>

AB Ray ABZ A Angle Am AB Measure of ABm ZA Measure of Z AZ.A— BC— D Dihedral angle A—BC—D1 Perpendicular

£ Not perpendicularii ParallelX Not parallel

Congruentm Not congruent

Similar= Equal

Not equal> Greater than< Less than> Greater than or equal to< Less than or equal to

Not greater than< Not less thanv ~ Square root{ } Set1 Such thath IntersectionAy Delta y

M Absolute value of aTherefore

A * -* H A corresponds to HP ~ * 9 If p then q~ p Not p

xiii

1

Definitions and Their Place in a Proof

T H E A R G U M E N T HAD R EA C H ED A STALEM ATE. . Finally, in disgust, the taller and heavier man shouted, “ If you don’t believe me, I ’ll prove it for you!” W ith that, he took off his jacket, rolled up his sleeves, clenched his fist, and laid his opponent low.

Well-, you may not w ant to consider this a “ proof,” but certainly the argument was forceful and, needless to say, the conclusion was no longer in doubt. Yes, this is one of m any ways in which decisions are arrived a t or “ arguments” are “ proved.” T hough most intelligent people would prefer not'to resort to violence in order to prove their point, w ith m any primitive people, with young children, and even with some nations, this seems only too often to be a way of imposing ideas on a weaker opponent. This method of proof is frequently referred to as proof by force.

I t is very likely th a t during the m any discussions you have had with friends, you may have tended to use various appeals to convince them of the correctness of your cause. In some cases you m ay have sought their pity. In others, if you were certain they were unfamiliar with the topic at h in d , you may have deliberately used their ignorance to drive your point home. And in still others, perhaps you clinched your argum ent By an appeal to authority such as, your teacher, your clergyman, your parent, your doctor, or, in desperation, possibly an advertisement on T V . All these methods of

1

2 DEFINITIONS AN D THEIR PLACE IN A PROOF“ p ro o f’ rely heavily on an appeal 10 the emotions of the listeners or to their respect for authority.

M a t h e m a t i c a l p r o o f s , o n t h e o t h e r hand, a r e d e s i g n e d s o t h a t t h e

c o n c l u s i o n s d r a w n b y m a t h e m a t i c i a n s a r e b u r a n o u t g i o w t h o f a r e l a t i v e l y

f e w s t a t e m e n t s t h a t ; h e y h a v e a g r e e d u p o n . D u rin g { h e c o m i n g y e a r y o u

w i l l ! c a m s e v e r a l m e t h o d s e m p l o y e d b y m a t h e m a t i c i a n s f o r p r o v i n g s t a t e

m e n t s . T h is u n d e r s t a n d i n g w i l l a l s o e n a b l e y o u t o d e t e r m i n e t h o s e c i r c u m

s t a n c e s f o r w h i c h i t is b e t t e r t o a p p l y o n e m e t h o d r a t h e r t h a n a n o t h e r . In a d d i t i o n , t h e s h o r t c o m i n g s o f e a c h o f t h e s e p a t t e r n s o f p r o o f w ill b e s h o w n .

■ Need (or Definitions

M ost of us rarely th ink twice about the words we hear during the course of the day. Perhaps this is as it should be. T here a re times, however, when a slight reflection on our p a r t m ay cause us to w onder w hat in the world was m eant by w hat we have ju st heard. T o illustrate, consider the incident, oft used by comic strip writers, of the m other asking her five- year-old son to wash himself. W ithin a few m oments the boy appears, proudly displaying his “ clean” fingers, bu t only to be m et by the indignant outburst from his paren t insisting th a t he return to the bathroom to wash himself. T o the boy, washing m eant cautious dipping of the finger tips in w ater and removal of the w ater as quickly as possible by rubbing vigorously w ith a towel. T o the m other, on the other hand, washing implied the application of soap from elbows to finger tips—from neck through forehead— certainly a grossly different interpretation of washing than had been giver to it by her offspring!

In a m ore serious vein, a great deal has been said recently about the n a tu re of our education. W ere we confronted with .the statem ent th a t .,

~ "** Ti ... • • *“ Most people receive an adequate education”

we would very likely declare, “ Why, of course it is so!” Yet, were we to reflect for bu t a moment, we would soon realize th a t there are several words in th e statem ent th a t are a b it vague. Thus, the word education means m any things to m any people.

T o some, education means occupying a seat in a building where there are other students and a teacher. To others, this word implies having a thorough understanding of mathematics, physics, chemistry, and a t least two foreign languages. Removing the word education from the original sentence and replacing it with each of the interpretations just stated, we obtain these two sentences:

(1) Most people occupy (or have occupied) a seat in a building where there are other pupils and teachers.(2)' Most people have a thorough understanding of m athematics, physics, chemistry, and at least two foreign languages.

WHO DETERMINES THE DEFINITIONS OF WORDS? 3

In all likelihood we would agree with the accuracy of the first sentence but question that of the second. Yet both are merely a rewording of the sentence, Most people receive an adequate education, wherein the word “ education” has been replaced by two of its definitions.

EXERCISES

1. Why did we reach two completely opposite conclusions concerning the tru th or falsity of the statement Most people receive an adequate education?

2. Which of the definitions of education would you say is the correct one?3. M ake up your own definition of the word “ education” th a t would make

the sentence true.4. W hat other words in the statem ent Most people receive an adequate education

need clarification?5. State one reason why people may have completely contradictory views

about simple statements.

I Who Determines the Definitions of Words?From the illustration in the preceding section we should

have learned that one of the reasons why people arrive at different conclusions is because they attach different meanings to the key words in the discussion. But who is to say whose definition is the correct one? Frankly, no one person can say. The people participating in any discussion must . agree on the meanings of the terms in question. O nce they have reached an understanding, however, it is im portant th a t no one change th e definitions of these terms without notifying the others. Not to notify the others would be quite foolish, for then this person- would arrive a t conclusions based on private definitions that would be meaningless to the others.

So, too, will be the case with the m any, m any words th a t will be considered in geometry. Once the definitions o f terms have been, agreed upon (they will be those th a t appear in this textbook), we can no t be fickle and; change them to suit ourselves as we m ight change our clothes. O n the o ther ’ han d there is nothing sacred about which definition has been agreed upon. W hat is im portant is that we do not change horses in m idstream. W hat would happen if part way through the course several students in the class, w ithout consulting the others, decided to go back and change the definitions of all the words they had learned to date?

Briefly, we can now say(1) Definitions are m ade to suit the needs of the people involved in any discussion.(2) Or.cc a definition has been agreed upon, it can not be changed by any single member without consulting the others of the group.

4d e f in it io n s a n d their p l a c e in a p R o o f

EXERCISES

D e f i n e t h e u n d e r l i n e d w o r d s i n t w o w a y s : ( a ) t o m a k e t h e s e n t e n c e t r u e a n d ( b ) t o TYiakc i t f a l s e .

1 . A t e l e v i s i o n s e t i s a p i e c e o f f u r n i t u r e .

2. A dog is m an ’s best friend.3 . A r e f r i g e r a t o r is a n i c e b o x .

4. T h e school nurse is a teacher.5. A m agazine is a textbook.

6. George W ashington was an intellectual.7. A cand y bar is food .

8. Clim bing the school stairs is participating in gymnastic activities.

■ Constructing a D efin ition

Although we have been discussing the need for clearly defined terms, in addition to the fact th a t we have been called upon to define several words, nothing, absolutely nothing, has been said concerning how to construct a good definition. This m ay appear as an oversight. Yet, on the o ther hand, it provided an opportunity to form ulate several definitions th a t can now be examined carefully. In this way we will be able to select the special ingredients from w hich definitions a re brewed. These ingredients are frequently called the characteristic s or properties of a definition.

I f each of us had written the definition of a teacher called for in Prob- lem 4 of the preceding Exercises, one of these would probably be

“ A teacher is a person who im parts information to h e r students,”

T here are two features of this definition that m ust be emphasized:

(1) T h e subject of the sentence is the word being defined.(2) T h e verb is sotnt form of the verb “ to b e .”

Equally as im portant, though, is the fact th a t the word following the verb pertains to a collection of objects which have sim ilar traits.N In this situation the collection is th e word person. O ther things th a t would have • sim ilar traits to the word teacher that would also belong to th e collection of things called person would be: policeman, fireman, engineer, man, wom an,

• taxi driver, and many, m any more. Can you name at least six o ther things th a t belong to the collection o f person? Should you exam ine some o f the o ther definitions you m ade for the preceding Exercises, you m ay find th a t th e collections to w hich each of the first three words belonged were

(1) A piece of furniture is a household article. . . .(2) A friend is a person. . . .(3) An icebox is a piece of furniture. . . .

|

CONSTRUCTING A DEFINITION 5

Quite apparently our definition did not end when we placed the word being defined into the collection of articles that had similar features. R eturning to the definition of teacher, you will notice that a modifying clause was placed after the word person. W hat is the purpose of this clause? Why is it that the definition could not have ended with the word person? If it had ended with the word person, name several o ther words whose meaning would be identical to that of the word teacher.

From your answers to the last four questions it should be clear that every good definition must contain a clause or phrase modifying the collection of objects. The purpose of this clause is to show how the word being defined differs from all the other words in the collection to which it belongs. If the modifying clause were not added, then all the words in the collection would have identical meanings. Thus, in the case of teacher, the modifying clause is

. . who imparts information to her students.”

Were this clause not in the definition, then the words teacher, policeman, fireman, engineer, m an, woman, and all the other words classified as person would have the same meaning. In each of the definitions that you wrote for the underlined words in the Exercises on .page 4, how did you distinguish the word you defined from the other words in that collection?

A third im portant property of a definition can be illustrated by asking you to criticize the following "definition”.:

“A teacher is a person who teaches.”

This definition certainly complies with the first two properties of a definition:

(1) The word is placed into a collection (persons) that contains objects having similar features.(2) The word is distinguished from the other objects in the collection by the modifying phrase who teaches.

But yet there is something confusing in this definition. T he person who did not know the meaning of the word teacher would certainly not know the meaning of the word teaches. Thus, he would be no closer to an understanding of the word teacher after he heard the definition than before he h ad heard it. Hence, w hat relationship should exist between the words in the definition and the word being defined?

The fourth and last property of a definition is not quite as apparent as the earlier three. R eturn again to the definition of a teacher:

“A teacher is a person who imparts information to her students.”

W ere we to reverse, this sentence— that is, interchange the predicate nom inative with the subject of the sentence—it would become

“ A person who im parts information to her students is. . . . ”

Quite apparently, it would be best if this person was a teacher, not a ikenian, nor a doctor, nor a lawyer.

6 DEFINITIONS AN D THEIR PLACE IN A PROOFThus, our definition of teacher is r eversible. Every definition m ust be

reversible. By this we m ean th a t when th e predicate nom inative and the subject of a definition are interchanged, the new sentence will be true. This is not so w ith all sentences. To illustrate, the reverse of

“ All kings are m en”is

“ A ll m e n a r e k in g s .”

Although most m en would prefer to believe that the latter sentence is true, most women know all too well that it is not! Can you m ake up five other statements that are not reversible?

Thus, we can now say th a t the properties of a definition are

(1) T he word being defined must be placed in to its nearest class. Earlier this class was referred to as a collection of words having sim ilar properties.(2) I t is necessary to show how the word being defined differs from theother words in its class. This was done by adding th e modifying clause or phrase.

(3) T h e words in the definition must be simpler than the word being defined.(4 ) T h e definition m ust be reversible.

EXERCISES

1. W hich of the properties of a definition are no t com plied w ith in each of the following “ definitions”?(a) An autom obile is a vehicle.(b ) A desk is that which is used to write on.(c) A history book is a book that contains history.(d ) An instrum ent used for keeping time is a clock.(e) If a rectangle is a square, then it has four equal sides.

2. Rewrite each of the sentences in Problem 1 so th a t they will conform w ith the properties of a definition.

3. Using the properties of a definition, define each of the following words:(a ) running shoes (b ) biography (c) football helm et(d ) w riting paper (e) chair (f) garage

■ Need for Undefined Terms

T h e m ethod w e have used for defining words is certainly no t the only way to form ulate definitions. By reflecting a m om ent, you w ould realize th a t no dictionary gives a definition in sentence form. T hedefinitions th a t you find in a dictionary are called synonymous definitions. W hy was this name chosen?

A very common type of definition is the one used when teaching a■ baby h o v to speak. I t would be a bit absurd to say to a young child;

NEED FOR UNDEFINED TERMS 7

“ A tabic is an article of furniture upon which food is placed.”

Not only would this approach be nonsense, bu t by insisting upon defining words in this manner to an infant we would probably retard his ability to speak by m any years! A parent would norm ally point to the object and repeat its nam e m any times. In due course the child would begin to m um ble something th a t vaguely resembled the word. No, this is not a scientific definition, but a definition nevertheless, as any proud paren t would vouch for. Definitions of this variety are called demonstrative definitions. W hen pointing to. a table and repeating the word "tab le ,” we never know whether the child m ay be thinking only of the four legs ra th e r than the entire object. Hence, to him, every desk, every chair, every stool, in fact every four-legged object

m ight be a table!T h e type of definition whose properties were listed on page 6 is

called a connotative definition. This m ethod for defining terms is used in most areas of elementary mathematics, including geometry. W ithout realizing it, in your study of algebra you defined such words as term , m onomial, binomial, trinom ial, and polynomial by employing the properties of the connotative definition. I f you are now studying biology, you will find th a t all the words in this science are defined in terms of the connotative definition. Similarly, if you study physics and chemistry, here, too, application will be made of

the connotative definition.T he fact that we insist upon the use of the connotative definition in

geom etry leads us squarely into a ra ther nasty predicam ent. Let us suppose th a t we did not understand the m eaning of the word “ person” a t the time we were writing a definition of the word “ teacher.” W ere we to look for the word “ person” in the dictionary, it would lead us to the classification of “ hum an beings.” Again, assuming th a t this phrase m eant little or nothing to us, we moved on undaunted to seek its m eaning in the dictionary also. And so we continued to investigate the m eaning of each new class that we encountered until finally our a rray of words resembled the picture below:

teacher4

person

hum an being >1

m anI*

prim atei

m am m ali

anim aliliving organism

anim al

8 DEFINITIONS AND THEIR PLACE IN A PROOFW hat criticism would you have of a dictionary th a t led us from “ an im al” to “ living organism ” and from “ living organism” back to “ anim al” ? W hich of the properties of the c c n n o ta tiv definition was not complied with by a dictionary that would lead us into a “ circular definition” such as this?

U nfortunately, were we to trace words from class to class, we would find th a t the authors of most dictionaries will lead us in circles as in the case above. They do, however, have another choice. At some point in this process of “ back stepping,” the authors can call a halt saying, “ I think th a t this word is so elem entary th a t its m eaning is known to all. Since there a re no sim pler words in my language w ith which to define this word, I will not define it a t a ll.” W ords such as this are referred to as first words. T h e m athem atic ian would call them primitive terms, or elemental terms, or simply undefined terms. In the diagram on page 7 “ living organism ” should have been considered as an undefined term, for in this case there appeared to be n o w ord in the language in which to classify it.

W ere we perm itted to use the demonstrative definition, it would have been possible to have pointed to the object and in that way “ defined” it. I n geometry, however, we insist that the connotative, and only the connotative , definition be used. Hence, in order to define a w ord, i t must be classified. S ince we must begin our language— and our language will be the language o f geometry—with some words, these first words will have no prior words b y w hich they can be classified. They must, therefore, of necessity remain undefined.

EXERCISES

1 . Prepare a diagram in which the word being defined is led from class to class to the undefined term that..appears a t the very end. Use the following group of words, where the first word is the undefined term : living organism, animal, person, citizen, teacher, m athem atics instructor, M r. Clark.

2 . D raw a similar d iagram for th e following terms where the undefined term is the last one: rhom bus, parallelogram, quadrilateral, polygon.

3 . In each of the following problems rearrange the terms in their proper order, then draw a diagram sim ilar to that found on page 7,(a) num ber, im proper fraction, fraction, nine-fifths .(b) furniture, moveable article, desk, table

4 . Using a dictionary, trace each of the following words to their original source. Draw a diagram sim ilar to the one on page 7, showing the class into which each word fell.(a ) fable (b) gold (c) rose (d) carpet (e) brick

5 . Will' it be possible to define the first word you learn in the subject of geometry? J ustify your answer.

THE LANGUAGE OF GEOMETRY 9

■ The Language of Geometry

PointT h e language of geometry is frequently opened with a

discussion of the term point. I t goes without saying that this word, of all words in geometry, can not be defined as it is the very first word in our language. Although it is not possible to define t this term , we are still faced with the problem of how to make its m eaning clear to all. This is done by listing some of the properties, or characteristics, of the word w ithout classifying it. W henever a word can not be defined, that word is described. Fundamentally, the difference between a description and a definition lies only in

the fact that

(1) W hen a word is defined, it is classified.(2) W hen a word is described, properties of the word are given without

classifying it.

To get some idea of why the properties of a point were so chosen, try to visualize the very fine end of a needle or a pin. Although you may consider that it is extremely pointed in its present condition, imagine what this end would be like if the refining or sharpening process were to continue indefinitely. I t is ju st such a notion that the m athem atician would like us to keep in m ind when we think of a point. He wants the point to indicate some fixed position such as the end of the needle or pin. And yet he does not want it to occupy any space as the needle end does! Hence, he asks us to consider this needle end in the process of refinement a t the moment it is vanishing. At that time, it represents a point to him . Thus, we obtain the

properties of a point.

A point has neither length nor width but indicates position.

Line■ ■ ■ ■ ■ ■ ■ ■ ■ W hen trying to define a line, we have two avenues of attack dpen to us. If the word is definable, there exists only one class into which i t can be placed. W hy? If it is no t definable, it will be the second of our undefined terms. T h e m athem atician prefers th a t it be undefined. T here are a num ber of ways, though, in which we can visualize a line:

(1) T h e fine edge of a formica counter.(2) T h e fine th read by which a spider lowers itself.(3) T he crease m ade by folding a piece of paper.(4) T he “ line of sight” of a gun.(5) A piece of elastic stretched to the breaking point.(6) T h e edge of a ruler.

t W henever the w ord “ d efine" appears henceforth in th is book, it wiii im ply th e use of. th e conno tative definition.

10 DEFINITIONS AND THEIR PLACE IN A PROOF

Perhaps one of the better ways of picturing what- the m athem atician conceives of as a line is the piece of elastic m entioned above. Assume that it will never break. As it is stretchcd more and more, it becomes thinner and thinner. Should this process be continued indefinitely, the elastic would become m inutely narrow and yet extend infinitely far in either direction. It is with this view in mind that the properties of a line were established.

A lin e has no w id th b u t can be exten d ed as far as d esired in e ith e r d ire c tio n ,^

At this stage of our work we have come to a roadblock. W e can not proceed with our work unless we recall some of the things we learned in algebra. O f great im portance to us now is an understanding of the term set.

Set

T he word set will be third of our undefined terms. As in th e case of the other two words we will try to make its m eaning clear by describing it. Thus,.

A set is a “ w ell-d e fin ed ” collection. )

Emphasis is placed on the words “ well-defined,” for

(1) If the members or elements of the set arc known to us, we should be able to describe how they were found.(2) Should we be given a description of a set of elements, it will be possible for us to list the members of this set.

To illustrate, given the set of elements

{a, e, i, o, w}

i t is possible for us to describe this set by saying that it consists of all the vowels in the alphabet. Notice that the letters a, e, i, o, and u are the elements o r members of this set and that they are enclosed in braces { }.

Now, secondly, were the members of a set described to us, this descript io n would have to be such that we could list these members. Thus, consider t h e description

“ T h e set of elements consisting of the names of the days of the week beginning with the letter S.”

T h is would be a “ well-defined” collection, for from this description it is possible to list the elements in this set :

(Saturday, Sunday}

R elative to the background that we will need, there are three concepts ■yet to be recalled. These are (1) subsets, (2) intersection of sets, and (3) union o f sets.

<*) Subsets: A set A is said to be a subset of the set B if every clement of A i s an elem ent of JS.

As an example, if the set of elements B is given as B = {2, 3, 4 , 5}

then, ifA = { 2 , 3}

A would be a subset of B. O ther subsets of B are

{ 3 ,4 ,5 } { 3 ,5 } {5}

Can you nam e a t least four other subsets of B?

(2 ) Intersection of Sets: The intersection of two sets A and B is the set of all

elements that are members of both A and B.As an example consider the sets

A = {1 ,3 , 5 ,7 ,9 }B = { 3 ,4 ,5 , 6, 7}

T h e intersection of A and B is the set

{3, 5, 7}

for this set consists of all elements th a t are members of both A and B,I f the two sets have no elements in common, then the set representing

their intersection will have no members. Thus, the intersection of the sets

C = {1 ,3 , 5 ,7 ,9 }D = {2, 4, 6, 8, 10}

will be th e empty set or null set { }, for there are no elements that are

comm on to both C and D.(3) Union of Sets: T h e union of two sets A and B is the set of all elemen.ts

th a t are m em bers of either A or B.As an exam ple consider the sets

A = {1, 2, 3, 4}B => {2, 4, 6, 8}

T h e union of A and B is the set0 , 2 , 3, 4, 6, 8}

for the m em bers of this set a re in either A or B.

EXERCISESl . F rom th e description given below, list the elements of each of the follow

ing sets:(a ) 'The set o f all integers greater than 10 and less than 20.

(b ) T h e set o f all odd num bers greater than 2 and less than 10.


12 DEFINITIONS AN D THEIR PLACE IN A PROOF

(c) The set of all m ultiples of 4 that are greater than 20 and less th an 30.(d ) The set of all prime num bers greater than 10 and less than 20.(e) The set of all the names of the days of the week that begin with the

letter “ M .”

(f) The set of all one digit integers greater than 9.

(g) The set of al! num bers greater than 0 and less than 10 that consist of '.he cubes of integers.

(h ) The set of all multiples of 9 that are greater than 12 and less than 15.(i) The set of all proper fractions whose num erator and denom inator

are members of {1, 2, 3, 4}.

2. W hat is the description of each of the sets below?(a) {2, 4, 6, 8}(b ) {5, 10, 15, 20,25}(c ) { 3 ,6 ,9 ,1 2 , t5 , 18}

(d ) {Tuesday, Thursday}(e) {a, b , c , d , . . . , x , y, z}(f) {1, 3, 5, 7, 9 , 1 1 , . . . , 2 * + 1 , . . . }(g ) . . . , ! / ( » + ! ) , • • •}

3. (a) List three subsets of {a, b, c, d, e}.(b ) List all the subsets of {1,2}. (Both the original set and the null set

should be included am ong the subsets of any set.)(c) List all the subsets of {1, 2, 3}.(d ) By analyzing problems (b ) and (c) can you state how m any subsets

there will be for {1, 2, 3, 4} w ithout listing them?

4. If K — (a, b, c, d) and M — {c, d, « ,/} , find a t least two sets of elements th a t will be subsets of both K and M .

5. If S = 0 , 2, 3, 4, 5} and T = {2, 4, 6, 8}, find(a) T he intersection of S and T.(b ) T h e union of S and T.

6. If R = {a, e, i, o, «} and Y = {a, b, c, i , e}, find(a) T he intersection of R and(b ) T he union of R and Y.

7. If A — {1, 3, 5, 7} and B — {2, 4, 6, 8}, then find the intersection of A and B.

8. If A is th e set of integers greater than 2 and B is the set of integers less than 10, then find the intersection of A and B.

9. A “ geometric figure” is often described as a set of points. D raw diagram s similar to the ones a t the top of page 13 and, in color, m ark those elements th a t represent the intersection of the two sets in each situation.


(d)

10. D raw diagrams similar to the ones below and, in color, m ark those elements that represent the union of the sets shown in each situation.

( a )

11. (a ) Is the intersection of two sets a subset of either of the two sets?Illustrate by using the sets S and T in exercise 5.

(b ) Is the union of two sets always a subset of either set? Illustrate by using the sets 5 and T in exercise 5.

(c) Give an illustration where the unio'n of two sets is a subset of one of these sets.'

12. Sentences 2 and 3 concerning the intersection and union of sets that 1 appear on page l l are actually the definitions of these terms. In w hat

way were these definitions made to comply with the properties of the

connotative definition?

Betweenness

■ ■ ■ ■ ■ ■ ■ ■ U ntil recently mathematicians showed little concern about the term “ between,” although they had used it quite often. Currently, however, efforts are being m ade to indicate precisely what this term shall m ean when used in a geometric discussion. If you are wondering why anyone would be confused by the use of the word ‘'betw een,” consider the diagram*' in Figure l - l . Is A , in each case, “ between” B and C; or is B “ between” A and C; or is C “ between” A and B?


Figure 1-1.

Allhough “ betw een” will be another of our undefined terms, confusion o f th e ;'nature shown above will not exist, for a t no tim e shall we ever use th is term unless the three points are points of the same line. A t this time it would b e well to state th a t points are named by using a single capital letter. In th e diagram s of Figure 1-1 the three points were named by the capital letters A, B , and C. Furtherm ore, by considering a line to be a set of points th e expression th a t points “ fall on” a line will m erely imply that these points a r e m em bers of th e set th a t comprise the points of the line.

Before we can speak of arranging things, it seems essential that these things be distinct. Thus, we would not think of “ arranging” a single book o n a shelf, for there is no other book with reference to which it can be a rranged . However, had we two books, we would have our choice of placing o n e first and the o ther second or in the reverse order. Should we be arranging th ree books— a history, an English, and a science book—on the shelf, then o n e of three situations m ight arise:

(1) T h e history book will be between th e English and the science books.(2) T he English book will be between the history and the science books.(3) T he science book will be between the English and the history books.

Figure 1-2. Figure 1-3. Figure 1-4.

I t would make no difference to us were the order in Figure 1-2 English, history, science or science, history, English, for in either event the history book w oulJ be between the other two.

So, too, is the case when referring to three points on a line. Were their order on the line either A, B, C or C, B, A, we would speak of B as being between A and C. As seen in the diagrams of Figure 1-5, it would make very little sense to speak of the order of the points as the first point in question, the second, and the th ird , for how we num bered the points would depend on


F ig u r e 1-5.

the m anner in which we approached them on the line. In all four cases, however, B is between A and C , for the order in which they appear on the line, no m atter how we approach them on the line, is either A, 3 , C or C, B, A.

In general, for any three points on a line one and only one of the following orders must exist:

A, B, C A, C, B B, A, C

R e m e m b e r , o f c o u r s e , t h a t t h e o r d e r A, B, C i s n o d i f f e r e n t t h a n C, B, A, f o r i n e a c h c a s e B is b e t w e e n A a n d C . »

Line Segment■ ■ ■ ■ ■ ■ H I We are now in a position where we can define the very first term in the language of geometry: line segment. As the word itself implies, we would like this term to denote a segment or piece of a line and exclude from it the remaining points of this line. To do this, we make the following definition:D e f i n i t i o n 1 : A l i n e s e g m e n t A B i s a s e t o f p o i n t s o f a l i n e c o n s i s t i n g o f t h e

p o i n t s A a n d B a n d a l l t h e p o i n t s b e t w e e n t h e m .

In Figure 1-6 the points A and B and all the points of the line between them is said to comprise the line segment AB. T h e points A and B themselves

F ig u re 1-6.

are called the endpoints of the segment. Oddly enough, the line of which A and B are points can also be named by the same two capital letters and, hence, be called the line AB. To distinguish the line AB from the segment A B , the following symbols are used:

line .4B : A B segment AB : AB

T he double arrow head over the A B helps to recall the property that the line can be extended infinitely, far in either direction.


Although a line segment must be named by using only those letters at its endpoints, a line m ay be named by referring to any two points of that

t t V-V > i— >line. Thus, in Figure 1-7 ihis line may be called A B , BA, AD, D A, o r any

A 8 C D EFigure 1-7.

one of a num ber of o ther ways. Can you name this line in a t least ten ways not already given?

Ray■ B H M We have often heard people speak of sun rays, o r m oon rays, or a ray of light. So, too, in geometry we would like to speak of a ray from a sim ilar point of view. We can picture a ray by visualizing th e th read o f light that would be seen if a flashlight were placed behind a dark sheet o f paper that had been punctured by a pin. As shown in Figure 1-8 th e ray

Figure 1-8.

•would sta rt a t the paper and go off to the right. No part of this ray w ould ex is t to the left of the paper.

This description is clear, but it is far from being m athem atically precise. T o m ake it so, we will have to clarify w hat is m eant by the statem ent that tw o points are on the same side of a third point. Thus, if B and C a re on the same side of A, then either B is between A and C as in Figure 1-9 or C is b e tw een A and B as in Figure 1-10. Were A between B and C, then B and C a re said to be on opposite sides of A as in Figure 1-11.

A b C ~ B ~r~ C A ~ 8~ a C ~

Figure 1-9. F igure 1-10. F igure 1-11.

D e f in it io n 9 • A rav is a se t o f p o in ts consisting of th e u n io n of a fixed p o in t o f a line a n d a ll th e poin ts o f t h a t lin e on th e sam e aide o f th e fjxed-poinl^

T h e fixed point is called the endpoint.oi the ray, while the ray itself is n a m e d by using two capital letters. T h e first of these roust be th e letter


a t the endpoint, while the second is the nam e of any other poin t of the ray.

T o illustrate, the ray in Figure 1-12 is ray A B , w ritten A B ] Figure 1-13 is

t f

F ig u r e 1-13. F ig u r e 1-14.

the ray CD (CD); whiie Figure 1 - 1 4 -is the ray FE (PE) or ray FG (FG). Notice that only single headed arrows are placed over the two capital letters. This is to indicate that a ray can be extended in one direction only. W hy is

—►it not possible to nam e the ray with endpoint F in Figure 1 - 1 4 as GE?

D e f i n i t i o n 3 : Opposite rays are two distinct rays of the same line thathave a comm on endpomT ---------——----------

—In Figure 1-15 A B and AC are opposite rays, for they are

(1) distinct

(2) of the same line (EF)(3) have a common endpoint (A)

F ig u r e 1-15.

--> . “4O n the o ther hand A B and A F would not be opposite rays, although they

- 4have a common endpoint and are on the same line, for A B and A F are merely two1 different names for the same ra y !

Angle■ ■ ■ ■ ■ ■ ■ M uch of geometry concerns itself w ith relationships th a t exist am ong line segments and am ong angles. I t is unlikely th a t this is the very first tim e th a t the word angle has come to your attention. However,

18 DEFINITIONS AND THEIR PLACE IN A PROOFas you realize by now, vague notions of w hat this word m ay mean can !eadto difficulties. Hence, by examining Figure 1-16 see if you can form ulate aclear definition of an angle. After having done this, compare your definition with that given here.

D e f in i t i o n 4 : An a n g le is th e set_of p o in ts c o n s is tin g of th e u n io n of tw n rays that have a com m on endpoint.

T he common endpoint is called the vertex oj the angle, while the. two rays are referred to as th e sides oj iht angle. An angle is nam ed by using three capital letters. T he letter nam ing the vertex m ust always appear as 'h e m iddle letter. T he other two letters are names of two points, one from each of the sides. Thus, in Figure 1-17 the angle may be nam ed either angle ACE ,

Figure 1-17,

written as /A C E \ or angle ECA ( /E C A ) ; o r /B C E ] o r any one of a num ber of other ways wherein C is always the m iddle letter. How would you nam e this angle in a t least four ways not already given?

EXERCISES

1.

8 C O

(a) Nam e th e line m arked 1 in two different ways.(b ) Nam e the line m arked 3 in two different ways.(c) Nam e the line m arked 4 in six different ways.(d ) Nam e the line m arked 5 in six different ways.(e) Nam e the line segment marked 1 in two different ways,(f) N am e the line segment marked 2 in two different ways.

2. (a) In the d iagram in Problem 1, at w hat point do the lines and DA intersect?

<—>■ (b ) In the diagram in Problem 1, what is the intersection of BD and

THE LA NG U AG E OF GEOMETRY

A

19

3.

(a) Name the(b ) Name the(c) Name the

(d ) Name the

(e) Name the(f). Name the

> C i - _C D

angle m a r k e d 1 in two different ways.angle marked 2 in two different ways.angle marked 3 in four different ways.-.1 --->

angle whose sides are A B and AC.► - 4

angle whose sides are CB and CA. two sides of /A .C B\ of / ACD ; of /.BC D .

4.

(a) Nam e the angle marked 1 in four different ways.(b ) Nam e the angle marked 2 in two different ways.(c) N am e the angle marked 3 in two different ways.(d ) Give two other names for the line BD.(e ) Give four other names for the line AE.(f) At w hat point do the lines AC and BD intersect?

(g ) Nam e the intersection of AB and EC.

(a) Nam e the angle formed by AB and AC.(b ) Nam e the rays that form the angle whose vertex is C.(c) Nam e two angles that have a common ray as a side of each 'of th e

angles.(d ) Nam e an angle whose sides are a pair of opposite rays.

(e) W hat is the intersection of AC and BD?

6. Using the properties of a connotative definition, explain w hy a line can not be defined as: “ A line is a set of points.”

7 . Using the diagram below, how m ight it be possible to describe ~A3

in terms of AC and BD?

D A S C

8 . Illustrate how two rays can have a point in common and yet their union will no t be an angle.

9. T h e set of points of a line on one side of a given point is called a half- line. How does a half-line differ from a ray?

1 0 . (a ) W hat is the intersection of A B and AB?

(b ) W hat is the intersection of A B and A B ?<— _____(c) W hat is the union of A B and AB?

■ Test

1. Explain why it is not possible to define the first word when building the language of a new science.

2. How is a “definition” distinguished from a “ description” ?

3. W hich of the properties of a connotative definition were no t com plied w ith when each of the following definitions was made?

(a ) A newspaper informs its readers of events that have recently occurred.'

(b ) An Englishman is a person.

(c) An isosceles triangle is a triangle that is isosceles,

4 . F rom th e descriptions given below, list the elements of each of the following sets:

(a ) T he set of all one-digit integers that are multiples of 3.(b ) T he set of num bers th a t a re the squares of the first five odd num bers.(c ) T he set of all prime num bers th a t are greater than 23 and less

than 29.

(d ) T he set of all fractions whose num erator comes from the set {1, 2, 3} and whose denom inator comes from the set {5, 7}.

20 DEFINITIONS AN D THEIR PLACE IN A PROOF

S . Give a description for each of the sets below,(a ) {4 ,8 ,1 2 ,1 6 ,2 0 }(b ) { 2 ,3 ,5 ,7 ,1 1 ,1 3 ,1 7 ,1 9 }

TEST21

( c ) { J a n u a r y , J u n e , J u l y )

(d ) {1 ,4 ,9 ,1 6 ,2 5 }6 . F i n d t h e i n t e r s e c t i o n o f t h e t w o s e t s i n e a c h o f t h e f o l l o w i n g p r o b l e m s :

(a) {Mary, Betty, Jane} and {Doris, M ary, Jane}

( b ) {1, 3, 5, 9} a n d (2, 4, 6, 8}7 . ( a ) E x p l a i n w h a t t h e c o n d i t i o n s w o u l d h a v e to b e s o t h a t t h e i n t e r s e c -

t i o n o f AB w i t h CD w a s AB.( b ) E x p l a i n w h a t t h e c o n d i t i o n s w o u l d h a v e t o b e s o t h a t t h e i n t e r s e c -

■<—*t i o n o f A B w i t h CD w a s t h e n u l l s e t ,

( c ) E x p l a i n w h a t t h e c o n d i t i o n s w o u l d h a v e t o b e s o t h a t t h e u n i o n o f

A B w ith CD was AB.8. a is the set of points of one line, b the set o f points of a second line, and c

the set of points of a third line.(a) Draw a diagram in which the intersection of a and 6 is an element

of c.(b ) D raw a diagram in which the intersection of a and A, the intersec

tion of a and c, and the intersection of 4 and c a re three distinct

elements.9. Each of the problems below should be answered in terms of this diagram.

— r -----------------

(a) Nam e the angle m arked 1.(b ) Name the rays that form the sides of Z.AED.

—> >(c) Nam e the angle whose sides are B E and BA.

' (d ) Name a pair of opposite lays.

(e ) Nam e the intersection of B E and CD.

(f) Nam e the intersection of AC and AE.10. (a ) I f two rays intersect, will their point of intersection be the vertex of

aii angle of which the rays are the sides of th a t angle? Justify your

answer.(b ) Two rays have a common endpoint and they are subsets of the

sam e line. Does this imply th a t they are opposite rays? Justify your

answer.(c) T he intersection of A T and BS is BY. Draw a diagram illustrating

this situation.

2 Z

Definitions of Geometric Terms

T H E T E R M S T H A T W E W IL L C O N SID E R IN T H IS chapter will enable us to examine the relationships th a t exist between angles and between line segments. These relationships concern themselves with the notion of equality—a concept that was examined qu ite thoroughly by you during your course in algebra.

As you recall, the equality

a = b

was a means th a t was used to express the fact.that a and b were bu t different symbols representing the same “ thing.” And, to a large extent, these “ things” in algebra were simply num bers. Thus, the equality

3 + 4 = 5 + 2

expressed the fact th a t the symbols on the righ t of the equality sign and thesymbols on th e left w ere m erely two different ways of representing the num ber seven.

In view of this, it would be both inconsistent and unwise to state a t any tim e th a t

A B = UD

for A B and CD a re the names of two different line segments and, as such, are no t symbols representing the same “ thing.” Similarly, to say th a t

22

THE MEASURE OF A LINE SEGMENT 23

/ .X Y Z = / .R S T

would be mathematically inaccurate, for the left mem ber of the equation is the nam e of one angle, while the right m em ber is the name of a completely different angle. Yet, there does exist an equality of some nature th a t can be examined between line segments and between angles. O f w hat character this equality is will be our concern for m uch of this chapter.

■ The Measure of a Line SegmentAn important u n i t of your study of algebra centered around

the “ one-to-one” correspondence that exists between the points on a line and the real numbers. Rather briefly, the real num bers consist of all the num bers you examined during that course except for the im aginary num bers

such as V —5. Specifically, these numbers did include positive and negative integers, positive and negative fractions, plus irrational num bers of the form V 2 , V 68, and ir. Furthermore, the term “ one-to-one” correspondence as used here merely implies that for each point on the num ber line there exists but one real num ber naming that point and, moreover, for every real num ber there exists but one point to represent it on the num ber line.

Such a correspondence is called a coordinate system. In addition, the num ber nam ing any particular point of the num ber line is called the

D w

- 2 0 l JT 2 3~jr "

Figure 2-1.

coordinate of that point. For the number line in Figure 2-1 the coordinate of A is 2. W hat are the coordinates of B and C? In the study of geometry, we are--yinterested only in that part of the num ber line consisting of the CM; th a t is, the point zero and all the positive points on the num ber line.

W hat we have tried to do is make the num ber line resemble an infinitely long ruler. By varying the position of the point named “ 1” we can in turn make the num ber line appear to be either the “ inch” ruler, the “ centim eter”

3"inch ruler

centim eter ru leri . ■ -1-------1------ >------ ------- ■--------------------0 1 2 3 4 5 6 7 8 9

Figure 2-2.

ru ler, the “ foot” ruler, in fact any dimension ruler we care to m ake it. Changing the position of the 1 changes the unit of measure. W hether w.e use the inch, the centimeter, the foot, the yard, the m eter, or any one of a m ultitude of other units in creating the num ber line, is not im portant. W hat

24 DEFINITIONS OF GEOMETRIC TERMS

is of great importance, however, is that we do not change the un it on the num ber Jine when part way through a problem. In fact, to avoid any misunderstanding, we will agree that the num ber lines encountered in any singie problem Will have identically the same unit.

We are now in a position where we can form ulate a very im portan t concept. Starting with any given line segment, '.ve can establish a coordinate system on the line containing that segment. By making the coordinate of one endpoint of that segment 0, the coordinate oj the other endpoint will be called the measure oj the line segment.2-3) we create a coordinate system whereby the coordinate of A , one end-

A 8------ a------------------------------ 1--------------------- 1---------------------1_____________________ i___________________ - ■ -

C 0 1 2 3 d

F ig u r e 2-3.

poin t of A B , is 0. Since the coordinate of the other endpoint, B, is 2, the num ber 2 is said to be the measure of AB. Using symbols, this is expressed as

m ~AB — 2 (1)

Q uite apparently, we are saying no more here thanIf a ruler is placed on line segment AB to determ ine its length,

this length would be 2.A_____________________ 6

1 ---------— I---------- T ~ r ~ -2 3 4

ru le r

F ig u r e 2-4.

Note th a t when we referred to the measure of "AB in (1) above, we did no t state that it was 2 centimeters, or 2 inches, or 2 feet, or 2 of any un it whatsoever. This is so since the measure of a line segment is the coordinate o f one of its endpoints (when its other endpoint is 0). And as a coordinate, i t is simply a real num ber and nothing else!

W ith the establishment of the measure of a line* segment, we have overcom e the difficulty raised in the opening paragraphs of this chap ter. I t ■was pointed out that it would be improper to say that

~AB = CD

fo r A B and CD are the names of two different line segments. Now, however, t o say that

m A B — mCD

'w ill simply imply that the symbols on the left of the equality sign and the sym bols on the right represent identically the same coordinate.

THE MEASURE OF A LINE SEGMENT25

Midpoint of a Line Segmenttm m m ttm im I Since we have a means of expressing equality, w e would like to examine several situations in which this will occur. T h e first o f these is

with reference to the m idpoint of a line segment,D efinition 5: T he midpoint of a line segment is a poin t of th a t line seg

ment such th a t the two segments formed have equal measures.

To illustrate, if C is the m id p o in t of AB, then by definition the measure

______________ i______________i--------------------- 1-------------------- -A C 6

Figure 2-5.

of AC will be equal to the measure of CB. This is expressed as m AC = m CB

There are m any, m any lines that contain th e m idpoint o f a line segment. Each of these, o ther than the one of w hich the line segment is a

subset, is called the bisector oj the line segmtnt.D e f in it io n 6: T h e bisector of a line segment is a line th a t intersects th e

line segment a t the m idpoint o f the line segment.^4 ——-

If, in Figure 2-6, A B is the bisector o f CD, it will im ply th a t B must be the midpoint of CD , Pursuing this further: since B is th e m idpoint of CD,

then as beforem C B ~ m BD

Frequently, inform ation in problems will .be given asLine segment X Y is the bisector of line segm ent R S (Figure

2-7).

Figure 2-7.

This may appear to be an as a line, ni

segment X Y

26

lu oe an inconsistency, for the bisector of a line segm ent was defined as a line, no t a line segment. I t is not in error, however, for the

W »T

DEFINITIONS OF GEOMETRIC TERMS

_ .44 noweveis merely a subset o f the line X Y . Hence, if XY contains- r

point cf RS, then so too must X Y.O ne m ore point to be noted is th'1

m ents are equivalent:

Statements such as these will be used same meaning.

the m id-

fact that the following tw0 state-

0 ) X Y is the bisector of fiS.

(2) X Y bisects RS.

interchangeably, f0r they will have the

EXERCISES

1. W hat conclusion can be draw n fromI-— l >lems below?

(a) £> is the m idpoint of AC.

A,

(e) F is the m idpoint of AB.

A.

(g) ^ a n d BD bisect each other. (Draw two conclusions.)

the d a ta given in each of the prob-

(b ) V T is the bisector of ■SM/.

(f) £>E bisects AB.

A

(h) A B and CD bisect each other.

THE MEASURE OF AN ANGLE. 27

2. Based on the property of a line, explain why it would not be possible for a line to have a midpoint.

3. If, in the diagram at the right,m BC = m CD = m DE

then C and D are called the trisection points of BE. How would you define the “ trisectiori points of a line segm ent” ?

__ __ ___4. In the diagram in Problem 3 where m BC = m CD = m DE, then AC

f4 _____and AD are called the trisectors of BE. How would you define the “ trisectors of a line segment” ?

5. “ If m EF = m FG, this will not necessarily imply that F is the m idpoint of E g .”(a) Draw a diagram justifying this statem ent.(b ) U nder what conditions will F be the m idpoint of £G?

6. W hat is the name of the point of intersection of the bisector of a line segment and the segment itself? Justify your answer.

I The Measure of an AngleIn trying to express an equality between angles we run

into the same difficulty as we did with line segments. T h at is, to say th a t in Figure 2-8

Z ABC = /D E F

C F

Figure 2-8.

would not be correct, for the name on the righjt of the equality refers to an angle that is different from the angle that is referred to by the nam e on the left. By now we realize that the equality sign can be used only if the symbols on both sides of it are names for the same thing. Thus, we are forced into a

position of having to create a m easure for an angle as we had created a measure for a line segment.

Whereas the num ber line, or infinitely long ruler, was used to enable us to express the measure of a line segment, to express the m easure of an angle we fall back upon the protractor. Ignoring the instrum ent itself and thinking only of its outline, we obtain Figure 2-9. T o point A a t the right


we assign the num ber 0, while to the end point, B, of the red figure— N O T the segment A B —we assign the num ber 180. For la te r inform ation, i t is im portant to notice that the outline of the p rotractor begins a t poin t A on

—¥ -—>ray KS and ends at point B on the ray opposite K9; th a t is, VP. T his outline is then divided into 180 equal parts. T h e point C a t th e end of th e first of these equal parts is m arked “ 1.” In the sam e way, each of the succeeding endpoints of every one of the equal parts is m arked with the consecutiveintegers 2, 3, 4...........The last of which, of course, is 180,

As a m atter of fact, in a m anner similar to that used .on the num ber line, we establish a pairing off, or one-to-one correspondence, between every point on this red outline of the protractor and the real num bers from 0 to 180 inclusive. W here will the point representing the num ber 1 j appear? The p o in t representing the num ber .25? T he point representing the num ber 179.99? Notice that once again we have created a(coordinate system^This time, though, the points lie on the outline of the protractor, wKIle the coordinates

THE MEASURE OF A N ANGLE 29

are the real, n.urnbers greater .Uxan.or equal to 0 and iess than or equal to l 80.Now, were we to place the vertex of an angle a t the point V and one

of the sides along the ray VA, the other side of the angle will intersect the arc at some point. T h e coordinate of this point is called the measure of the angle.

In Figure 2-10 the measure of Z.RVS is 50, for the side VR intersects the> outline of the protractor ai the point whose coordinate is 50. W ith symbols,

this is expressed ass m Z R V S = 50 N

A nd, in the same way a$.ye>tt4}ad interpreted this in-previous worjc in m athematics, we do say thatj’Z.RP.S' an angle nf 50 decrees rso0*). Inst as the inchand the foot are nam e^for emits of m easure for a ljne segm ent/so the degree _ is the name fqr the u n il oTmeasure of an angle. ' ' '----- Retamfrig to 'Flgure 2-id, we notice tEat ------

m Z W V S = 75

andm Z M V S = 149

W h at is the measure of ZN VS? O f ZTVS? O f ZPVS?Before we leave this topic, two features must be stressed. T he first of

these is the fact th a t the measure of an angle is merely the coordinate of the point on the arc . As such it j s a num ber and no more. Hence, we should never express the measure of an angle as, "let us say, 15 degrees, for the coordinate oTa point is the num ber itself w ithout.the word “ degree.” Secondly, by limiting the outline of the protractor we drew, we have restricted ourselves to angles whose measure can be no greater than 180.~This will exclude angles such as the one (ZC B A ) pictured in Figure 2-11. Although angles of

Figure 2-11.

this nature—and others m uch larger, too—occur in m ore advanced courses in mathematics,, they do not arise in our work. We shall, therefore, ignorethem . ------------- ---------""

With the understanding of the measure of an angle a t ou r disposal,we are in a position to define a great m any new terms.

Right Angle and Straight Angle9EBBMEBSH T he angles discussed most frequently in the study of geometry a te the angles about to be defined. They have further im portance since m any other terms are defined in terms of these angles.

30 DEFINITIONS OF GEOMETRIC TERMSD e f i n i t i o n 7: A righ t angle is an angle of 90 degrees.

Equally as often, a right angle is referred to as being an angle whose measure is 90. This, too, is correct, for the statem ent th a t

“T h e m easure of an angle is 90” is equivalent to the statem ent that

“ A certain angle is an angle of 90 degrees.”

T o help fix this in your m ind, these statements will be used interchangeably in this book.

D efin itio n 8: A straight angle is an angle of 180 degrees.

H ow m ight this definition have been worded if the w ard “m easure” appeared in th e definition?

At

r i0 h ,O n 8 1 9 . J tr a ig h t a n g leF ig u r e 2-12.

T h ere is a very im portan t property that follows from this definition th a t m ust not be overlooked. In the discussion of the measure of an angle on p ag e 28 it was iipplied th a t an angle whose measure was 180 would be such that its sides would form a pair of opposite rays. Being opposite rays,i t is possible to say th a t the sides of such an angle fall on the same line. H ence, it follows that

The sides of a straight angle fall on a line.

T his , of course, gave rise to the use of the word straight in the name straight ■angle.

Acute Angle and Obtuse AngleH M B B M A t present we have names for angles whose m easure ise ith e r 90 o r 180. I t is quite apparent that it would be impossible to havedifferent names for every angle as there are infinitely m any angles of differentm easures. W hy is this so? T o simplify this naming process, angles other thanth e right or straight are grouped so as to belong either to the set whosem easures fall betw een 0 and 90 or to the set whose measures fall between 90 a n d 180.

D e f i n i t i o n 9: An acute angle is an angle whose measure is greater than 0 and less th an 90.

D efin it io n 10: An obtuse angle is an angle whose measure is greater than 90 and less than 180.

a n a c u te a n g le a n o b tu ie a n g le

F ig u r e 2-13.

With these two definitions we have completed the process of naming all angles whose measures are greater than 0 but less than or equal to 180. You may have noticed that no name was given to the angle of 0 measure. This was done deliberately, for reference to this angle does not occur in the study of geometry. W hat can be said concerning the sides of an angle of 0 measure?

Complementary and Supplem entary AnglesThroughout the study of m athem atics we often find refer

ence to quantities being treated in pairs. T h e first illustration of this nature that occurs in geometry is the pairing of two angles wherein the sum of their measures is 90. Two such angles would be those whose measures are 63 and 27, or 15 and 75, or 89 and 1. Pairs of angles such as these are called complementary angles.

D e f in it io n 1 1 : Complementary angles are two angles, the sum of whose measures is 90. •

W hen two angles are complementary, one is said to be the complement of the other. Thus, an angle whose m easure is 10 is the complement of one whose measure is 80; an angle of 40 degrees is the complement of one of 50 degrees. W hat is the complement of an angle of 70°? 25°? 1°? 54°? £°r

It m ust have been apparent to you th a t if a special nam e was given to £ pair of angles the sum of whose measure was 90, th a t a special nam e would also be given were the sum 180. Tw o angles having this property are called supplementary angles.

D e f i n i t i o n . 12: Supplementary angles a re two angles the sum of whose measures is 180.

If one of two supplementary angles has a measure of 150, what is the measure of the other? Each of these angles is said to be the supplement of the other. In this illustration the angle whose measure is 150 is the supplem ent of the angle whose measure is 30. W hat is the supplement of an angle of 25°? 40°? 2°? 179°?

Since the measure of a straight angle is also 180, the definition of supplem entary angles m ight have been stated as

D e f i n i t i o n 1 2 a : Supplem entary a n g le s a r e tw o angles t h e s u m of w h o s e m e a s u r e s is the measure oj a straight angle.

This statem ent is equivalent to th a t of Definition 12, for the mr-asurr. oj a straight angle and ISO are the same num ber, the num ber 180!

Similarly, the definition of com plem entary angles is equivalent to the statem ent

D e f i n i t i o n 1 1 a : C om plem en tary angles a re tw o angles th e sum o f w h o se m easures is the m easure o f a r ig h t angle.

EXERCISES

t. Classify each of the following angles as to whether they are acute, right, obtuse, or straight angles.

(a ) 124° (b ) 56° (c) 90°(d ) 12J° (e ) 180° (f) 179}°


(a ) O n the basis of your observation, w hat would you judge the m easureof Z A B C to be? In view of its measure, w hat nam e can be given to ZABC?

(b ) Angle ACD is an angle of approxim ately how m any degrees?(c) Approximately w hat is the measure of /B A C ? W ere this so, by w hat

nam e should / BAC be called?—►

(d ) I f CD and CB are a pair of opposite rays, w hat is the nam e of /B C D ? W h a t is the measure of /B C D ?

(a ) Approximately w hat is th e m easure of ZDCB? W hat is th e nam eof this angle? '

(b ) Approximately w hat is the m easure of ZA D C ? W hat is the nam e of this angle?

(c) Nam e two straight angles in this figure.

THE MEASURE OF AN ANGLE 33

(d) Name four angles that appear to be acute angles.(e) Name two angles that appear to be obtuse angles.

4. (a) W hat is the complement of an angle of

15°, 48°, 1°, 5J°,

(b) W hat is the supplement of an angle whose measure is

126, 57, 38J, 129?, A + B

5. If the measures of an angle and its supplement are equal, what is the measure of each?

6. If the measure of an angle is five times as large as its supplem ent, what are the measures of the angle and its supplement? (H in t: Let * equal the measure of the supplement, then 5x will be the m easure of the angle.)

7. If the measure of an angle is 46 more than its supplem ent, then what is the measure of this angle?

8. (a) When a right angle was defined, into what classification was thisfigure placed?

(b) How was a right angle distinguished from the o ther members of its class?

9. Criticize the following statements as “ definitions.”(a) The measure of a straight angle is 180.(b ) A right angle is the union of two rays having a common endpoint.

10. If the num ber 200 had been assigned to point B in Figure 2-10 on page 28 rather than 180, how would each of the following terms have been defined?(a) right angle (b) straight angle(c) acute angle (d) obtuse angle

11. State your answer and then justify it for each of the following questions:(a) Is it possible for two obtuse angles to be supplementary?(b) Is it possible for an obtuse angle to be com plem entary to an acute

angle?(c) Can a right angle be one of two supplem entary angles?(d) Can a right angle be one of two complementary angles?

Perpendicular Lines

■ B i m B I S W hen two lines intersect, the measures of the angles formed may be any real numbers between 0 and 180. Thus, in Figure 2-14, Z D B C appears to be acute, while Z A B D is apparently obtuse. If. however, a t least one of the angles formed when two lines intersect is a right angle, then the lines are said to be perpendicular.


F ig u re 2-14.

D e f i n i t i o n 1 3 : Perpendicular lines are two lines th a t intersect and form right angles.

F igure 2-15. F igure 2-16. F igu re 2-17.

In F igure 2-15 the rays BA and BC form a right angle, LABC . Hence, these rays are said to be perpendicular. Being rays, however, they are subsets of lines, and as such, their opposite rays can be drawn. I f in Figure 2-15

—Vthe ray opposite to BC is drawn, then two right angles will be formed as in Figure 2-16. Can you justify why L.ABD will have to be a righ t angle? If

—y —>in Figure 2-15 the rays opposite both BA and BC were drawn, then fourright angles w ould be formed as in Figure 2-17. Since angles A B C and A B Dare righ t angles, can you justify why angles D B E and E B C should be right angles also?

T h e symbol used to represent the word “ perpendicular” is ± . In view

of this, the expression A B X CD is read as, A B is perpendicular to CD.

EXERCISES

U sing the information given, nam e the right angles in each of th e figures below.

1 < - +. A B X BD R T X S T 2.

tI

. - t

3. AC 1 BD

THE MEASURE OF AN ANGLE

7. A C X C D

RS X R W 4.35

OP X A B 8 .

Bisector of an Angle■ ■ ■ ■ ■ ■ ■ ■ Earlier we learned that it was possible to have a point, the m idpoint, forming two line segments of equal measure on a given line segment. So, too, is it possible to have a ray whose endpoint is the vertex of an angle and creating two angles of equal measure from this angle. A ray such as this is called the bisector of an angle.

D e f in it io n 14: T he bisector of an angle is a ray such that its endpoint is the vertex of the angle and it forms two angles of equal measure w ith the sides of the angle.


W ere we given the information for Figure 2-18 that BD is the bisector of (or bisects) /.ABC , the definition of an angle bisector would perm it us to conclude that m /A B D = m /D B C .

Frequently, students beginning their study of geometry find it a bit difficult to select the two angles of equal measure when an angle has been bisected. This is particularly true when the figure contains m any lines. Should this occur to you, the following points may help clear up your difficulty.

(1) T he vertex of the angle that has been bisected and the vertices of the angles of equal measure are identically the same point.(2) T he bisector of the angle will be a side of cach of the two new angles th a t have been formed.(3) Each side of the bisected angle will be a side in each of the new angles th a t have been formed.

As a last resort, ignore everything in the diagram other than the three rays consisting of the angle bisector and the two sides of the angle th a t has been bisected. Thus, in Figure 2-19 with the data given, think only of these

—4 "4rays: R W , RS, R T . By so doing, it should soon become apparent th a t the angles of equal measure must be / W R S and /T R S .

RS is the bisector of / W R T .

EXERCISES

In Problems 1 through 8 what conclusion can be draw n in te rm s of the da ta given?

*1. D B is the bisector of /A D C . I AC is the bisector of /B A E . 2 .

THE MEASURE O F A N ANGLE

3 . BE bisects /A B C .

7 , WS is the bisector of / R W T .

BD bisects /A B C .

A

4.

31

5 C is the bisector of /A D B . 6 .

9.

, “ C D - 4 ■

If in the diagram above m /BAC = m /CAD = m /DAE, then AC and AD are said to be the trisectors of /.BAE, How would you define the trisectors of an angle?Congruency of Angles and Line Segments■ ■ ■ H i The symbols that ws have used in expressing the equality of measures between line segments or between angles have been found to be rather cumbersome. To overcome this feature, a new symbol was invented, this being It is read as the word “congruent” and appears in relations

such as

(1) A B SS CD and (2) Z X Y Z S Z R S T

These relations are read as

( t ) A B is congruent to CD. and (2) Z.X YZ is congruent to Z R S T .

T o make this symbol operational, the following two definitions a re necessary:D e f i n i t i o n 15: Congruent line segments are line segments having equal

measures.

D e f i n i t i o n 16: Congruent angles are angles having equal measures.

In view of these definitions, where formerly we spoke of the equality of the measures of line segments, now we can refer directly to the congruence of these segments. And a similar relation will hold with angles. Hence, the following statem ents are said to be equivalent:

m A B — m CD is equivalent to A B = CD. m Z X Y Z = m Z R S T is equivalent to Z X Y Z S Z R S T .

Furtherm ore, the definitions of the m idpoint of a line segment and bisector o f an angle can be restated in terms of congruence ra th e r than in terms of tqual measures. Thus,

T h e m idpoint of a lin e segment is a point on that segm ent such that the two segments formed are congruent.

Com pare this definition with the definition of the m idppint that appears op page 25. Can you write a similar definition for the bisector of an angle?

The following problem will illustrate the m anner in which conclusions will henceforth be made.


O n the basis of the data given, we shall say that

Z B A D S Z D AC

a n d justify our conclusion by quoting the revised definition of the bisector o f an angle:

The bisector of an angle is a ray such that its endpoint is the vertex o f the angle and it forms two congruent angles w ith the sides o f the g-iven angle.

THE MEASURE OF AN ANGLE39

H ,d £ » F « „ , . 2 .® b « » „ive„ . . U *

drawn would be EC

< o r « * « » » « “ ‘iPOin,of a line segment.

EXERCISESW hat conclusion about congruence can be d raw n from the

d ata given in each of the problems below?

1 . D is the midpoint of AC.

0

3 . BD bisects AC.D

5 . E is the m idpoint of BD.

7 . D B is the bisector of EG.

G

I 1E

D.B bisects ZADC. 2.

AE is the bisector of Z D A B .

__ oM is the m idpoint a( AB. V ■

H Drawing a Conclusion Based on the Reverse of a DefinitionYou may recall that the fourth property given to a con

notative definition was the fact th a t it was reversible. This was interpreted as

I f th e su b ject and p red ica te n om in ative o f a d efin itio n a r e in te r c h a n g e d , th e n th e n ew se n ten ce w ill b e a tru e statem ent.

Thus, the reverse o f the definition of the bisector of an angle will be

A ray w hose en d p o in t is the vertex o f a n a n g le a n d th a t form s co n g ru en t a n g les w ith the sid es o f th e a n g le is th e b isector o f th e a n g le .

W hat is the reverse of the definition of the bisector of a line segment? O f the m idpoint of a line segment? O f perpendicular lines? O f a righ t angle? O f a straight angle?

T o illustrate how the reverse of a definition can be applied, let us sup-—■'Vpose that Z A B D ~ ZD B C in Figure 2-21. In order th a t this can be so, BD

Figure 2-21.

w il l have to be the bisector of ZABC . Hence, we say th a t BD is th e bisector o f ZA B C . T o justify this conclusion we merely refer to the reverse of the de fin ition of the bisector of an angle that appears in red above.

You m ay be questioning this procedure by saying, "W hy isn 't it poss ib le to justify this conclusion by simply referring to the definition of the b isec to r of an angle rather than the reverse?” . T he definition of the bisector o f an angle can be used as justification of a conclusion only if th e d a ta given s ta te d th a t the ray was the bisector. In this problem, however, this was not s ta te d . You knew only that there were two congruent angles, and from this y ou, yourself, concluded th a t the ray BD must be the bisector of the angle. W h e n such a conclusion is m ade, it is justified by quoting the reverse of the d e fin itio n . This is so since it is the reverse of the definition that, states that

CONCLUSION BASED ON REVERSE OF DEFINITION 41the ray-ihat fo rm congruent angles with the sides of the angle is the bisector of the angle.

As another illustration, in Figure 2-22 AD S DC. From this it can be

concluded that D is the m idpoint of AC. T o justify the conclusion, we would

say “ A point on a line segment that forms two congruent line segments is the m idpoint of th a t line segm ent.”

T he reverse of the definition is given since it is known th a t point D forms two congruent line segments on~A5. T he definition of the m idpoin t of a line segment was not given as the reason, for th e fact th a t D is th e m idpoint was no t known to us a t the outset of the problem,

EXERCISESW hat conclusion can be draw n on the basis of th e da ta given

in each of the problems below? Justify your conclusion by stating the reverse of one of the definitions you have had. In order th a t your work will follow the pattern th a t will be used throughout the year, arrange your conclusion

and reason as it is shown below. D

Illustration.*

A Given: ZA B C is a straight angle.

CONCLUSION REASON

Z A B D and ZD BC are supplem entary angles.

Two angles the sum of whose measures is the measure of a straight angle are supplementary angles.

1. Given; Z B A D ^ Z D A C Given: ZA C B is a right angle. A-

3 . C-iven: AD S DC A,:


Given: Z A B C is a right angle. A . (Give two conclusions.)

A ------------------------------------------

Given: m Z A B C = 62

A,

• Given: m Z A B C = 134

A ----------------—

. Given: ZA C B S ZACD

Given: Z A B C is a straight angle.

C

Given: Z A B C is a straight 6 . angle.

- O -

Given: Z A D B is a right 10.angle.

Given: A E S EC

A12.

DEFINITIONS AND REVERSE OF DEFINITIONS 43

| Drawing Conclusions on the Basis of Definitions and the Reverse of Definitions

At the outset of this text it was pointed out that there are a great many ways in which arguments can be “ proved.” O ur objective this year is to show how the m athem atician justifies the conclusions th a t he makes. Thus far, you have learned that he can justify these conclusions through one of two ways :

(1) He can refer to the accepted definitions of the words that appear in the data that have been given.(2) He can refer to the reverse of the definitions that have been agreed upon.

O ur attitude toward mathematics is determ ined largely on how well we understand the work we are called upon to do. T h e following set of problems is designed to help you gain confidence in draw ing conclusions and then justifying these conclusions through either the definitions of terms or the reverse of the definitions. Arrange your work as illustrated in the problem below.

Illustration:

Given: BD JL AC

CONCLUSION REASON

Z A D B and ZC D B are right angles. *

Perpendicular lines are two lines th a t intersect and form right angles. (Def.)

EXERCISES

Draw a conclusion in each of the problems below. Your conclusion should be based only on the data th a t has br.en given, not on w hat you believe the diagram may imply. Im m ediately following your reason, indicate whether this reason is a definition (Def.) or the reverse of a definition (Rev. of Def.).

44

1 . Given: E is the m idpoint of AC.

DEFINITIONS OF GEOMETRIC TERMS

Given: CA bisects ZD C B. 2.

3 . Given: Z A C B is an acute angle.

Given: Z A B C and ZC B D are supplem entary angles.

4.

* B 0

Given: CE bisects ZAC D . 6 .

7. Given: A is the m idpoint of BC.

angle.Given: Z A B D and Z D B C 10.

are complementary angles.

DEFINITIONS AND REVERSE OF DEFINITIONS

11. Given: “a B bisects CD.

13. Given: Z A B C is a right angle. (Give twoconclusions.)

15. Given: Z R V S is a straight angle. (Give twoconclusions.)

17. Given: A B =. BC

A.

19. Given: FC bisects BE.

Given: AD X BC 12.

45

Given: m ZC A B — 55; m ZC B A =■ 35

14.

G iven: AC is the perpendicu- 16. lar bisector of BD.(Give two conclusions.)

angle.


21. Given: B and C are the trisection points of AD (See page 27. Problem 3.)

A___ B C

23. Given: L A C B and ABCD are complementary angles.

A v ----------------------------------------- .C

25. Given: /.A D C is a rightangle.

■ Test

Given: Z Y ± X Y 22.

Al

Given: R T and S fV bisect 24.each other. (Give two conclusions.)

R

Given: Z B E C ^ Z D E C 26. 8

IAJ1. (a) W hat is m eant by the measure of a line segment?

(b ) Is it possible for the same line segment to have two different measures? Justify your answer.

2. (a) W hat is m eant by the measure of an angle?

(b ) Is it possible for the same angle to have two different measures?■ > • — ^ <—»

3. I f VR and TO are subsets of RS, then w hat are the two possibilities concerning the m easure of Z R V S ?

4. If ZABC is an acute angle, what can be said concerning the measure

of its supplement?

5. (a) If A B = BC, does this imply that B is the m idpoint of *4C? Justify

your answer.

(b) If Z.ABC = ZCBD, does this imply that BC is the bisector of

ZA B D ? Justify your answer.

6. U nder what conditions only would it be possible to write the following

equality?A B = CD

7. Classify each of the following angles as to whether they are acute, right,

obtuse, or straight.(a ) 90° (b) 156° (c) 84°

8. (a) W hat is the measure of an angle if the measure of its complement

is 82?(b ) W hat is the measure of an angle if its supplement is an angle o fy

degrees?

9. (a) W hat is the measure of an angle th a t is four times as large as its

complement?(b ) T he measure of an angle is 15 more than twice its supplem ent. How

large is the angle?

10. T he definition of a reflex angle is

A reflex angle is an angle whose measure is greater than 180 and less th an 360.

Answer the following questions in terms of this definition: r (a) W hat is the reverse of this definition?

(b ) Explain why a reflex angle is no t discussed in plane geometry?(c) How was a reflex angle classified?(d ) How was a reflex angle distinguished from the o ther m em bers of its

class?(e) W hat can be said concerning the supplement of a reflex .angle?

0Draw a conclusion in cach of the problems below. Im m e

diately following your reason for this conclusion, indicate w hether this reason is a definition or the reverse of a definition.

TEST 47


1 . Given: F is the m idpoint of AE.

3. Given: / .D A E i s /C A E

5 . G iven: B E = CE

7. Given: /C D E and /A D E are com plem entary angles,

G iven: A D _L ED

Given: /B D A is a right angle. 4.

Given: A f bisects/4C.

A B

Given: /A B C is a straight angle.

9. Given: D E bisects /B D C .

TEST

Given: BD is the _L bisector 10. of AC. (Give two conclusions.)

49

3Assumptions and Their Place in a Proof

E A R L IER WE DISCU SSED T H E IM P O R T A N T role definitions play in helping to mold the decisions we make each day of our lives. T h ere are, however, factors other than the interpretation of words th a t lead us to the conclusions we make or the acts we do. T he decision of whether Jo h n or Bill is the better candidate for president of the student council frequently has little bearing on how terms are defined. O u r attitude toward questions such as this is shaped more by the convictions we hold about these young men than by the clarification of terms they may have used. T his chap ter is devoted to showing a t least a few of the ways in which convictions are born and, when related to m athem atics, their significance to the developm ent of this subject.

■ How Do the Blind Draw Conclusions?Among the poems you m ay have read and enjoyed is the

one below about the blind men and the elephant. As you read it through now, ask yourself why a poem such as this should appear in a geometry book. Is there any connection between this poem and the basis for conclusions that we, as hum an beings, reach each day of our lives?

H C W D O THE BUND DRAW CONCLUSIONS'?

T he Blind M en and t h e E leph a n t by John Godfrey Saxe

It was six men of Indostan,T o learning m uch inclined,

Who went to see the elephant(Though each of them was blind,)

T hat each by observation M ight satisfy his mind.

The first approached the elephant.And happening to fall

Against his broad and sturdy side,At once began to bawl:

“ God bless me! but the elephant Is very m uch like a w a ll!”

The second, feeling of the tusk,Cried: “ Ho! w hat have we here

So round, and smooth, and sharp?To me ’tis very clear

This wonder of an elephant Is very like a spear!”

The third approached the animal,And happening to take

The squirming trunk within his hands, Thus boldly up he spake:

“ I see,” quoth he, “ the elephant Is very m uch like a snake!”

The fourth reached out his eager hand, And fell upon the knee:

“ W hat most this wondrous beast is like, Is very plain,” quoth he;

“ ’Tis clear enough the elephant Is very like a tre e !”

The fifth who chanced to touch the ear Said: “ E’en the blindest man

Can tell w hat this resembles most: Deny the fact who can,

This marvel of an elephant Is very like a fa n !”

The sixth no sooner had begun About the beast to grope,

Then, seizing on the swinging tail T h at fell within his scope,

52ASSUMPTIONS A N D THEIR PLACE IN A PROOF

“ I see,” qu o th he, " th e elephant,Is very like a rope!”

And so these men of Indostan Disputed load and long,

Each in his own opinionExceeding stiff and strong,

T hough each was partly in the right,And all were in the wrong!

U pon w hat evidence d id the first blind man draw his conclusion th a tthe elephan t resembled a wall? W hy d id the second blind m an arrive a t thedecision he had made? W hat had led each of them to the conclusion he h ad drawn?

A r e W e So D iffe re n t fro m th e B lin d ?

■ ■ ■ ■ ■ ■ H i You were probably a little amused as you read the poem, for you knew th a t each “ saw” bu t a sm all p a rt of the elephant. W ere these m en capable of seeing the anim al in its entirety, they would no t have com e to such completely different points of view. A nd yet, perhaps we should no t trea t this poem too lightly. W e norm al hum an beings m ay be wearing blinders w ithout realizing it. In terms of one interpretation th a t some scientists give to light rays, the graph below presents a picture of the relative sizes o f the wavelengths of “ light” coming from the sun.

viiibl* lightlight' Wavelengths

-J--- !_ Hi III

W o o ~oW a v e le n g th {c e n tim e te r* )

T o be exact, a wavelength of light that can be seen by the hum an eye ranges from 4 hundred-thousandths to 7 hundred-thousandths o f a centim eter! T h e length of each “ wave" is the length of the line segment jo in ing

Figure 3-1.

tw o peak's such as A and B. T h e centim eter itself is a very small distance, while 4 hundred-thousandths of a centimeter is far, far sm aller than the

POSTULATES IN GEOMETRY 53

naked eye might ever see. Thus, it appears that of all the “ light” waves that strike the earth, only a very small fraction of these are visible to the hum an eye. Hence, we may be just a little removed from the blind men of the poem. Were we suddenly granted the ability to observe “ light” whose wavelengths varied between 1 millionth of a centimeter in length and 1 hundredth of a

1 cen tim eter

centimeter, then the things that seem “ obvious” to us now m ight appear rather silly under our newly found “ vision.” Many of the things we say and many of the ideas we have are based on the observation of things as. we see them. And this vision of ours may not only be faulty but is certainly lim ited !

The Faulty Eyesight of Man■ ■ ■ ■ ■ ■ ■ Since m an first began to think, his eye-sight led him to create many beliefs that he felt were justifiable. At the present we realize that a great num ber were completely groundless. Among these was the notion that the earth was flat. For, so it was reasoned, if the earth was round, how might a m an stand a t the south pole without tum bling off into nothingness! Yes, there were some scientists who realized that the earth was “ round .” In fact, an early Greek mathematician gave a ra ther close approxim ation for the circumference of the earth, These men, however, were in the minority, a very small m inority a t that. The great m ultitude of people “ realized” that the earth had to be flat, for did not their eyes tell them so !

EXERCISES

1. List three beliefs in the field of aviation that had been held for m any years yet now have been discarded.

2. List three beliefs in the field of medicine that were discarded because of •medical research.

3. List five beliefs in general that people held for m any years bu t now feel that they are probably not so.

H Postulates in GeometryThe early Greek m athem atician laid the foundation for his

work on certain “ beliefs” that he referred to as “ self-evident tru ths,” for to him they reflected the world about him as he saw it. T he m ere sophisticated modern m athematician, however, recognizes the principles from which he evolves his subject for what they really are; that is, no more than a a t of mutually agreed upon properties about figures that he himself has created. T o illustrate, he created the line and the real num ber system; he then linked the

54 ASSUMPTIONS AND THEIR PLACE IN A PROOFtwo together by the property th a t to every real num ber there exists one and only one point on this line and further, th a t to every point there exists one and only one real num ber. A statem ent such as this the m athem atician accepts in order to lay the foundation for further discussion. This discussion can not take place w ithout some core of ideas or frame of reference from which to begin. These fundamental principles from which m athem atics springs are the postulates or, as they are often called, axioms or assumptions.

Current interpretation of m athematics has divorced a postulate from its original in terpretation as a “ belief.” W hen considering a belief, we are sometimes left w ith a feeling of vagueness as to the tru th or falsity of the statement. O n the other hand, a postulate carries with it no such squeam ishness; it is like asking, “ Are the rules of baseball true or false?” T he question, obviously, makes little sense. These are the rules established to create the game; they are neither true nor false! So, too, are the postulates, the “ rules” to create the subject of mathematics.

I t would be inaccurate to leave you with the impression that creating mathematics is com parable to creating a game where the rules are established at the whim of the creator. Far from this! M athem aticians frequently form ulate their postulates as a model o f w hat they perceive in the w orld th a t exists about them. I t is true that some aspects of m odern m athem atics appear to bear no remote resemblance to any practical application. However, as has happened almost always in the past, future scientists will very likely discover a need for these branches o f m athematics in their work.

In view of the foregoing analysis, it would seem wise if we, too, placed our cards face up on the table and admitted to a num ber of postulates we had failed to establish in Chapters 1 and 2, although we had m ade use of them. Briefly, they were concerned with these points:

(1) T h e possibility of extending a line as far as desired in either direction.(2) T he notion of "betweenness” th a t implied th a t between any two points on a line there existed a third point.(3) T he existence of a pairing between the points on a line and the real numbers.

These principles we will now state formally as our first three postulates:

P o s t u l a t e 1: A line may be extended as far as desired in either direction. P o s t u l a t e 2: For any two points on a line, there exists a third poin t that

is between them.P o s t u l a t e 3: T here exists a one-to-one correspondence between the points

on a line and the real numbers.

T he last of the principles concerning a line th a t we w ant to consider at this time is one that you have used each time you drew a m argin on a piece of paper. In draw ing a “ half-inch” margin, it is likely th a t you placed a point \ inch from the edge near the top of the paper and repeated this process at the bottom. After which you laid the ruler along these two points

POSTULATES IN GEOMETRY 55

and drew the line. In so doing, you were using the principle stated here:

P o s t u l a t e 4: There exists one and only one line through two points.

T h e combination of Postulate 3 and the interpretation we have given to the measure of a line segment as being the coordinate of one endpoint when the coordinate of the other endpoint is zero is but a modified form of two postulates that are often called T he R uler Postulate and T he R uler Placem ent Postulate. Had these two postulates been a t our disposal earlier, we could have created a m ore forma) definition of "betw eenness.” Thus,

Point B being between points A and C means th a t the three points are different elements of the same line and,

m A B -f m ~BC = m AC

By examining Figure 3-2 it becomes evident that in a ra ther elaborate way this definition says no more than if it so happens th a t when the “ distance” from A to B is added to the “ distance” from B to C the sum turns

Figure 3-2.

out to be the "distance” from A to C, then B is between A and C.At the time we established the measure of a line segment we in some

ways trapped ourselves into an awkward position. Consider the coordinate system below. There would be no problem in determ ining the measure of

-5 -4 - 3 - 2 - 1 0 1 2 3 4 5 6 7 8 9 10 1112 13 14 15 I t 17 18 19 20

______ I___I___I ... I___I___I__________I I - I l I I___1 1 , 1 , 1 ___I___I______^

B S P /t S C D

F ig u r e 3 -3 .

TA for the coordinate of P is zero and hence, the coordinate of A which is 5 represents the measure of PA. Similarly, finding the measures of ~FB, TC, and TT) also presents no difficulty for in each case the coordinate of one of the endpoints, P, is zero. However, finding the measure o l A B does create a problem for the coordinate of neither A nor B is zero. Hence, either we set up a new coordinate system with A o r B as the zero point or we develop another device for determining the m easure of a line segment. T h e latter of these alternatives frees us from constantly having to shift the zero point,

By simply counting the num ber of units from A to B we can see th a t m AB = 3, Doing the same for the num ber of units from B to C, we discover th a t tr.'SC = 7 and, similarly m 7tP = 2 while m HB = 10. Obviously, counting the units between the two endpoints of a line segment is not very practical and by now you must certainly have discovered that the 3 which is the measure of A B can be determ ined by subtracting the coordinate, 5, of A from the coordinate, 8, of B. In the sam e way,

56 ASSUMPTIONS AN D THEIR PLACE IN A PROOF171 'EC = 1 5 — 8 = 7

and m~AB = 17 — 5 = 12

How '.vould you find the measure of 5 5 ? T he measure of UB? W e do run into a slight difficulty, though, in determ ining the m easuie cf S 3 for the coordinate of R is a negative num ber. By recalling the technique for com puting the difference between two signed numbers, this problem is easily elim inated. Thus,

m K ? = 5 - ( - 2 ) = 7 and m 315 = 17 — ( — 1) = 18

W hat would happen, however, if we inadvertently interchanged the positions of the coordinates in finding th e difference? Then,

m ~B = 5 — 8 = —3m ~E€ = 8 — 15 = —7

and m Z D = 5 — 17 = —12

Hence it appears th a t each of the answers turns out to be the negative ofw h a t w e would w ant it to be. T o elim inate this from occurring we m ake useof the concept of the absolute value of a number th a t we had learned in our study of algebra.

|12 - 3| = |9| = 9|3 - 12) = | —9| = 9

Recall th a t the absolute value of either a positive or a negative num ber was the num ber itself devoid of its sign. Thus, in the situations above we will say,

m W = |8 - 15| = | —7| = 7and m A B = |5 - 17| = | - 1 2 | = 12Similarly, m TlB — | — 2 — 8| = | —10) = 10and m TJR = |15 - ( - 2 ) | = |17| = 17

Now we are in a position where we can conceive of th e m easure of a line segment:

T he measure of a line segment is the absolute value of the difference of the coordinates of its endpoint?.

EXERCISES

1. Express each of the following in terms of a single num eral,

. (a) |8 — 2| (b) |15 - 6| (c) ]12 - 5](d ) |3 — 7 | (e ) |5 - 6 | (f) |0 - 2|(g ) 1 - 2 - 5 | (h ) | - 6 - 3| (i) | - 1 - 7 |G) |6 - ( ~ 2 ) | (k ) |5 - ( -4 )1 : (1) |8 - ( —9)|(m ) | - 2 - ( - 3 ) | (n ) | - 7 - ( - 7 ) | (o) |0 - ( - 5 ) |

THE SUM AND DIFFERENCE OF TWO LINE SEGMENTS 57

j ' f A and B are given in each of the problems2. T he coordinates of points A ana 6below. Find the measure of I S .

(a) A: 10; B: 7 (d) A: 6; B: - 5 (g ) A: - 5 ; B: 8

(b ) A: 26; B: 5 ( e ) A: 0; B: - 8 (h ) A: - 3 ; B: 3

(c)(f)(i)

8; B: - 1 - 2 ; B: 6 - 1 ; B: - 5

3. -3 -2

Using the coordinate system above, determ ine the measure of each of

the following line segments.(a) Z5is (b) ~UQ (c) b B(d ) W (e ) T5S (f ) F R(g) T J U (h) TO (i) W0‘) ~EA (k ) AT) (1) U B

4. Given the information that B is between A and C, determ ine the answer

to each of the following problems.(a) m Z § = 6, m BU = 2, m JSU = ?(b ) m ~Kf3 — 5, m EC = 9, m 1C = ?(c) m A fi = 9, m EC = 2, m AB = ?(d ) m Z C = 16, m EC = 1, m ~AB = ?(e ) m A B — 5, m AC *= 12, m EC = ?( f ) m A S — a, m ISC ~ b, m Z C = ?

5. I f m TQ - 7, m = 12, and m 1FR = 5, and P, Q and R are points of the same, line, then which of these points is between the other two?

6. (a) The coordinate of A is 4 while m Z B = 15. If the coordinate of Bis a positive num ber, w hat is that number?

(b ) In a coordinate system the coordinate of A is 2 while the coordinate of B is 7, If the same unit is used for a coordinate system where the coordinate of A is changed to 0, then w hat is the coordinate of B

in this system?

B The Sum and Difference of Two Line SegmentsT he postulates used most often in geom etry are called the

“ operational” postulates, for they are concerned w ith the operations of addition, subtraction, m ultiplication, and division. Before they are investigated, it is apparent th a t we will have to in terpret these operations w ith

reference to geometric figures.D efinition 17: T h e sum o f two line segments, I B and BC, is J C if, and

only if, B is hetween A and C.

Figure 3-4.

58 ASSUMPTIONS AND THEIR PLACE IN A PROOFBy using symbols this relationship will be expressed as

AB + BC = AC

Having used the equality sign, we mean tc imply no m ore or no less than that the nam e used on the left (AB + BC) and the name used on the righ t (/ft?) are but two different names for the same segment (~AC). This situation is similar to saying that 4 -j- 3 ■-* 7 for both the 4 + 3 and the 7 are merely two different names for the num ber seven.

The definition, furthermore, states that the sum of two line segments exists only if the two line segments lie on the same line and their intersection is one and only one point. In Figure 3-4 I S and ~BC lie on line AC, while B is the only po in t they have in common.

W ith reference to the definition explain why the sums given below can not be found.

8

A________ C B A 1 C D

XI + sc Si + CB

F i g u r e 3 -5 .

T h e diagram s we will encounter will frequently have many, m any lines in them. W hen trying to find the line segment th a t represents the sum of two line segments, direct your attention to that line on which the two segments fall. All o ther lines will have no im portance a t th a t time. Thus, if you are

- 1>looking in Figure 3-6 for A F + FE, focus your attention on A E and ignore

F ig u re 3-6,all other lines. T he definition enables you to say that

I F + T E = A E

In this same diagram w hat is th e sum of B £ and £ £ ? ZD and T>C? "BP and FZ5? Explain why the sum of A F and FD does not exist.

D e f in it io n 18: T h e difference between two line segments, AB and BC, is ~AC if, and only if, C is between A and B.

. . . A '__________ ________________ C___________ 8

Figure 3-7,

THE SUM AND DIFFERENCE OF TWO ANGLES 59

By using symbols this relationship w illb e expressed as

I B - W = AC

Here, as in the addition of line segments, the use of the equality sign signifies merely that the two names (I B — BC) and (ZC) are but different names for the same segment (AC). Also, as before, subtraction of line segments can exist only if the two segments are subsets of the same line and their intersection is one of these segments. In the d iagram above, A B and ~EC are subsets of line AB, while their intersection consists of all the points of one of these segments (BC).

EXERCISESIn each of the problems below, you are asked to find the

sum or difference of two line segments. If it is hot possible to find this line segment, simply write “ no answer” after the problem number.

1 . (a) I B + BC = ?

(b ) AC - BC = ?

2. (a ) 2Z5 + Z5C = ?(b ) EC - BC = ?

(c) IB +BU = ?

(d) CE - EZ5 = ?

3. (a) aE + EC = ?

(b) 1C - EC = ?

(c) W + UD = ?

•, (d) J B - BE = ?

.(e) W . + M » ?

(f) a D - BC = ?

■ The Sum and Difference of Two AnglesIn the same way that we defined a point as being between

two points so, too, is it possible to define a ray as being between two rays.Thus,

D e f i n it io n 19: The ray P S being between two rays PA and PC means that,

m Z APB + m LB P C = m LAPQ

And here again we seem to be saying no more than th a t we w ant the

60 ASSUMPTIONS AND THEIR PLACE IN A PROOFsum of the measures of / APB and ZBPC to be the same as the m easure of /A P C . In Figure 3-8 this would imply

• /

F igu re 3-8. F igure 3-9. F ig u re 3-10.

an d 40 of the two sm aller angles are added, their sum of 70 m ust be them easure of ZA P C . In Figure 3-9 we see th a t this definition holds equallywell w here the ray PB happens to be between the pair of opposite rays PA and PC.

—V ■- ■> - ■")As seen in Figure 3-10, PB is not considered to be between PA and PCfor th e sum of the measures of Z A P B and ZB P C does not equal the m easure of ZAPC .

W ith Definition 19 a t our disposal, defining the sum or difference of two- angles will present little difficulty.

D efin it io n 20; T h e sum of two angles, ZA B C and ZD BC , is Z A B D if,

and only if, BC is between BA and BD.

F ig u r e 3-11,

By using symbols this relationship can be expressed as

ZA B C + ZD BC = ZABD

H e r e again, the equality sign is used to imply the fact that although the nam e t h a t appears on the left ( ZA B C - f ZD BC ) is different from the name a p p e a r in g on the righ t (ZA B D ), both represent the same angle (ZA B D ).

In Figure 3-12 the sum of ZB A D and ZCAD is ZBAC. In this figure,—> —V —><AD is between A B and AC. W hat is the sum of ZA B D and ZCBD? How w o u ld you express ZA C D .-h ZBC D by naming a single angle?

THE SUM A N D DIFFERENCE OF TWO ANGLES

A

61

in stio n 2 1 : i n e u m c i u ^ . . --------- 0 ,—> > # ►

if, and only if, BC is between BA and BD.

Symbolically we express this asZA B D - ZABC = ZCBD

where the two names (Z A B D — ZABC ) and (ZC B D ) are bu t different

names representing the same angle (ZCBD),

EXERCISESIn each of the problems below, you are asked to find the

sum or difference of two angles. I f it is not possible to find this angle, simply write “ no answer” after the problem number,

(a) LE A D + ZD AB = ?(b) Z A B D y ZCBD — ?(c) Z A D E + ZA D B = ?(d ) ZA E D + ZBCD = ?(e) ZE A B - ZD A B = ?(f) Z A B C - / C B D •= ?(g) ZAEC — ZB C E — ?(h ) ZED C - ZA D C = ?

(a) ZBAC + /C A E = ?(b) ZACB + ZA C E = ?(c) ZC A E + /B E A = ?(d ) /B D C + ZE D C = ?(e) ZABC - / A B E = ?(f) ZBC E - / B A E - ?(g) Z B A E - ZC A E = ?(h) /A D C - /A D E = ?

A

2.

62 ASSUMPTIONS AND THEIR PLACE IN A PROOF

| The Addition PostulateHaving defined what is m eant by the “ sum ” and “ differ

ence” with reference to line segments and angles, we a r t now prepared to introduce the first of the “ operational” postulates. These are the postulates that were needed in determ ining the solution set. or roots, cf an equation at the time you studied algebra. As an example,

If (1) x - 2 = 7then (2) x ~ 2 + 2 = 7 + 2

The reason enabling you to write step (2) based on the information given in step (1) was the " law ” stating that

If a = 1 (A)then a + c = 6 + c

In the illustration above, a — x — 2, 6 — 7, while c = 2. Perhaps youlearned this postulate as

“ If equals are added to equals, the sums are equal.” ra ther than with the symbols given in (A). In either event, we shall need this postulate, and the o ther operational postulates, as a foundation for the geometry work. Hence, it will be repeated here, in a slightly different form.

P ostulate 5: The Addition Postulate If a = band c = dthen a + c = b + d

O r as a statem ent:If equals are added to equals, the sums are equal.

W hen this postulate was applied in algebra, the symbols a, b, c, and d were placeholders for num bers. Furtherm ore, the statem ent a — b implied th a t the symbol a and the symbol b were merely different names for the same num ber. T his will still be true in our work in geometry. Now, however, we m ust realize th a t since the. statements

m l B = m CZ5 and ~AB — V B

are equivalent, we can and will treat the congruence symbol ( ~ ) in identically the same w ay as we had treated the equality symbol («=) in algebra. Hence, were we to apply the Addition Postulate in the following situation:

Given: J B ^ T J E

W s i ' E ?

Figure 3-14.

we could conclude th a t(1) JB + W'S*7TE + W

M oreover, since ~AE + 5 C is b u t an o th e r nam e for "AC, w hile TJE + E F is

an o th e r nam e for U P , it is p refe rab le to w rite (1) as

J C ^ T J F

T his can be expressed in quite an elem entary way as follows:

If the measure of ~A& is 9 (inches), then D E must have a measure of 9 (inches), for the two segments by being congruent have the sam e measure. Similarly, if the measure of BU is 5 (inches), then the measure of F F must

THE ADDITION POSTULATE 63

9* b 5“________ _— - o -

9"-V -

D £ FF igure 3-15.

be 5 (inches). The conclusion th a t ~XC S ~BF can be interpreted here as merely implying that both segments have a measure of 14 (inches).

I t frequently happens that there are m any lines in the diagram that are completely irrelevant to the data given. T ry to ignore these lines. T o illustrate, Figure 3-15 m ight have been draw n as

C

G iven : A S — B E

W ^ T Z F

F ig u r e 3-16.

A^ before, the conclusion would still be A C ~ B F and for the reason stated; that is, the Addition Postulate. Nam e five lines in the d iagram th a t are inno way related to the information stated.

T h e Addition Fostulate as applied to angles would occur under condi

tions such as the following:

Given: /LABD — Z.ACD ZD BC — ZD G B

Figure 3-17.


W hat conclusion can be drawn on the basis of the given data? T he use of num bers may help make the picture a little clearer to you. If /.A B D has a m easure cf 20, then /A C D will also have the measure of 20, for as siated in the given data, these two angles are congruent. So, too, if the m / DBC = 30, w hat can be said of / DCB? W hat is the measure of /A B C ? O f /A C B ? ■Since the measure of both is 50, w hat conclusion should follow?

T his same conclusion could have heen arrived at by resorting to the A ddition Postulate ra ther than to specific numbers. T he formal arrangem ent of the problem is identical to th a t used on page 41. This time, however, the reason that justifies the conclusion is a postulate rather than a definition or the reverse of a definition.

CONCLUSION REASON

/A B C S /A C B The Addition Postulate: If congruent angles ( /A B D and /A C D ) are added to congruent angles ( /D B C and /D C B ), their sums will be congruent angles ( /A B C and /A C B ).

EXERCISES

By using the Addition Postulate only, what conclusion can b e d raw n in each of the following problems? State your conclusion and re a so n exactly as shown above.

1. Given: /A B G — /D E H /G B C £* /H E F

A / 0 /

3 • Given: A B = ~AE 1

Given: A E £= ~BF ~ E D S iT €

A

- V -S

Given: /D AB'-. /D A C ',

/D B A ; /D B G

4 .

the SUBTRACTION POSTULATE

5 . Given: S U■ p E & W

7 G iv e n : /E E C £ /C D F /D E C 9 Z Z-EDF

9 . G iv e n : T J E s C E

65

Given: AF■pE c x W

Given: /A C B ~ /E C D 8./A C F S /E C F

G iv e n : X E s E E

M & ’SC

| The Subtraction PostulateT h e second of the operational postulates was first encoun

tered a t the tim e it was necessary to determine the solution set of the equation

(1 ) * + 5 = 17

T o find this num ber we first w rote (A)(2 ) x ■+• 5 — 5 = 17 — 5

In justifying step (2) on the basis of the inform ation given in step (1), we

applied the following postulate:

P o s t u l a t e 6 : The Subtraction PostulateI f a = band c — dthen a — t ~ b — d


O r as a statement'.

If equals are subtracted from equals, the differences are equal.

In illustration (A) above, a = x + 5, b = 17, c = 5, and i - 5. Kence a — c is x + 5 — 5, while b — d is 17 — 5.

In the study of geom etry this postulate wiil be applied to line segments and angles very m uch the same way we had used the Addition Postulate. Consider Figure 3-18.

Given:E E ^ F D

i ----------------------------- L V - ■? Given: I B S UD

C F 0t------------------ * --------------------}

F igure 3-18.

If the measure of IE is 8, while th a t of AS is 20, w hat can be said of the measures of FB and CD? W ith this in mind, w hat would the measures of both AE and UF have to be? W hat should follow then concerning AE andUF?

W ithout resorting to num bers, the steps leading to the conclusion that AE ~ UF would be

I f (1) A B ^ U Dand (2) ~EE FDthen (3) AS - IB US - FDor 'Z 2 S UP

If (1) a = band (2) c = dthen (3) a — c — b — d

T h e steps were arranged in parallel columns to point out the similarity between the Subtraction Postulate and the segments in this problem. W hat segments represent a, b, c, and d! Since A l and UF are bu t other names for AS — EE and US — 7<T> respectively, the conclusion is written in the preferred form of

A E S Z U F

EXERCISES

By using the Subtraction Postulate w hat conclusion can be draw n in each of th e following problems? W hen you write your conclusion and your reason, pu t the conclusion a t the left and the reason at the right as shown in the illustration on page 64.

THE SUBTRACTION POSTULATE67

1 . G i v e n : S P C

A E ^ D F

3 , Given: /.ABC S /A C B / ABD SS /A C D

5 • G i v e n : AD £= SU W S 5 F

7 • Given: U S —W & 'E F

G i v e n : A C ~ ~ A B

m ^ E B

G i v e n : /.B A D ~ . /C D A / B A E S* Z CDE

Given: /.B A D = /C D A 6 ./E A D S /F D A

£ F

Given: /A F E == /.C D E / BFA S /B D G

8.

9. Given: AC BD Given: /LEAD S ZC A D 10.A E S Z D E Z B A E ^ Z C A F

68 ASSUMPTIONS AND THEIR PLACE IN ' A PROOF

^ /B F r

H The Multiplication and Division Postulates

T he need for the third of the operational postulates rarely arises in the study of geometry. I t will be stated here so th a t you m ay see th e com plete pattern of all the operational postulates. Furtherm ore, there a re several, instances that arise later in our work when we will w ant to refer to this postulate.

P o s t u l a t e 7 : T he M ultiplication Postulate

I f a = ba n d c — ith e n ac = bd

O r as a statem ent:

If equals are multiplied by equals, the products will be equal.T h e last of the operational postulates is

P o s t u l a t e 8 : T he Division Postulate

I f a = b

c = d (where c and d are no t zero) a _ b c d

O r as a statem ent:

If equals art' divided by nonzero equals, the quotients will be equal.

In actual practice this postulate has widespread use in geometry, b a t . 'o n ly w here th e divisors c and d are the num ber “2.” In this case the conc lu s io n becomes

a _ b 2 ~ 2

w h e re a /2 and b/2 are but halves of the original equal num bers a and b. U n d e r this special circumstance, the postulate is stated as

Halots of equals an equal.

I t is only natural th a t this form of the Division Postulate would find

a n d

th en

THE MULTIPLICATION AND DIVISION POSTULATES 69

most frequent application. Situations,calling for its need would be those in which the bisector of an angle, the bisector of a line segment, or the m idpoint of a line segment were found in the problem. Let us illustrate by using the m idpoint of a line segment. This point will separate the segment into two congruent segments. T h e measure of each of these segments will be half of

the original segment. Thus,

A ____________ 8 Given: M is the m idpoint of AB.

F ig u r e 3-19.

m AM = AB

alsom M B = \m AB

Similarly, if BD is the bisector of ZABC , then by definition

Z A B D £ ZD B C

O r this relation can be stated asm Z A B D — ZA B C

andm ZD B C = \m ZA B C

Two illustrations of the application of Postulate 8 appear below. W hat conclusion based on all the given data can be drawn in each case?

Illustration 1:

Given: A B ^ A CD is the m idpoint of A B .

'^•C £ is the m idpoint of A C ..

F ig u re 3-21.

CONCLUSION R EA S O N

a d ^ l a E Division postulate: Halves {AD and AE) of congruent segments (A B and AC) are congruent.


T o avoid confusion, it m ight be well to point out that the term "halves”

as used in these examples refers to the fact that the measures of both AD and A E are halves of the measures of A B and AC. The Division Postulate will also enable uo to conclude th a t BD S? CE, for their measures, too, are halves of those of the congruent segments AB and ~A£. W hat two other conclusions can be draw n in this problem?

illustration 2:A F

Given: / B A E ^ / D E A

AC bisects /B A E .

l,B c D EC bisects /D E A .

tFigure 3-22.

CONCLUSION REASON

/ B A C = /D E C Division postulate: Halves ( /B A C and /D E C ) of congruent angles ( /B A E and

i/D E A ) are congruent.

IEXERCISES

In each of the following problems draw a single conclusion based on all the da ta given. Do your work as shown in the two illustrations above.

1. Given: AB ~ DC

E is the m idpoint of AB.

F is the m idpoint of DC.

Given: A B £ AC

D E bisects AB.

D F bisects AC.

2.

THE MULTIPLICATION AND DIVISION POSTULATES 71

3, Given: /.A D C £ /-BCD

D F bisects /A D C .

CE bisects /B C D .

5 . Given: / A B E ^ / D C E

BG bisects /A B E . _►CF bisects /E C D .

7 . Given: /A D B £ /B C A

D E bisects /A D B .

CE bisects /B C A .

G iven: CA == CB

D E bisects CA and CB. (H int: W rite these d ata as two pieces of information.)

Given: AC = BD

AC and TlD bisect each other. (H int: W rite these data as two pieces of information.)

G iven: PB £ PC

A is the midpoint

of PS .D is the midpoint

of PC.

8.


9 . Given: OF bisects DC._

OE bisects AB.

A B==UC

Given: AC bisects /.B A D .

CA bisects /.BC D .

/B A D S /B C D

10.

■ The Postulates of EqualityWe have been m aking some ra th e r vague references to the

fact th a t in the relationa = b

the symbol on the left and the symbol on the right are but two different nam es for the same “ thing," where most frequently this “ th ing” has been a num ber. To express this concept formally, the m athem aticians created three postulates setting forth what they call the properties of an equality. P o s t u l a t e 9 : Reflexive Property of Equality

P o s t u l a t e 1 0 : Symmetric Property of Equality

Ifthen

a — b b = a

P o s t u l a t e 1 1 : Transitive Property of Equality

I f a = band b — cthen a - c

T h e first of these properties has-a. very long history, having been first noted by the Greek phi!osopher*Aristotle. Hj) referred to it by saying that a quan tity m ust be identically eqvral to itself and noted this as his first law of logic. I t was called the Law of Identity. TJiere are two other laws of logic th a t Aristotle established. Both of these will play an im portant role in our w ork la te r in th e course.

T he Symmetric Property establishes the idea that an equality between num bers will hold in both directions, T h a t is, if a is equal to b, then so, too, m ust b be equal to a. Finally, the third, or transitive, property of equality enables us' to discover two num bers th a t are equal by showing th a t they are

equal to the same th ird num ber. Although the inform ation in the T ransitive Property was given as

a — b

andb = c

by applying the Sym m etric Property to the second of these equalities, it can

be w ritten asc = 6

T hus, Postulate 1X can be written as

I f a = 4and c = bthen a - cW hen stated in this way, we can easily recognize that both a and c are equalto b. T hen by the Transitive Property we will conclude that

a = c

T he equivalence of the following statements was noted several times in

the p a s t:m A B = m CD is equivalent to AB — CD.

Since it follows from the Symmetric Property that

APPLICATIONS OF THE PROPERTIES OF EQUALITV 73

if m AB ~ m CDthen m CD = m AB

Therefore, it can be said that

if A B S CD

then CD — AB

T he same relations will hold with reference to the Reflexive Property and '.he Transitive Property. T h at is, henceforth,

T he Reflexive, Symmetric, and Transitive Properties of Equality 1 will also be interpreted as Reflexive, Symmetric, and Transitive

Properties of Congruence of line segments and Congruence of angles.

B Applications of the Properties of EqualityT he reflexive property of equality has some very special

applications when used in conjunction with the addition and subtraction postulates, T h e four illustrations below wili help point up .those situations under which this occurs. Before reading the “ Conclusion” and “ Reason” that appear lo» each problem, try to draw your own conclusion and justify it in terms of the postulates.

74 ASSUMPTIONS AN D THEIR PLACE IN A PROOF

F ig u re 3-23,

CONCLUSION REASON

CD £ CD

BD £ CE

Illustration 2:5 -

Figure 3-24.

Reflexive property of congruent segments.Addition postulate: I f congruent segments (CD and CD) are added to congruent segments (BC and D E), the sums will be congruent segments (BD and C l) .

Given: S W £ T V

| CONCLUSION REASON

T F £ T W

S r ^ i W v

Reflexive property of congruent segments.Subtraction postulate: If congruent segments ( T W and T W ) are subtracted from congruent segments (S W and TV ), the differences will be congruent segments (S T and W V ).

Illustration 3:

APPLICATIONS OF THE PROPERTIES OF E Q U A L S 75

C O N CLU SIO N

/.CAD £ LCAD /B A D £ /C A E

REASON

Reflexive property of congruent angles. Addition postulate: If congruent angles ( /C A D and /C A D ) are added to congruent angles ( /B A C and /D A E ) , the sums will be congruent angles ( /B A D and /C A E ) .

Illustration 4:

B Given: /A B D £ /C B E

CONCLUSION REASON

/E B D £ /E B D / A B E £ /C B D

Reflexive property of congruent angles. Subtraction postulate: If congruent angles ( /E B D and /E B D ) are subtracted from congruent angles ( /A B D and /C B E ) , the differences will be congruent angles ( /A B E and /C B D ).

An illustration of the transitive property of congruence is given below Before reading the conclusion and reason th a t appears for the problem , try to,form ulate your own.

Illustration 5:

/ k

/ ........................_

Given: Z1 £ / 2 \ Z3 £ Z2

- p i

Figure 3-27.

t A ngles a re frequently nam ed w ith num bers ra th e r th a n w ith th ree cap ita l le tters.

76ASSUMPTIONS AND THEIR PLACE IN A PROOF

CONCLUSION

a? Z3REASON

Transitive property of congruence: If two angles ( Z l an d Z3) are congruen t to the same angle { / 2), then they are congruen t to each other.

e x e r c is e s

reason ju tions above.

In each of these problems, state one conclusion a n d th e justifying this conclusion. Do

1. Given: A G ~ E E

your work as shown in the five illustra-

Given: /A B C S Z /.B A G 2. / A C B ^ / B A C

A

Given: /A E C S Z / D E B

A.— E

7. Given: / I S / 2 I Given: B E S = F C 8.Z3 S Z2

APPLICATIONS OF THE POSTULATES OF GEOMETRY 77

9. Given: BC = /!£ A E S G D

1 1 . G iven: £>£ = BF FC = D E

Given: / R Y T ^ . Z W Y S 10. y

Given: / B A D = /E A C 12.

H Applications of the Postulates of GeometryThough we will learn other postulates throughout the year,

the few we have studied in this chapter have wider application than any of the others. If you understand these thoroughly, there is little likelihood th a t you will encounter any great difficulty with the work th a t follows. O n the other hand, should you still feel unsure of yourself w ith the application of these postulates, it would be best to go back and redo each of the sets of exercises in this chapter. T o help you decide whether this review is necessary, do all the problems in the exercises that follow. If you have no difficulty in determining which postulate should be used to justify each conclusion, then it is very likely that you understand how and when to. apply the operational postulates.


EXERCISES

Unless otherwise stated, draw only one conclusion on the basis of the d a ta given in each of the following problems. T o the righ t of each of you" conclusions, state the pustulate you have used to justify thisconclusion.

1. Given: m / \ = 35

m / 2 = 35

0 E H

5 . Given: / \ £= /A / 2 = Z3

7 . G iven : /.C A D ^ /C B D

/ D A B S /D B A

C

Given: A B S D E

B C ^ E FA

o E

Given: M N = 0 ?

Civen: Z1 S Z2

4.

6.

G iven: S PRm n ^ F q

t _ M N

8 .

APPLICATIONS OF THE POSTULATES OF GEOMETRY

11. Given: BC £= BD B E = S F

.A

15. Given: Z l £ ^2 / I £ / i

1 7 . G iven : / i — / 2 Z4 £ Z2

Given: /A B C = . /A C B 1

Z1 S / 2

Given: / C E A ^ A BE D

A B C

Given: Z1 S= / 'b Z2 = Z4

Given: Z1 ^ 2 ZCCD S

80 ASSUMPTIONS

21, Given: Z l ^ Z2

Z 2S* Z3

23. Given: A B ^ D CE is the m idpoint of AB.F is the m idpoint of CD.

A----------------------------.0

25. Given: ZA B C £ ZA C B

BD bisects ZABC .

Given: ZA B C £= Z D C B 20. BE bisects ZA B C .

CE bisects ZD C B.

A N D THEIR PLACE IN A PROOF

Given: A B ^ D C 2 2 .

G iven: E C ^ A B 26.EDS* AS

TEST

29. Gitfen: Z A B C ^ ZE F C Z \ “ Z l

31. Given: Z A B C = Z A D C Z l = Z l

M Test

Given: C A ^ C B<—► . .. -

D E bisects CA and

CB.

Given: BA £ BC A D ^ C E

81

2 8 .

30.

Given: ZA E C = Z D E B 32.A D

1. In what way does a “ postulate" differ from a “ belief” ?

2. Two m athem aticians in different parts of the world develop subjects that they call plane geometry. T he postulates used by one are completely different than those used by the oth^r. Does this imply tn a t one of these men is right while the other is w r|8grjfu$tify your answer.

82 ASSUMPTIONS AND THEIR PLACE IN A PROOF3. State the postulate that was used in concluding Step B on the basis of

the information given in Step A in each of the following problems.(a) Step A: 3x = 12 (b) Step A: x + 4 = 10

Step B: j — 4 Step B: x = 6(c) Step A: 2x - 5 = 13 (d) Step A: 5x + 10 = 5

Step B: 2a = 18 Step B: x + 2 = 1

4. I f points A, B , and C are elements of RS, then under what condition will AC + CB = AB?

i—► <—>5. If A B and CD intersect in points X and Y, then w hat conclusion can be

drawn?

6. Answer each of the questions below in terms of this diagram. If an answer is not possible, simply write “ no answer” after the problem num ber.

A —~____________________________ C

6

(a) AD + D B = ? ■ (b) CE + ED = ?

(c) A E + ED = ? (d) B S - FC = ?

(e) AC - A E = ? (f) ZB A C - / FAC = ?

(g ) Z BCD + /B C A = ? (h ) / A E F - Z D E F = ?

7. Express each of the following in terms of a single numeral.

(a ) \1 - 3| (b ) |4 — 9| ' (c) |3 - (—7)|

8. I f the coordinate of A is 9 and the coordinate of B is 15, w hat is themeasure of A~E?

9. If P is between R and Q, w hat is m S ? if m RQ = 17 and m FQ = 9?

10. T h e m / A P B = 60 and m ZB P C = 20,

(a ) If PB is between PA and PC, what is the m ZAPC1—y — ■■ >

(b ) If PC is between PA and PB , what is the m /APC ?

T ¥ 1f D raw a conclusion in each of the problems below. Im m e

diately following your reason for this conclusion, indicate whether this reason is a postulate, a definition, or the reverse of a definition.

ji1

TEST

1 . Given : AD SD C ^ E C

3 , Given: /.D B A = ZE B C

5 , G iven : EF -L AC

A

7 . Given: AD S AE AB AC

Given: CE bisects ZACB. 2.

83

B

Given: E D = BC 4.F is the m idpoint of ED.G is the m idpoint of BC.

Given: / E D C ~ / F D B 6>

Given: AD bisects /.B A F . 8 .

A

84ASSUMPTIONS AN D THEIR PLACE IN A PROOF

$

9 . Given: AC £ BD

11. Given: AB and CD bisect

each other.

15. Given: A E ^ E B

G iven: Z B J F ^ Z C G E 10.J K bisects Z B JF .

C H biscots z 'CGE.

C / GF

Given: RA £ RC

R C = iR B12.

G iven: ZA F D ^ Z B E C 14.s ZC£/1

A f _____ n

E C

Given: ACSsZDB 16.<■4 _

£ H bisects JC and~BB.

r

—__D

u/ F G - \

----- \ h

4The “Simple”Theorems

IN T H E F IR S T T H R E E CHA PTERS W E LAID T H E foundation blocks of the course in geometry. These blocks are our ckfinitians-- and postulates. Now we are prepared to build a large (superstructure; based on these few primitive ideas. W hat, however, is the m ortar that binds the superstructure to these definitions and assumptions? It is known either as the “ Rules of Reasoning' 1 or the “ Formulas for Drawing Valid Inferences.” Only one of these rules is im portant to us at this time,

, Consider the following two statements:

(1) “ If I receive a passing grade on my exam, then I shall pass for the term .”

(2) “ I received a passing grade on my exam.”

O n the basis of these two statements, what conclusion do you believe should follow? Q uite apparently, by virtue of the.fact that I received a passing grade on my exam, I shall pass for the term.

A statem ent of the form

“ If I receive a passing grade on my exam, then I shall pass for the term”

is called a conditional statement. T h a t part of the statem ent following the word “ if” is referred to as the antecedent, while the clause following the word “ then” '

is t h e consequent o r conclusion. If , a s in th e i l l u s t r a t i o n a b o v e , w e a s s e r t th e t r u t h o T b o t h t h e c o n d i t i o n a l s t a t e m e n t

“ I f I r e c e i v e a p a s s i n g g r a d e o n m y e x a m ,

t h e n I s h a l l p a s s f o r t h e t e r m ' '

a n d t h e a n t e c e d e n t

“ I r e c e iv e d a p a s s in g g r a d e o n m v e x a m ”

t h e n i t w ill fo llo w t h a t t h e consequent w il l also be tr u e :

“ I shall pass for the term .”

If this relationship is expressed with the symbols fi and q w here/; rep resents the antecedent and q the consequent, w hat we have ju st discovered w ould take the form of

P o s t u l a t e 12: Accepting the conditional statem ent: I f p then q.And asserting th e tru th of p : Given p Affirms the tru th of q; Then q follows.

T h e application of Postulate 12 to geometric situations will occur as follows:

A ccepting the conditional statem ent: If an angle is a right angle, then itsmeasure is 90.

And asserting the tru th of the antecedent: / ABC is a right angle.Affirms the tru th of the consequent: T he m easure of / .A B C is 90.

In this illustration th e conditional statem ent is nothing m ore th an another form of the definition of a right angle. In fact the entire problem was encountered earlier in our work as

86 THE "SIMPLE" THEOREMS

T he conclusion we drew was identically the same as the consequent-in this illustration. A pparently, each time we drew a conclusion in the first three chapters we were accepting Postulate 12 w ithout being aware of this!

A reasonable question fo ask a t this point is, “ If we assert the tru th of the consequent in a conditional statem ent, will this in turn affirm the tru th of the antecedent?” T o answer this question, let us examine the conditional sta tem ent

“ If a person is a king, then that person is a m an .”

Should we assert the truth of the consequent,“Joe Smith is a m an”

it would surely not infer the tru th of the antecedent,

“Joe Smith is a king.”

THE "SIMPLE" THEOREMS

F ig u r e 4-2.

This same situation can be illustrated ra ther clearly through the use of a Venn diagram . Here we see that both “Joe Sm ith” and “ Kings” are subsets of the universal set called “ M en.” Joe Smith, however, is not necessarily a m em ber of the set called “ Kings.” Thus, reasoning of the form

Accepting the conditional statem ent: If p , then q.And asserting the tru th of q : Given qAffirms the truth of p: T hen p follows.

is FA LSE, or IN C O R R EC T, REA SO N IN G .Similarly, if we were-to deny the antecedent, this would not imply the

tru th of a denial of the consequent. Thus, with reference to the conditional statem ent '

“ I f a person is a king, then that person is a m an .’’

if we deny the antecedent

"Jo e Sm ith is not a king”

this denial does hoi imply the tru th of a denial of the consequent

“Jo e Smith is not a m an”

By exam ining the Venn diagram above, we can see that although Jo e Sm ith is not an element in the set “ Kings,” he is an elem ent in the set “ M en.” Hence, reasoning following this pattern is incorrect.

Incorrect Reasoning Accepting the conditional statem ent: I f / i . t h e n fAnd asserting the truth of a dental of p : Given not-/)Affirms the tru th of a denial of q : T hen n o t-f follows.

EXERCISES.


Ai___In cach of the following problems state whether the reason

ing is correct or incorrect. Justify your answer in either event.

1. If M r. Strong is elected senator, then our taxes will be reduced.M r. Strong was elected senator.

f O ur taxes will be reduced.

2. You wiil have no th roat irritation if you smoke R obin cigarettes.You smoke Robin cigarettes.

You will have no th roat irritation.

3. I f it rains, we shall not go to the dance.I t is not raining.

W e shall go to the dance.

4. I f you are a good citizen, then you will vote a t election tim e.You voted a t election time.

You are a good citizen.

5. I f the price of goods is, not increased, it will not be necessary to raise the salaries of employees.T he price of goods was increased.

I t will be necessary to raise the salaries of employees.

6. I f A B _L CD, then Z l and Z 2 are right angles.

G iven : A B JL CDZ l and Z 2 are right angles.

7. If a ray is the bisector of an angle, then two congruent angles will be formed.

G iven: BA bisects Z B .Tw o congruent angles are formed.

8. You will be able to write quickly if you use R A PID pencils.You do not use R A PID pencils.

You are not able to write quickly.

9. I f a = 6 and c — d, then a + c = i> + d.a does not equal b, and c does not equal d,

a + c does not equal b + d.

t The symbol .'. is used to represent the word "therefore,"

THEOREM ON RIGHT ANGLES

11Granting the truth or the conditional statem ent and the

statem ent that follows it, what further statem ent will be true? If no further statem ent can be m ade by “ correct’’ reasoning, sta te why this is so.

1. If a substance is an acid, it will tu rn blue litm us paper red.This substance turns blue litmus paper red.

2. If a student is not a senior, he can not run for office in the senior class. Fred Williams is a senior.

3. If x = —2, then xt = 4.But = l

4. If two angles are congruent to the same angle, then their measures are

equal./LA and /-B are not congruent to the same angle.

5. A B = 'BC if B is the m idpoint of AC.

B is the m idpoint of AC.

6. Z l is not a right angle if A B is not perpendicular, to CD.

A B X CD.

7. Two angles are supplem entary if the sum of their measures is 180. ■

m Z A + m / .B = 1 8 0 .

8. x is an element of B if x is an element of A. x is an element of B.

4— 4—♦ v < ■}9. If the intersection of A B and CD is the empty set, then A B and CD a re

, not equal sets.

The intersection of A B and CD is not the empty set.

10. 11 a point is the vertex of an angle, then it is a com m on elem ent to each

of the sides of the angle.P is an element common to each of two rays.

H Theorem on Right AnglesW e spoke earlier of the superstructure th a t we p lanned to

build on the foundation of definitions and postulates. This superstructure consists of statements, m any of them conditional, whose tru th we propose to justify. T o do this, we will make frequent use of Postulate 12, for it is th e

90 THE "SIMPLE" THEOREMSonly m eans we have at our.disposal for ' ‘reasoning correctly.” Quite often the words “ rei'soning correctly” are used interchangeably with “ reasoning logically” or “ draw ing valid conclusions.” T hroughout this book these three p h nses will have exactly the same meaning.

To a very large extent the nature of our work will be to show why we mus; accept q in the statement

“ If p, then q”

if we accept the tru th of p. Then, with the knowledge that both p and q are true, we will agree that the statement “I f p, then q ” is also true.

M any of the statem ents that we prove— that is, justify by reasoning correctly—will have little consequence in the developm ent of our work in geometry. Those statem ents that are important will be singled out and referred to as theorems. Theorem s are used in the justification o f the proofs of o ther statem ents.

All of this can be m ade a great deal clearer by showing an application to a specific illustration.

T H E O R E M 1: I f tw o angles a re r ig h t angles, th en th e y a re congruen t.

At the outset we m ust realize that the/> of this statem ent is “ Two angles are right angles”

w hile the q is

“ These two angles are congruent”

O u r objective now will be to accept p and show why q m ust also be accepted. T h is is done by m aking several applications of Postulate 12. T o simplify o u r work, let us call the two right angles LA and LB .

P R O O F

First Application of Postulate 12

If an angle is a right angle, then its measure is 90. L A is a righ t angle,

m L A = 90

Second Application oj Postulate 12

I f an angle is a right angle, then its measure is 90. L B is a r ig h t angle,

m L B = 90

Third Application oj Postulate 12

I f a - b and b = c, then a = c. m /-A = 90 and m /LB = 90

m /LA - m /LB

Fourth Application of Postulate 12

If two angles have equal measures, then they are congruent,/ .A and L B have equal measures..'. /LA and /LB are congruent.

Thus, accepting the statem ent that /LA and /LB were right angles led us to the conclusion that /.A and L B were congruent. T he justification for this was based on the method of “ correct reasoning” given to us b y Postulate12. In addition, we fell back upon the definition of a right angle, the reverse of the definition of congruent angles, and the transitive property of equality. W here did these three statements appear in the proof?

I t is quite apparent that this m ethod of proof is not only lengthy but also ra th e r tedious. In view of this fact, the “ Two-Colum n” m ethod was developed to shorten and simplify the proof of a statem ent, T he proof p resented above will now be repeated in its “ Two-Colum n” form.

THEOREM ON RIGHT ANGLES vt

Given: L A \ is a right angle./LB is a right angle.

C oncl.: /LA = /LB

F i g u r e 4-1

frR O O F 1 STATEMENTS

1. /LA is a right angle.2. m /LA = 903. L B is a right angle.4. m L B — 905. m /LA — m /LB6. L A S L B

F orm al p.t W-f ■

REASONS

1. Given2. Definition of a right angle3. Given4. Sam e as 25. Transitive, property of equality6. Reverse of the definition of congru

ent angles

f A n ang le can be nam etf w ith th e single leUcr a t its vertex if there are no o th e r angles in the d iag ram having the same vertex.

92THE "SIMPLE" THEOREMS

T here are several features abou t the preceding proof that should be called to your attention, for they will be repeated in most proofs.

(1) Notice that the Conclusion you are working toward is stated directly below the Given Data. Your objective will be to try to reach this conclusion by m aking a series of Statements, Each of these statem ents will have to be

justified either by the fact th a t it is pa rt of the Given B a ta or by virtue of the definitions or postulates th a t have been agreed upon.(2) In developing the proof it is best not to leave any piece of the Given D ata w ithout pointing out its value in the proof. Thus, before leaving the statem ent th a t " Z A is a right angle,” it was shown th a t because o f the definition of a right angle, the m easure of Z A m ust be 90. This w as im portant, for arriving a t our conclusion was dependent upon th e fact th a t the measures of both L A and Z B be shown equal to the same num ber, 90. (3) W henever the same reason appears m ore than once in a proof, it is unnecessary th a t it be repeated. In the proof illustrated, the reason forstatement 4 was the same as that for statement 2. This was signified by writing, “Same as 2.”

T h e proof of the theorem , “ If two angles are right angles, then they a re congruent” was based on two definitions and the postulate, the transitive property of equality. You m ay say, “W hat if we had decided no t to accept this postulate?” Then, we would very likely not have been able to prove this theorem. In the same vein, should a postulate be discarded a t any tim e during the period in which you are learning this subject, it would be necessary, too, to discard those theorems th a t were based upon this postulate. All is no t lost, however, for an equally elegant superstructure can be built on th e postulate that replaced the discarded one!

Along the same line, the question is often raised, “W hich has greater ‘tru th ,’ the postulates th a t we have accepted or the theorems we prove?”A moment's reflection on how the theorems were justified will make you realize that the question is meaningless.

O nce a theorem has been proved, it can be used thereafter as justificatio n for statements in the same m anner as definitions and postulates have b een used in the past. But do, do be careful—never use them as a reason befo re they have been proved. As an example, refer to statem ent 6 in the p ro o f on page 91. Could the reason for this statem ent have been given as,“ If two angles are right angles, then they are congruent” ? Justify your answ er.

O ne final point m ay still be annoying you. W hy bother to prove th a t r i g h t angles a re congruent or, in fact, why prove any statements? W hy no t m a k e things a great deal easier for ourselves by accepting this statem ent and a l l others! T he answer to this lies in the na ture of mathematics. Although t h e m athem atician wants the postulates and definitions to be clearly set F o rth a t the outset, he prefers to keep their num ber a t a m inim um. T h a t is,

h e will call no new statem ent a postulate if it is a t all possible to prove this

statem ent in terms of the previous postulates that are already a t his disposal. T here will be occasions in the future when we shall make statem ents whose proofs are possible, and yet we shall postulate them. We do this so as not to bewilder you with small details at this early stage of your m athem atical

development.Now, on to an application of Theorem 1'.

THEOREM O N RIGHT ANGLES 93

Given: AB J- CD Concl.'. Z \ = Z2

PR O O F STATEMENTS

1. AB JL CD2. Z l and Z2 are right angles.3. Z 1 S Z 2

REASONS

1. Given2. Definition o f perpendicular lines3. If two angles are right angles, then

they are congruent. (Theorem 1)

Note that we d id not have to prove Theorem 1 over again in arriving a t Z l = Z l . Since the two angles were shown to be righ t angles, then Theorem 1 enables us to conclude that they are congruent.

EXERCISESShow how to arrive a t the conclusion in each of the follow

ing problems. Use the same method as illustrated above.

Given: Z B is a right angle. 2.1 . Given: Z B is a right angle.

ZD is a right angle.

Concl.: Z B S ZD Concl.: Z BDC X BC

ZC

94

3. G iven: A B ± CD

Concl.: Z A B C ^ /A B D

5. G iven: A B 1 BC <-*

ED J . AC C oncl.: Z A B C g x /A D B

7. Given: Z1 and Z 2 are com

plem entary angles. L 2 and Z 4 are complem entary angles.

Concl.: / A B C £~ ZD C B

Given: A B A. BC 4 *«-» <->

EC i . BC Conci.: Z B 9 z Z C


Given: Z \ and Z2 are com- 6. plementary angles.ZD C B is a right angle.

Concl.: Z A B C S Z Z D C B

Given: A B X CD 8 .Z 1 complementary to Z 3

Concl.: Z A C B S Z Z 1

SUPPLEMENTARy AN D COMPLEMENTARV ANGLES 95

9. Given: AB and DC J_ BC Z C = Z D

Concl.: ZD

Given: Z \ and Z3 are com- 10.plem entary to ZZ.

Concl.: Z D B F S= Z C B E

■ Theorem on Straight AnglesI t should have been apparent to you th a t ju st as we had

been able to prove that if two angles are right angles, then they are congruent, so, too, is it possible to prove that if two angles are straight angles, then they will be congruent.

T H E O R E M 2: K two angles a re s tra ig h t angles, th en th ey a re co n g ru ent.

The proof of this statem ent is very m uch the same as the one given on page 91. In w hat m anner will the diagram have to be altered in order to conform with the information given in this theorem ra th e r than in the one on right angles? How would you change the Given Data? W hat word in reason 2 will have to be changed? W hat other changes will have to be m ade in the proof? W rite out the complete proof of Theorem '2.

■ Theorems on Supplementary and Complementary AnglesIf Z B was supplementary to Z A and the m Z A was 20,

w hat would be the measure of Z B ? W hat operation did you perform to arrive a t 160? If Z C was also supplementary to Z A , then w hat would be the measure of Z C i In view of w hat you found the measures of both Z B and Z C to be, what conclusion can be drawn? How were both Z B and Z C related to ZA? W hat relationship existed between Z B and Z C based on the fact that they were both supplementary to ZA? In general then, w hat do you think we will be able to prove concerning two angles that are supplem entary to the same angle?

THEOREM 3: If two angles are supplem entary to the same angle, then they are congruent.

A n a ly sis: T he proof will be pa tterned after the m ethod used to show th a t two right angles are congruent. T h a t is, we will show th a t both angles have the same m easure; hence, they will be congruent to each other.

• Since Z B and Z A are supplem entary, then m Z B -|- m /LA = 180. If this is so, then what is the m easure of ZB?. In a similar m anner we can show th a t m Z C is also 18Q — m Z A . Hence, it follows th a t Z B == ZC.


Given: Z B is supplem entary to Z A , Z C is supplem entary to Z A .

Concl.: Z B = Z C

Figure 4-5.

P R O O F STATEMENTS

1. Z B is supplem entary to ZA .2. m Z B + m ZA = 1803. m Z B = 180 - m ZA4. Z C is supplementary to ZA.5. m Z C + m ZA = 1806. m Z C = 180 - m Z A7. m Z B = m Z C8. Z B S * Z C

REASONS

1. Given

2. Definition of supplem entary angles3. Subtraction property of equality4. Given5. Sam e as 26. Sam e as 37. T ransitive property of equality8. Reverse of definition of congruent

angles

Before exam ining the theorem below, you should realize th a t there will b e ano ther theorem whose wording will be very m uch the same as th a t of T h eo rem 3. W hat do you think this theorem will be?

'THEOREM 4: I f two angles are com plem entary to th e same angle, then they are congruent.

A n a ly sis: R efer to the proof above, W hy will the d iagram for the proof o f this theorem have to be changed from the one used in the proof of T heo re m 3? W hat information in the Given D ata will have to be changed? W h y is no change necessary ift th e Conclusion? Exam ine each step of the p r o o f to determ ine w hich words and w hich num bers will have to be changed.

Using the proof above as a guide, write the complete proof of Theo r e m 4,

T h e next two theorems sound very m uch the same as Theorem s 3 and 4. T h e i r proofs, however, will be quite different.

T H E O R E M 5: I f two angles.^are supp lem entary to tw_fl_congruent angles, then they are congruent.

SUPPLEMENTARY-AND COMPLEMENTARY ANGLES 91

A n a l y s is : In order to arrive a t our conclusion, we will try to show that both Z \ and Z l are congruent and also A ABC and EFG m ust be congruent (see Figure 4-6). Once this has been accomplished, Z A B D and Z E F H will have to be congruent by th e subtraction property o f congruence.

Given: Z A B D is supplem entary to Z l . Z E F H is supplem entary to Z l . Z l S Z 2

Concl.: Z A B D S Z E F H

Figure 4-6.

PRO OF STATEMENTS REASONS

1. ZA B D is supplementary to

Z l.2. ZA B C is a straight angle.

3. ZE F H is supplementary to Z l .

4. ZEFG is a straight angle.5. ZA B C '= ZEFG

6. Z 1 S / 27. ZA B D = ZEFH

1. Given

2. Supplem entary angles are two angles the sum of whose m easures is the measure of a straight angle. (See Definition 12a, page 32.)

3. Given

4. Same as 25. If two angles are straight angles, then

they are congruent. (Theorem 2)

6. Given7. If congruent angles ( Z l and Z 2 ) are

subtracted from congruent angles (Z A B C and ZE F G ), the differences will be congruent angles. (Subtraction property of congruence)

T he theorem th a t follows differs from Theorem 5 only by the fact th a t the word “ complementary’’ will replace the word “ supplem entary."

THEOREM 6 : I f two angles are com plem entary to tw o congruent angles, then they are congruent.

Using the proof of T heorem 5 as a guide, write the com plete proof of

Theorem 6.T he illustration th a t follows is an example of how T heorem 6 can be

applied in the proof of a problem .

98

Illustration:


Given: A B ± BD<—V ►

CD 1 BD z i s z z

Cone!.: Z3 ^ Z4

PR O O F STATEMENTS REASONS

1. A B ± BD 1. Given2. Z A B D is a right angle. 2. Def. of _L lines3. Z i is com plem entary to Z l. 3. Rev. of def. of com plem entary angles

4. CD 1 BD 4. Given5. Z C D B is a right angle. 5. Same as 26. Z 4 \s com plem entary to Z2. 6. Same as 37. But, Z l S Z l 7. Given8. Z 3 £■! Z4 8. If two angles are com plem entary to

two congruent angles, then they arecongruent. (Theorem 6)

T here is an alternate method for arriving a t the conclusion in this illustration. Since Z A B D and ZC D B were.shown to be right angles, then by Theorem 1 we can conclude that they are congruent. From the Given D a ta we know th a t Z l = Z2. Hence, why should it follow that Z 3 = Z4?

EXERCISES

1 • Given Z \ is comp, to ZZ. Z i is comp, to Z t .

Concl.: Z \ 9* Z l

G iven: Z l is comp, to ZZ.

CB 1 A B Concl.: Z i ~ Z \

3. Given: ZA B C is a siraigln G iven: A B X BC

angle- _ D E L EF/.D E F is a straight ^

angle' Concl.: Z 3 S Z 4Z l S* ZZ

Concl.: Z3 = Z4

UPPLEMENTARy AND COMPLEMENTARY ANGLES

4.99

5. Given: Z l is supp. to Z l .Z i is supp. to Z l .

Concl.: Z l = Z i

7. Given: ZEG H is a straight angle.

< ZFHG is a straightangle.Z l m &

Concl.: Z 3 S / 4

G iven: Z i is supp. to Z l , 6 . ZE F G is a straight angle.

C oncl.: Z i S l Z Z

G iven: BCDE is a straight line.Z l S ZZ

C oncl. : Z i & Z 4

8.

100

9 . Given: Z C is comp, to Z l .

A B 1 BC Cone!.: Z C S Z l

11. Given: BC extended to A

D E extended to F Z l ^ Z A

Concl.: Z \ S Z2

13. Given: AC J . BDZ l S Z 2

C oncl.: Z 3 S ZA

Given: Z \ is supp. to Z2. 1 0 .

ZA B C is a straight j anele.| Cone!.: Z l ^ Z l


Given: Z \ is comp, to Z l . Z l is comp, to ZA. --►

B E bisects ZABC . C oncl.: Z l ^ Z l

Given \ DB ± AC

BD bisects ZE B F. C oncl.: Z l ^ ZA

12.

14.

15. Given: AB 1 AC <-> <-> AD 1 AE

Concl.: Z l S Z l

VERTICAL ANGLES

Given: AC and BD are straight lines.

Concl.: Z l ^ Z l

16.

101

■ Vertical AnglesIn doing Problem 16 of the preceding set of exercises, you

have actually proved a very im portant theorem. T he angles 1 and 2 in that diagram are called vertical angles. Notice that in order to obtain the sides of Z l , it was necessary to extend the sides of Z \ back through the vertex E. Similarly, had Z l been drawn first, then to obtain the sides of

Z l , it would have been necessary to have extended the sides of Z l , ED and

Figure 4-8.

EC, through vertex E. In defining vertical angles we make use of this property. Into w hat class would you place vertical angles? How would you show hoi# these angles differ from the other words in its class?

D e f in it io n 2 2 : Vertical angles are two angles such th a t the sides of one are rays that are opposite to those of the sides of the other.

W ere Problem 16 expressed as the statem ent of a theorem, it would have been:

THEOREM 7: If two angles are vertical angles, then they are congruent.

A n a l y s is : The diagram in Problem i6 suggests that we make use o f Z l . Since Z B E D is a straight angle, then Z2 will J>e supplementary to Z l . Similarly, Z A E C is a straight angle. Hence, what relation exists between Z l and Z3? If Z l and Z 2 are both supplementary to Z3, then w hat sh o u ld ' follow?


G iven: Z l and Z2 are vertical angles. Concl.: Z l S Z2

PROOF STATEMENTS

1. Z l and Z2 are vertical angles.—» —y

2. EA and EC are opposite rays.3. Z.AEC is a straight angle,

4. Z l is supplem entary to Z3.

5. E B and ED are opposite rays.6. Z BED is a straight angle.

-7. Z 2 is supplem entary to Z3.8. Z l SS Z2

REASONS

1. Given

2. Definition of vertical angles3. Reverse of definition of a straight

angle4. Reverse of definition of supplem en

tary angles

5. Same as 26. Same as 37. Same as 48. I f two angles are supplem entary to

the same angle, then they are congruent. (Theorem 3)

T h ere is another p a ir of vertical angles in th e diagram above. C an you nam e this pa ir of angles? Using the same m ethod as in the proof ju st given, prove th a t these two angles a re also congruent.

Illustration of an Application of Theorem 7:

Given: Z 3 S Z 2 Concl.: Z l £= Z3

A nalysis : Since Z 3 is already congruent to Z2, if we can show th a t Z l is also congruent to Z2, then our conclusion would follow. Why?

VERTICAL ANGLES 103

PROOF STATEMENTS REASONS

1. Z3 = Z2 1. Given

2. Z l ^ Z2 2. If two angles are vertical angles, \then they are congruent.

3. / I Z3 3. Transitive property of congruence

A few of the problems in the next group of exercises involve the necessity of applying the transitive property of equality several times before the conclusion can be inferred. Since this same approach is found in the proofs of quite a num ber of the problems in the development of geometry, it seems only wise that a theorem be established now to cover this situation.

(1 ) (2) (3)

THEOREM 8 : If andbut also,th e n

a = x b = V x = y a — b

A n a l y s is : By applying the transitive property to equations (1) and (3 ), we find that a - y. Using this information and the fact that b — y (2), we discover that another application of the transitive property will enable us to say that a = b.

G iven: a = x b = y x = y

Cone!.: a = b

PROOF | STATEMENTS REASONS |

1. a = x 1. Given

2. x = y 2. Given

3. a — y 3. Transitive property of equality

4. -But, b = y 4. Given

5 a — b 5. Same as 3

This theorem will most frequently be applied to line segments and to angles. Therefore, we will use the following statements for these special situations.

THEOREM 8a: If two lin e segments are congruent to two congruent line segments, then they are congruent.

THEOREM 8b: If two angles arc congruent to two congruent angles, then they are congruent.

t It is unnecessary lo point out in the proof that / I and i l are vertical angles.


J Illustration of an Application of Theorem 8:

Given: Z l S= Z2 Z4 3 ; Z3

AB bisects ZC AD . Concl.: Z l = Z4

Figure 4-11.

A n a ly sis: If it were possible to show that Z2 — Z3, then by T heorem 8, Z \ and Z4 would also be congruent. Consideiing the Given D ata, do you see any reason why Z2 should be congruent to Z3?

PR O O F | STATEMENTS REASONS

1. Z l £ Z2 1. Given2. Z 4 S Z 3 2. Given

- > 3. GivenI . AB bisects ZCAD.4. Z2 S Z3 4. Definition of the bisector of an angle

5. Hence, Z l S Z4 5. If two angles are congruent to twocongruent angles, then they are congruent. (Theorem 8)

EXERCISES

T he proofs of each of the following problems is based prim arily on Theorems 7 and 8, definitions, and postulates.

1 . Given: Z \ £ Z3 Concl.: Z \ SS Z 2

Given: Z D — Z2 Concl.: Z D S Z l

2.

VERTICAL ANGLES 105

3. Given: Z A — Z l Z Z ) S Z2

Concl.: Z A S ZD

G iven: Z B £= Z \ Z C S Z2

C oncl.: Z B £= ZC

5 . Given: Z2 S Z3 C oncl.: Z \ = ZA

G iven: Z3 Z4 Concl.: Z \ = Z2

EXERCISESReview of Theorems 1 through 8.

1 . Given: Z \ = Z l C oncl.: Z 3 = ZA

3 . Given: Z l is supp. to Z2. Concl.: Z l S Z l

Given: Z l == Z2 Concl.: Z3 •= Z4

Given: Z3 is supp. to Z2. Concl.: Z l = Z3


ZA C B is a right angle. Concl.: Z3 b? Z l

5 . Given: Z3 is comp, to Z2.

7 . G iven: Z3 £ Z lZ4 £ Z2

C oncl.: Z3 •= Z4

9. G iven: Z t is supp. to Z2.Z3 is supp. to Z4.

C oncl.: Z i £ Z3

Given:

ZD C B 3s Z2 Concl.: ZA BC £= ZD C B

Given: Z D is comp, to Z2. 8 .Z E is comp, to Z l.

Concl.: Z D ^ Z E

Given: Z A B C is a right angle.ZD C B is a right angle.Z \ =* Z2

10.

11. Given: Z A C F = ZD BEConcl.: Z l £* Z2

TEST

13. Given: ZA is comp, to Z l.

Z C is comp, to Z2.

BD bisects ZABC. Concl.: ZA = ZC

Given: AB _L BCZ2 is comp, to Z l .

Concl.: Z2 = Z3

Given: Z l is supp. to Z3.

Concl.: Z2 ■= Z4

107

12.

14.

Test

1. In each of the following problems state whether the reasoning is correct or incorrect. Justify your answer in either event.(a) If Jo e M oran plays basketball this season, our team will win the

pennant.O ur team won the pennant.

Joe M oran played basketball this season.(b ) If I do not pass this geometry test, I .shall receive a failing grade for

this cycle.I passed this geometry test.

I shall receive a passing grade for this cycle.

108 THE “SIMPLE" THEOREMS

(c) If I brush my teeth with W hite Tooth Paste, my teeth will have no cavities.I do no? brush my tcelh with W hite Tooth Paste.

My teeih have cavities.

(d ) If M is the m idpoint of AB, then A M = M B .

M is t h e m i d p o in t o f AB.

a m = W b

2. G ranting the truth of the conditional statem ent and the statem ent th a t follows it, what further statem ent will be true? If no further sta tem ent can be m ade by “ correct” reasoning, state why this is so.(a ) If the field is m uddy, our team will not win today.

T h e field is not m uddy.(b ) If a ray is not the bisector of an angle, it will not form two congruent

angles with the sides of the angle.

A B does not bisect Z D AC.(c) If the measures of two angles are not equal, then the angles are no t

rig h t angles.m Z A — m Z B

(d ) W e will enjoy our stay in high school if we participate in the school activities.W e enjoyed the years we spent in high school.

W rite the proof for each of the following problem s:

1 , G iven: Z l = Z4

Concl.: Z2 = Z3

Given: A B ± BC <-> *->

AD X DC Concl.: Z A B C S Z Z A D C

TEST109

3 . Given: Z A is comp, to Z l.ZC is comp, to Z2.

Concl.: Z A ^ Z C

5 . Given: Z A is comp, to Z l .

A B 1 ED Concl.: Z I S Z ^

7 . Given: AP JL CP

BP L D P C oncl.: Z l S Z l

Given: CA is extended to E. 4.Z l Z2

Concl.: Z 3 ~ Z4

Given: Z C S Z l 6.Z D S Z2

BA bisects ZE B F.Concl.: Z C ^ Z D

£ |A F

C B D

Given: Z l is supp. to Z4. 8 .Concl.: Z 2 = Z3

5Congruence of T riangles

A . VERY LARGE AND IM PO R T A N T U N IT O F work in geometry is concerned with the relationships that exist among triangles. Although the word “ triangle” has been part of your vocabulary for some years, it should be quite apparen t to you by now that launching into a discussion of this figure before clearly defining it would be foolhardy. W hat m ay seem rather odd, however, is the fact th a t to define this term we exam ine n o t th e triangle bu t the m uch m ore complex figure called the polygon. W e do this, surprisingly, not to create greater difficulty for you in learning this subject but rather to establish a general class of figures. Not only will the triangle be classified as a polygon but m any other geometric figures will be so grouped too.

We would like the definition of a polygon to be so designed th a t Figure 5-1 will be a m em ber of this set, while Figures 5-2, 5-3, and 5-4 will be

Figure 5-1,

110

CONGRUENCE OF TRIANGLES 111

Figure 5-2. Figure 5-3. F igure 5-4.

excluded from the set. Notice that Figure 5-1 consists of line segments; while Figure 5-2 does not. Hence, if we insisted that the polygon consist of line segments, Figure 5-2 would be eliminated. But this still leaves us w ith Figures 5-3 and 5-4, since both of these consist of line segments. Note,

though, that in Figure 5-3 two of the line segments (AC and B E ) have a point in common other than one of the points A, B, C, D, or E. Perhaps this feature will help us to elim inate Figure 5-3. Finally, the ap paren t difference between Figures 5-4 and 5-1 lies in the fact that the latter is “ closed,” while the former is “ open.”

These properties are now combined into the formal definition of a polygon.D e f i n i t i o n 23: A polygon is the union of the set of points

A t , f t , Pi, . . ' . , Pn-n P«with the line segments

P\Pi, PlPit • • ■ , Pr>-lP„, PnP\such that if any two of these line segments intersect, their intersection will be one of the points

P\) Pi) Pi, • - - , Pn—l, Pnand no other point.

f This is read as tlP one” or “P sub-one.”

Figure 5-5.

112 CONGRUENCE OF TRIANGLES

Now let us examine this definition to see how Figures 5-2, 5-3, and 5-4 w ere eliminated.

(1) Figure 5-2 is elim inated by the fact th a t a polygon consists of the unior. of a set of points and the hr.c segments joining those points.(2) Figure 5-3 is elim inated by the fact that the line segments have no point of intersection other than the points Ph P;, . . . , P.,.(3) Figure 5-4 is elim inated by the fact that a line segment m ust be drawn between the last point P„ and the first point Pi, thus “ closing" the figure.

T he set of points P., P2, P3, . . . , P„_,, P„ are called the vertices of the polygon, while the line segments T j \ , 7\Pj, ■ ■ . , KP> are the sides ofthe polygon. And, lastly, the angles Z P h ZP 7, . . . , ZP„_,, Z P n are the angles of the polygon.

A polygon is named by simply referring to the letters at its vertices in either a clockwise ( ^ i ) or counterclockwise order (r- \). Several ways of nam ing the polygon below are polygon ABODE, or polygon AED CB, or polygon CDEAB. Can you nam e this polygon in a t least three o ther ways?

B CorrespondenceIn C hapter 2 we discussed a one-to-one correspondence

when examining the points on the num ber line and the real num bers. At this time we would like to examine the notion of correspondence a bit more intensively. Consider the two sets of elements

{David, H arry, James} and {Doris, H arrie t, Jane}

T he boys in the first set can be paired, or m atched, w ith the girls in the second set as follows:

David with Doris H arry with H arriet Jam es with Jan e

W hen each elem ent of one set is m atched with one and only one elem ent of a second set, we say that a one-to-one correspondence exists between the

elements of these two sets. And a matching such as the one above would be read as

David is matched with Doris or David corresponds to DorisH arry is matched with Harriet or H arry corresponds to H arrie tJam es is matched with Jane or Jam es corresponds to Ja n e

By using symbols this is simplified to

David *-* Doris Harry <-> H arriet James <-* Jane

The doubleheaded arrow, however, implies two things. It is not only true that “ David is matched with Doris,” but also th a t “ Dovis is m atched with D avid.” T hat is, the matching holds in both directions.

R ather than express the matching with the three separate statem ents

David <-* Doris Harry <-> H arriet James <-+Jan e

it is m uch preferred to write this in the single form of

David Harry James <-* Doris H arrie t Jane

This will be read as either

(1) David H arry James corresponds to Doris H arriet Ja n eor

(2) T he correspondence of David H arry Jam es and Doris H arrie t Jane

The order in which the elements appear in a correspondence is very im portant, for there exists a number of correspondences, or m atchings, between the elements of two sets. Thus, a correspondence of the form,

David Harry Jam es H arriet Doris Ja n e (1)

is vastly different than

David Harry Jam es <-> Doris H arrie t Ja n e (2)

W hile the first implies a matching of

David with Harriet Harry with Doris Jam es with Ja n e . (1)

the m atching of the second is

David with Doris Harry with H arrie t James with Jan e (2)

Any three couples will tell you that whereby the first m atching m ay be the ingredients for a very pleasant evening, the second may invite disaster!

CORRESPONDENCE 113


EXERCISES

1. E x p r e s s t h e c o r r e s p o n d e n c e

Jo e «-* M ary Fred *-* Aun Bil! «-► Carol

by a single double an ew ,

2. List the matchings that exist in the correspondence n b «-» xy.3. For the two sets of elements {a, b, c} and {1, 2, 3}, there are exactly six

correspondences. Tw o of these are a b c <-» 1 2 3 and a b c *-> 2 1 3. N am ethe rem aining four.

4. “ A one-to-one correspondence exists between the students in your geom etry class and the seats in your geometry classroom.” In terpret this sta tem ent. (Exclude the seats th a t are vacant.)

5. Using the replacem ents below for the equality x + y = 5, list the one- to-one correspondence that exists between the replacements of x and y such th a t each pair of values in this correspondence will make this equation true.

x replacements: {0, .1, 2, 3, 4, 5} y replacements: {0, 1, 2, 3, 4, 5}

6. T h e 2, 3, and 4 of Clubs are removed from a deck of cards. I f these three cards are shuffled thoroughly and then placed face up one a t a tim e on a table, there is only one chance in six that these cards will appear in the order 2— 3— 4. By using correspondences can you explain why this is so?

■ Correspondence Related to PolygonsShould two polygons have the same num ber of vertices,

then it is possible to draw up one-to-one correspondences between these

vertices. T hus, for the two polygons in Figure 5-7 we m ight investigate correspondences as

(1) A B C D ^ E F G H or (2) A BCD <-> HEFG

or any one of a num ber of others. It is only in terms of a correspondence th a t such words as corresponding sides and corresponding angles can take on

CORRESPONDENCE RELATED TO POLYGONS 115

meaning. To illustrate, a pair of corresponding angles of two polygons will be two angles whose vertices are a pair of corresponding elements. Thus, in correspondence (1)

A *-> E, B <-+ F, C *-* G, D <-» H

/.A and Z E are considered to be corresponding angles and the same can be said of Z B and ZF, ZC and Z C , and Z D and ZH . For correspondence(2), however, since

A <-*H, B E, C ^ F , D ~ G

then Z A will correspond to Z H , Z B to ZE , Z C to ZF , and Z D to ZG.In the same way, a pair of corresponding sides are two sides whose end

points are pairs of corresponding elements in a correspondence between the vertices of two polygons. Before retreating in despair, let us exam ine the meaning of this sentence in the light of correspondence (1) in the preceding paragraph. Since

A *-» E and B «-> Fthen

side A B would correspond to side E F

This can be pictured as4 B C D ^*E F G H

Similarly,A <-> E and D «-* H

thereforeside AD would correspond to side E H

W e can illustrate this asA BCD <-> EFGH

Two pairs of corresponding sides have been named in correspondence (1). Can you name the remaining two pairs of corresponding sides? W hat are the four pairs of corresponding sides in correspondence (2)?

D e f i n it io n 2 4 : Corresponding angles of two polygons are two angles whose vertices are a pair of corresponding elements in a correspondence between the vertices of two polygons.

D efinition 25: Corresponding sides of two polygons are two sides whose endpoints are a pair of corresponding elements in a correspondence between the vertices of two polygons.

Before leaving this topic, it is im portant to clarify two m ore points. T he first.of these is concerned with the equivalence of two correspondences. T he

correspondenceABCD <-> EFGH

is said to be equivalent to the correspondence CDAB <-> CHEF

116 CONGRUENCE OF TRIANGLESfor in both

A*-* E, B <-> F, C<-*G, £><-* H

Thus, there are a great m any equivalent correspondences between the elements in two sets. Correspondences that are equivalent are those that preserve the same one-to-one correspondence between the elements. Can you explain why the first three correspondences below are equivalent, while the fourth is not equivalent to any one of the first three?

A BCD <-» EFGH AC D B *-» EGHF BCD A *-* FGHEA BD C HGFE

The other point to be raised is the fact that we can not select a t random two pairs of vertices in a correspondence and say that they will be the end points of a pair of corresponding sides. To illustrate, again refer to the diagram on page 114 and the correspondence

A BCD <-* EFGH

Although A «-* E and C <-+ G, the “ side” AC does not correspond to the “ side” EG, for neither is a side of the polygons!

EXERCISES

1. List the six different correspondences that exist between the vertices of the polygons below.

2. In the correspondence R S T <-* Y X Z for the two polygons below, nam e(a) T he three pairs of corresponding angles.(b) T he three pairs of corresponding sides.

3. If in the diagram in Problem 2 the correspondence had been R S T <-> Y Z X , then(a) W hat are the three pairs of corresponding angles?(b) W hat are the three pairs of corresponding sides?

CORRESPONDENCE RELATED TO POLYGONS 117

(c) Are there any corresponding sides or corresponding angles in this problem and in Problem 2 that did not change?

4. If ZA = Z B , Z.C<= ZF, and Z E = Z D , then write a correspondence between the polygons ACE and BDF so that the corresponding angles will be congruent.

5. If &? = W Y , 3 T = YX, and T t i S X W , then write a correspondence between polygons R ST and X Y W so that the corresponding sides would be congruent.(a) Draw a diagram for the polygons R S T and X Y W where the corre

sponding sides are congruent. Do the corresponding angles appear to be congruent also?

(b ) Draw a diagram for the polygons ACE and BD F of Problem 4 where the corresponding angles are congruent. Do the corresponding sides appear to be congruent also? If the corresponding angles are congruent, can the polygons be so draw n that the corresponding sides are not congruent?

6. For the polygons below, write a correspondence so that the corresponding angles will be congruent as m arked in the diagram .(a) Are the corresponding sides in this correspondence also congruent?(b ) If the corresponding angles in a correspondence between two poly

gons are congruent, then are the corresponding sides also congruent?

7. For the polygons below, write a correspondence so that the corresponding sides will be congruent as marked in the diagram .(a) Are the corresponding angles in this correspondence also congruent?(b ) If the corresponding sides in a correspondence between two polygons

are congruent, then are the corresponding angles also congruent?

A ______________ D E,----------------------- e---------------------- ,h

1

-tt-

B. (a) In terms of the markings of the two polygons on page 118, write two correspondences which are not equivalent bu t in which the corresponding angles will be congruent.


(b ) In terms of the markings of the two polygons below, w rite two correspondences which are not equivalent but in which the corresponding sides will be congruent.

■ Congruent Polygons

One of the principal topics with which the study of geom etry is concerned is the discovery of those conditions that must exist before two polygons can be placed “ on top of one another and made to fit exactly.” Polygons that do “ fit exactly,” or coincide, are called congruent polygons. Yet to use this as the definition of congruent polygons would severely handicap us, for it is difficult to prove that polygons “ fit exactly.” In order to develop a m ore usable definition of congruent polygons, it is necessary to fall back upon the concept of correspondence.

D e f i n i t i o n 26: Congruent polygons are two polygons in which there exists a one-to-one correspondence between the vertices such that

(1) All the corresponding sides are congruent.(2) All the corresponding angles are congruent.

From a simple point of view this seems to say m uch the same as the earlier statem ent that congruent polygons are polygons th a t can be m ade to fit exactly when placed on top of one another. For, if they did “ fit ex-

' actly,” then their corresponding sides and corresponding angles would have to be congruent. In our formal definition, however, we have no need to concern ourselves with such vague terms as “ fit exactly,” “ no overlapping,” or “ placed on top of one another.” Furtherm ore, complicated as this definition m ay seem, it will make the development of new work far simpler th an had we fallen back upon congruent polygons as meaning “ to fit exactly.”

For the polygons in Figure 5-8, as we know, there are m any correspondences th a t exist between the vertices. N ot all of these will lead to the

CONGRUENT POLYGONS 119

information that the polygons are congruent. Several of these correspond- > ences are

ABCD *-* EHGF and its equivalent forms

orABCD <-> GFEH and its equivalent forms

If, however, there is at least one correspondence in which all the corresponding sides are congruent and all the corresponding angles are congruent, then by the reverse of the definition of congruent polygons the polygons will be congruent. In this case, that correspondence is

ABCD <-» EFGH

for from the markings in the diagram the corresponding sides and the corresponding angles of this correspondence are congruent.

Quite apparently, there will be many polygons for which there will exist no correspondence such that the corresponding sides and corresponding angles will be congruent. These polygons will not be congruent. This is the case with the two polygons in Figure 5-9.

F ig u re 5-9.

Although there are some corresponding sides and some corresponding angles that are congruent in the correspondence

ABCD EFGH

not all are congruent. In fact, no correspondence exists between these two polygons in which all the corresponding parts, sides and angles, are congruent. Hence, these polygons can not be shown to ,be congruent.

Of the many types of polygons that exist, the most im portant in geometry is the one having the fewest num ber of sides. As you are probably aware, this polygon is the triangle. How would you classify a triangle? W hat is the


fewest num ber of sides th a t a polygon may have? Why? How would you distinguish the triangle from all the other polygons?

D efin itio n 27: A triangle is a polygon that has three sides.

To prove triangles congruent by resorting to the reverse of the definition of congruent polygons would necessitate proving three pairs of corresponding sides congruent and three pairs of corresponding angles congruent. This, apparently, would require a 'great deal of work. Hence, the rem ainder of this chapter will be devoted to ways of proving triangles to be congruent w ithout the need of showing th a t all the corresponding parts are congruent. In addition, we will also learn w hat conclusions will follow once the tr iangles are congruent.

■ Postulates (or Proving Triangles CongruentSo that you m ight have some justification for the reason

ableness of the next postulate, you will need a ruler and protractor. O n a piece of paper draw a triangle similar to the one in Figure 5-10. W e call this A ABC', th e symbol A represents the word triangle.

O n a second piece of paper draw a line segment whose measure is the same as that of BU. Call this DE.

* _ h •

Figure 5-11.

Using your protractor, find the measure of Z B and m ark off an angle a t point D whose measure is the same as that of Z B .

EFigure 5-12.

POSTULATES FOR PROVING TRIANGLES CONGRUENT 121

W ith your ruler find the measure of BA. Then by using D as one endpoint of a line segment, mark off D F so that the m easure of D F will be the same

as that of BA.

Figure 5-13.

Now complete A FDE by drawing the line segment joining the points F and E.

Figure 5-14.

In the correspondence A B C <-* F D E we know by our m easurements th a t f l C S M , Z B = Z D , and A B = FD. From the appearance of the two triangles, w hat other corresponding parts do you believe will be con

gruent? Using your ruler, find the measure of FE and compare it w ith that of AC. W ith your protractor find the measure of Z E and com pare it w ith th a t of ZC . Finally, find the measure of Z F and com pare it w ith th a t of ZA . If your drawing was accurate, you should find that AC ~ FE, Z A — ZF\ and Z C = ZE , Although originally we knew of only three pairs of corresponding parts that were congruent, now it appears that all six pairs of corresponding parts are congruent. W hat conclusion can be drawn if the corresponding sides and the corresponding angles are respectively congruent? Why?

You will notice that the two angles that were congruent were not ju st any angles in the triangles but the angles formed between the corresponding sides whose measures were also being m ade congruent. Thus, Z B was formed by the sides AB and BC, while Z D was formed by the sides FD and DE. I t was through our efforts th a t the measures of AB and ¥ 5 were m ade equal, as were the measures of BC and DE. An angle formed in this m anner is referred to as an included angle.


In Figure 5-15 Z S is the included angle between the sides S T and SR; Z T is the included angle between the sides T S and T R . W ith reference to w hat sides would Z R be considered the included angle?

K

F i g u r e 5 -1 5 ,

W ith these terms at our disposal, we can now say

P o s t u l a t e 1 3 : Two triangles are congruent if there exists a correspondence between the vertices in which two sides and the included angle of one triangle are congruent respectively to those corresponding parts in the second triangle. (The symbols used to express this entire statem ent are

W hat is the distinction between a theorem and a postulate? In view of the discussion presented on the preceding few pages, should we not have called Postulate 13 a theorem rather than a postulate?

Illustration: f

\ I Given: A B = DF Z B — Z F

F i g u r e 5 -1 6 ,

In order to prove A A B C = A D F E \ by the use of Postulate 13, w hat re m aining parts will have to be congruent? If we knew that B U was congruent to FE, why could we conclude that A ABC = A DFE? Rather than BC = FE, suppose we knew that AC = DE, could we still conclude that A A B C was congruent to A DFE?

Assuming th a t the Given D ata included the fact that BC -= FE, we would know by Postulate-13 that & ABCS= A D F E . Quite often the letters A BC and D E F are arranged in any order whatsoever. Although m any m athem aticians see nothing wrong in this, it would be best if we were a bit m ore careful. Specifically, to say th a t A B C A — A D F E will imply the congruency correspondence of BCA <-* DFE. In turn, the definition of congruent polygons enables us to say that A B £ ED and Z B = Z D . This is

t Read this as: triangle ABC congruent to triangle DFE.

POSTULATES FOR PROVING TRIANGLES CONGRUENT 123

so by virtue of the fact that corresponding sides and corresponding angles of congruent triangles are congruent. O n the other hand, the Given D ata states that AB is congruent to DF, not to ED, and th a t Z B is congruent to ZF, not to Z D \ Hence, writing the letters carelessly in the congruence may set up a correspondence in which the corresponding parts are not congruent!

There is yet another postulate through which triangles can be proved to be congruent. Showing that this postulate is plausible can be done in the same manner as we had for Postulate 13. This will be left as an exercise for you to do.

P o stu la te 14: Two triangles are congruent if there exists a correspondence between the vertices in which two angles and the included side of one triangle are congruent respectively to those corresponding parts in the second triangle. (The symbols used to express this entire statem ent are A S./l.)

A side is said to be included between two angles if the vertices of the angles form the endpoints of the side.

Illustration: ,

If we were to apply Postulate 14, what rem aining parts would have to be congruent before we m ight conclude th a t A A C B = A ftS T ? If Z C Sr ZS, would A A C B ~ A R S T ? Why or why not? If Postulate 13 were to be applied, what rem aining parts would have to be shown congruent in order that A A C B be congruent to A RST? A rrange the letters in the correspondence ACB <-> R S T so th a t they will represent a correspondence equivalent to this one. Will the parts in the Given D ata be corresponding parts in your correspondence? W rite a correspondence in which the parts in the Given D ata will no t be corresponding parts.

EXERCISES

0In each of the problems below, nam e th e parts that will

still have to be shown to be congruent before the triangles will be congruent

124 CONGRUENCE OF TRIANGLESby the £./!,£. postulate. T he markings on the triangles indicate those parts th a t are congruent on the basis of the Given Data.

1. Oonci.: A A B C S i A D F E

3. Concl.: A A B D ^ A A C D

A_

5. Concl.: A A C D A BCD

7. Concl.: A A E B ^ A D E C

Concl.: A R S T — A Y X W 2, k w X

Concl.: A A B C 3 * A A E D 6.

Concl.: A A E C S Z A A B D 8 .

POSTULATES FOR PROVING TRIANGLES CONGRUENT

9. Concl.: A A E B = ZCFD | Concl.: A E A C ~ A E D B 10.

125

BIn each of the problems below, name the parts th a t will

have to be congruent before the triangles can be shown to be congruent by the A.S.A. postulate. T he markings on the triangles indicate those parts th a t are congruent on the basis of the Given D ata.

1. Concl.: A A B C — A F E D ’

A . f

B C D £

3. Concl.: A C D B == A A E B

A -

5. Conci.: A A D B S Z A C D B

Concl.: A A B F — A C D E 2.

Concl.: A A C B — A F D E 6.

7. Concl.: A F A C ^ A E D B

A 8 C o

126

9. Concl.: A A D F S t A C B E

A o

11. Concl.; a A BC S i A DCB

CONGRUENCE OF TRIANGLES

C oncl.: A AEC S A D EB 8.

Concl.: A AG F& . A C G E 10.A _ _ _ _ 0 F

C

Given: A B ^ A S 12.ZC EA S Z DBA ? •= ?

Concl.: A A D B z= A A C E

| Applications of the Postulates on Congruence to Formal Proofs

W e are now in a position where it is possible for us to prove th a t triangles are congruent by using a formal proof. As before, this will im ply th a t each tim e we make a statem ent, we will have to justify its use by v irtue of th e fact that

(1) I t is a piece of information stated in the Given D ata .(2) I t is the property of a word as stated in the definition of that word.(3) I t is the property implied by the reverse ot the definition of a word.(4) I t follows from a postulate we have wade.(5 ) I t follows from a theorem th a t we have already proved.

POSTULATES ON CONGRUENCE TO FORMAL PROOFS 127

Illustration 1 '■ 1

Given: AD and BC bisect each other at E.

Concl.: A A E B Z Z A D E C

A nalysis: The very first thing to do is to separate the Given D ata into two<—f ___ i—> , _ ..

distinct pieces of information. T h a t is, AD bisects BC and. BC bisects AD.O n the basis of the first piece of information, we can conclude th a t CE S EB, while the second piece tells us that D E ~ EA. This inform ation is marked in the diagram as was done above. We note that there is still one piece of information lacking before it is possible to conclude th a t the triangles are congruent. From the fact th a t two sides in one triangle are congruent respectively to two corresponding sides in the other triangle, the apparen t postulate to try to apply to show that the triangles are congruent is the S.A.S. postulate. Thus, the only part lacking is the included angle in each triangle.

The diagram can not be used to conclude th a t certain sides o r angles may be equal because " they appear so in the d iag ra m "! I t is helpful, however, in calling our attention to the fact that

(1) There are vertical angles'in the figure.(2) The reflexive property of equality can be applied.(3) There are pairs of supplem entary angles in the figure.

In this problem we discover by examining the diagram th a t the angles A E B and DEC are vertical angles. Hence, from the theorem on vertical angles we can conclude that they are congruent.

PfcOOF STATEMENTS REASONS '

<-4 __1 . AD bisects BC.

1. E is the midpoint of BC.

3. CE = EB (i) jjj,. •

4. BC bisects AD.

5. £ is the midpoint of AD.

6. D E = EA (j)

7. ZAEB ^ ADEC [a) '

8. A A E B ^ A D E C .

1. Given

2. Def. of the bisector of a line segment

3. Def. of the m idpoint of a line segment

4. Given

5. Same as 2

6. Same as 37. If two angles are vertical angles, then

they are congruent. (Theorem)8. S.A.S. (Postulate)

£

Figure 5-18,


< 4 __Giveii: AD is the _L bisector of BC Cone!.: &ADC £= A A D B

An a ly sis: A s in the previous illustration, rewrite the Given D a ta as two pieces of information. Since the lines are perpendicular, there will be right angles a t point D, and from this fact we can conclude th a t the angles will

be congruent. Since AD bisects BC\ a side of one triangle will be congruent to a side of the other. W e have exhausted the Given D ata bu t do not yet have enough information to say that the triangles are congruent. Again we resort to an examination of the diagram . T o prove the triangles congruent by the A.S.A. assumption would require showing th a t Z B — ZC . T here is no means of doing this. T o prove the triangles congruent by the S.A.S,

assumption would require showing that AD ~ AD. This we know by the reflexive property of congruency! Hence, the conclusion follows.

P R O O F STATEMENTS REASONS

1. AD X BC 1. Given2. Z A D B and ZA D C are right 2, Def. of perpendicular lines

angles.3. Z A D B S ZA D C (a) 3. If two angles are right angles, then

they are congruent. (Theorem)

4. AD bisects BC. 4. Given

5. D is the m idpoint of JSC. 5. Def. of the bisector of a line segment

6. BD S DC (s) 6. Def. of the m idpoint of a line segment

1. ~A D ~ AD (s) 7. Reflexive property of congruence8. & AD C SZ & AD B 8. S.A.S. (Postulate)

Ju s t a word of suggestion before you tackle the proofs of the problems in the exercises that follow. Each piece of Given Data is usually designed to lead you to a pair of congruent sides or a pair of congruent angles. In most cases, therefore, it is best no t to leave any piece of information until you have shown a pair of sides congruent or a pair of angles congruent. T o keep track of w hat you have done and where you are going, m ark your diagram each tim e you prove a pair of sides congruent or a pair of angles congruent. Do the same for the statem ents in the proof by using a small (s') or a small (a).

POSTULATES O H CONGRUENCE T O FO R M A L PROOFS

1. L i v e n : AB S AD Z l S Z2

C o n c l . : A A B C ^ ^ D C

y ' \/ I ■ )

/ 3 . / Given: B D is the bisector

1 J o{ ZABC.

D B is the bisector

of ZAD C .Concl.: CxABD S A C B D

i ^ i v e n : I C and BD bisect / each other a t K-

^ Concl/. A A E 3 S A ®

G iv e n : AD = BC Z 1 S / 2

C o n c l . : A ^ D B S A C B D

Given: AC bisects ZB A D .

AC L BD

Concl.: A ^ C S A ^ C C

Given: Z l = /Lla b & F e

BD t= C E

Concl.: h A B C & ^ E D

I A.

c 0

k o a lv jU ta the write-up of your proof.t Do not include an


7. Given: A B S DE

Z E

R F ^ C E C oncl.: A A BC ~ A D E F

9. G iven: M is the midpoint o iB C .Z \ S Z l

C oncl.: A A B M S A DCM

11, G iven: CA bisects ZDCE.

BA bisects Z D B E . C oncl.: A D B C S A EBC

G iven: C is the m idpoint 8.. of BD.

Z 3 S Z4 Z l S Z l

C oncl.: A .4BC ~ A E D C

Given: A B S A E 10.B D & E C Z \ S Z l

Concl.: A A B C — A A E D

Given: A E = D E 12.E is the m idpoint

of BC.ZA E C SZ Z D E B

Concl.: A A B E S DCE

PROVING LINE SEGMENTS OR ANGLES CONGRUENT 131

13. Given: A B SZ D CM is the midpoint

of BC.

AB J. BC <-> «->DC ± B C

Concl.: A A BM == DCM

15. Given: B E = B C

T b i c dZ l S Z l

Concl.: A A B C ^ A D B E

Given: B F S lC E 14.Z l S Z l

A B _L B E

D E 1 BE Concl.: A A B C ~ A D E F

Given: Z l £= Z l 16.Z 3 Z 4

AC bisects Z B A D .A B — AD

Concl.: A A B C ~ A A D C

■ ’Proving Line Segments or Angles Congruent Through Congruent Triangles

In view of the definition of congruent triangles, w hat conclusions can be drawn if A ABC = A D E F ? Thus, it appears th a t if it were necessary to prove that ZC = ZF, one m ethod of atta.ck m ight be to find

132 CONGRUENCE OF TRIANGLESa pair of triangles that contain these angles and prove them to be congruent. It would follow from the definition of congruent triangles that the corresponding angles were congruent, t Similarly, if we had to show that A B £= DE, it wuuld merely be a m atter of proving A -.15C — A DEF, and, again, fiom the second property of congruent polygons we would know th a t the corresponding sides were congruent, t

In the above diagram and in m any of the problems we will encounter, it will be quite simple to select the triangles that are to be proved congruent. In other situations, however, the selection of triangles will no t be quite so simple. T o illustrate, exam ine Figure 5-21 and assume that it was

necessary to prove Z B = ZC. Since there are m any pairs of triangles containing Z B and ZC, deciding which pair to prove congruent m ay be quite difficult. Name three pairs of triangles we might try to show congruent in order to prove that Z B = ZC.

Illustration:

Given: C is the m idpoint of AE.Z l S Z \

Concl.: A B S £ D E

A n a ly s is : In order to prove A S = 5 E , we will try first to prove that A ABCZ= A E D C . Once these triangles are congruent, it will follow from the definition of congruent polygons th a t AB = DE. C being the m idpoint of A E will give us AC ~ CE. Angles ACB and ECD are vertical angles, hence they are congruent. Therefore, the problem simply reduces to showing that Z 2 is congruent to ZBAC . However, since both Z B A C and Z2 are congruent to Z l , then by the transitive property of congruence it will be possible to conclude that ZB A C =. Z l . Thus, the triangles are congruent by the A.S.A. postulate,

t See page 118. _

r

PROVING LINE SEGMENTS OR ANGLES CONGRUENT 133

PRO OF STATEMENTSREASONS

C is the midpoint of AE-

2. AC S CE W3. Z2 S Z \4. Z B A C = Z \

5. Z2 S ZB A C 0 )6. ZA C B S ZECD {a)7. A A B C ^ A ED C

8. A B = DE

1. Given2. Def. of the m idpoint of a line segment

3. Given4. If two angles are vertical angles, then

they are congruent. (Theorem)5. T ransitive property of congruence

6. Sam e as 47. /i.S ./i. (Postulate)

, 8. Def. of congruent polygons

EXERCISESt

1. Given: CD is the 1 bisector

o f AB.Concl.: 54

3 , Given: Z \ S Z l Z3 S Z4 BD = CE

Concl.: I C S F O

Given: AB J . BD

ED 1 BD C is the m idpoint

of BD.Z l S Z l Z A & Z E

Given: Z \ £= ZSL Z3 S ZA

C oncl. Z A = ZC

4.

t .Do not include an AaalysU In <** ” rit‘ -uP of ^ p r°° f'

'r-vw

atm

'

5 . Given; A B ^ E F

A D ^ C F Z l £ Z2

C oncl.: A B S Z E

134

7 . Given: Z l S Z2> __

bisects AC. Concl,: £ £ ? £ ?G

9 , Given: /)C and S i) bisect

each other a,t E. Concl.: Z l S Z2

A*

G iven: Z l S Z2 6.Z 3 S Z4

C one'.: A B — CD


Given: B E ^ B Fi--7

AB is the _L bisector of CD.Z l £ Z2

Concl.: U E ^ m

8.

Given: Z! S Z2

BC S EF

1 AC <-> +->

M l

10.

11. Given: Z l S Z2Z3 is comp, to Z l. Z4 is comp, to Z2.

Concl.: LA S ZC

PROVING LINE SEGMENTS OR

1 3 . Given: j4C = AD B D SZC E Z l S Z 2

Concl.: Z3 = Z4

A

15. Given: Z B ^ ZCE is the m idpoint

of AB.G is the m idpoint

of D C .___A B S iD C Z l £ Z2

Concl.: W ^ H C

A,______________.0

Given: BD £ CE 12.Z l £= Z2Z B is comp, to Z3.Z E is comp, to Z4.

Concl.: A B = EF

ANGLES CONGRUENT 135

Given: F is the m idpoint 14. of BC.Z B ~ Z C Z E F B £Z ZD FC

Concl.: D F = EF

Given: I B £* AD 16.TiB S CD Z l £ Z 2 Z3 = Z4

C oncl.: Z5 = Z6


SI Further Conclusions That Can Be Drawn on the Basis of Congruent Triangles

Frequently it is neces.'ary to prove lhat a lay is the bisector of an angle, or that a certain point is the rpidpoipt of a line segment, or that a line is the biscctor of a line segment. T o illustrate, w hat will

have to be true in Figure 5-23 before it can be concluded th a t BA bisects

Figure 5-23.

ZCBjD? W hat is one m ethod of proving that /.CBA ~ /.D BA? Hence, if A A B C = A A B D , it will follow that /C B A = /.D B A . In view of this congruence and the reverse of the definition of the bisector of an angle, we

can say that BA bisects /.CBD .Apparently, then, congruence of triangles can lead us to the congruence

of certain line segments or angles, and this fact, in turn, will lead us to the conclusion we were hoping to draw.

Illustration:

JGiven: AD bisects /B A C .

<-> <-»AD 1 BC

Concl.: D is the m idpoint of BC.

Figure 5-24.

A n a ly sis: In order to prove D to be the midpoint of BC, it will be necessary to show that BD — DC. These line segments, however, will be congruent if A A BD S A ACD. Hence, the problejn reduces to one of showing that these triangles are congruent. '

CONCLUSIONS DRAWN ON CONGRUENT TRIANGLES 137

[ p r o o f - STATEMENTS REASONS

1. AD bisects /B A C .2. /B A D S /C A D (a)

3. AD 1 BC4. /.A D B and /A D C are right

angles,5. /A D B = /A D C (a)

6 AD ^ AD (s)7. &.ABD ~ A ACD

8. B D S Z DC9. D is the m idpoint of BC.

1. Given2. Def. of the bisector of an angle

3. Given4. Def. of perpendicular lines

5. If two angles are right angles, then they are congruent. (Theorem )

6. Reflexive property of congruence7. A.S.A. (Postulate)

8. Def. of congruent polygons9. Reverse of def. of the m idpoint of a

line segment

EXERCISES

1. Given: A B ^ C B

BD bisects /A B C .

Concl.: DB bisects /ADC.,

aft

3 , Given: AD bisects BC.

AB ± BC

DC 1 BC

Concl.: BC bisects AD.

Given: A B = BC / I S /2 -

Concl.: D E bisects /A D C .

A .

2.

Given: B is the m idpoint

of AC.

. BD J_ EF Concl.: / \ £* / ^

E A 8 ______ C F


5 . G iven: L B ZC

A E ^ A D

a S ^ a c<-*

F E X AC

GD X A B

Concl.: B F 9 iG C A

■7 _ . < - ►< - */ . G iven: CD X BF

L \ ^ Z2 S Z £

a D ^ e d

C oncl.: D is the m idpoint of ~BF.

9 _ . < - > + - >. G iven: £ C 1 CD

<-> <->FZ> 1 CD ...

A B 1 CD

m ^ i cB is the m idpointo l W .—¥

Concl.: BA bisects LE B F.

Given: B E ^ I ) E 6 .<-> <-►

CD 1 AD

A B X BC C oncl.: Z l S* Z 2

Given: A B X BC and AD 8. < - > < - ► <->

D C X B C and AD E is the m idpoint o ( W .L D E B S L A EC

Concl.: L E A D = L E D A

Given: BD bisects LABC . 10.BD X AC

DA &TTB=~DC Concl.: Z l £* Z2

(Hint: Prove A A D B == A B D C \ then note the correspondence.)

OVERLAPPING TRIANGLES 139. ' ”b

■ Overlapping TrianglesOccasionally the difficulty of the proof is increased by the

fact that the triangles to be proved congruent overlap one another. As a case in point, consider the problem below.

Given: A B X DC <->

D E X AC l i e s c S

Concl.: D U ^ l C

Figure 5-25.

T he very first thing to consider is “ W hat triangles contain the line segments DC and AC as sides?” I t is these triangles that will have to be shown to be congruent. A DCE has the segment DC as one of its sides, while AC is a side of A A C B . Should it be possible to prove these triangles congruent, then DC will be congruent to AC.

I tA <-' '

Given: AB X DC *-> <-»

D E X AC U S = <JE

Concl.: m ^ A C

Figure 5-26.

PRO O F STATEMENTS REASONS

1. A B X DC2. L A B C is a right angle.

3. £ £ JL 4C4. LD E C is a right angle.5. L A B C £ LD EC (a)

6. ~SG S C £ (j)7. L C = L C8. A D C E = A A C B

9. ~DC = "Z C


3. Given4. Same as 25. I f two angles are right angles, then

they are congruent. (Theorem )

<5. Given7. Reflexive property of congruence8. A.S.A. (Postulate)

9. Def. of congruent polygons

140CONGRUENCE OF TRIANGLES

W ith experience you will have little difficulty in keeping your attentionon those triangles in which you are interested, although they m ay overlap.A t present, however, i t is advisable to use two differently colored lead pencils to distinguish one triangle from the other.

EXERCISES

1- Given: D B ^ D A

d c ~ d eConcl.: Z B ^ Z A

3 . Given: A B x BC

A D JL D E Z l S ZZ

a b ^ z dC oncl.: Z C s Z E

5. Given: BE & ad Z l S* Z2

Concl.: S H ^ a E_C

Given: AC ^ A E

A B ^ a S Z l £ * Z 2

Concl.:

A

Given: A B ^ ACE is the m idpoint of AC,

D is the m idpoint of

Concl.: Z \ Z2

G iven: Z l ^ Z 2 <r* «-*

BC ± A B and CD Concl.: D B ^ AC

Ap

THE ISOSCELES TRIANGLE 141

7. Given: A B _L BC

DC ± BC Z l S Z2

Concl.: A B — DC

9. Given: Z l = Z2 Z3 S Z4

Concl.: Z.A £= Z B

11. Given: A E S D E

B E £= CEV-+ «-+

A E J_ B E <-» *-*

D E I . EC Concl.: Z A ^ Z D

G iven: Z3 = ZA Z l S Z2

C oncl.: Z S £= Z6

Given: Z A B E = Z A E B 10.B C 9 Z E D

Concl.: B D S -E C A

Given: Z E = ZC Z l S Z 2 5 is the midpoint

of EC.C oncl.: A B = D B

■ The Isosceles TriangleThus far we have exam ined triangles in which there existed

no special features about the triangles. T here are, however, certain triangles that contain elements th a t m ake them distinguishable from the general run of triangles. These peculiarities revolve about the relations that m ight exist between the sides of a triangle o r perhaps the angles of the triangle. Thus, a triangle with three congruent sides is. called an equilateral triangle, while one w ith three congruent angles is an equiangular triangle. A complete list of these special triangles is given below.

/✓

142 CONGRUENCE OF TRIANGLESDistinguishing Features of Special Triangles

. rSides Xar,\e

E q u i l a t e r a l

T r i a n g l e

Iso sc e le s T r i a n g i c

j congruen t Sides

2 Congruent Sides

No C ongruent Scalene Triangle Sides

IAngler

3 C o n g r u e n t

A n g l e s

1 R i g h t A n g l e

Name E q u i a n g u l a r

Triangle R ight Triangle

1 Obtuse Angle Obtuse T riangle

3 Acute Angles A cuteT riang lc

T h e definitions for these triangles are very similar.

D e f i n i t i o n 2 8 : An equiangular triangle is a triangle having three congruent angles.

D e f in it i o n 2 9 : A right triangle is a triangle having a right angle. D e f i n i t i o n 30: An equilateral triangle is a triangle having three congru

ent sides.D e f i n i t i o n 31: An isosceles triangle is a triangle having two congruent

sides.

I t is the isosceles triangle to which we will turn our attention a t this time. T h e two congruent sides of the isosceles triangle are called the legs

(A B and AC), while the third s id e 's called the base (BC). T h e two angles whose common side is the base (Z B and ZC ) are called the base angles, while the angle formed by the congruent sides is the vertex angle. In terms of th e words given in this paragraph, w hat property about an isosceles triangle is given in its definition? Examine the base angles of an isosceles triangle. W hat do you believe we will be able to prove about these angles? W hat p roperty of the equilateral triangle is given to i t by its definition?

A lthough our ultim ate objective is to prove that the base angles of an isosceles triangle are congruent, we will have to make a slight detour to pick up certain properties th a t are needed to do this.

D efin itio n 32-A: The interior of an angle is a set of points such that if a ray whose endpoint is the vertex of the angle is drawn through any one of th e points in the set, the ray will be between the sides qf the angle.


Now we would like to go further and define the interior of a triangle. Almost intuitively we get the feeling that points P. Q, and R of Figure 5-28 belong

to the interior of h A B C , whiie X , Y, and Z do not. Notice, furtherm ore, th a t from the definition of the interior of an angle the points P, Q, and R are in the interior of both Z A and Z B . T he same can not be said for point X , for although it is in the interior of Z B , it is no t in the in terio r of ZA . How is point Y related to the interiors of angles A and B? How is point Z related to the interiors of angles A and B? W ith this property in m ind the following definition was made.D e f in it io n 5 2 $ : T he interior of a triangle is the set of points th a t arc com

mon to the interiors of any two angles of the triangle.

In reference to the isosceles triangle, if we are to prove th a t the base angles are congruent, we are im m ediately confronted with a difficulty.

O u r principal method for proving angles to be congruent is through congruent triangles, yet in Figure 5-29 there is only one triangle! W e are in need of two triangles, one that will contain Z B and the o ther ZC . Perhaps the best way of obtaining two such triangles would be by drawing a ray through point A. Any ray, however, as seen in Figure 5-30, would not give us triangles that are congruent.

Since AB is already congruent to AC, and AD is congruent to itself by the reflexive property of congruence then, obviously, it seems advisable that the ray we draw should be the bisector o f ' Z B AC, thus m aking the included angles congruent. “So,” you say, ‘Vlraw the bisector and be on with the proof!” This would be nice, and this is exactly w hat was done for, m any hundreds of years. M athem aticians were disturbed by this, too, for


equally as long a period. "H ow do we know,” they wondered (see Figure 5-31), “ th a t when we draw the bisector of L A , this ray m ight not get lost w ithin the interior of the triangle and never emerge? O r, should it intersect a second side of the triangle, why m ust it be BC and no t either A B or AC?" Overcoming the first question was their greater problem and they did this by postulating th a t

P o s t u l a t e 15: A line th a t intersects one side of a triangle and enters theinterior of the triangle m ust intersect a second side of the triangle.

This postulate is known as Pasch’s Axiom f after the m an who first stated it. Now the proof of the second question that was raised follows ra ther readily. Restating it, it becomes, “ M ust the bisector of L A intersect side BC, or can it intersect A B or AC and therefore bypass BC?” If the bisector were to intersect AC as shown in Figure 5-32, it would have to do it

a t some point E. This would imply that there exists two lines through points A and E, the side AC and the bisector AD. This, however, is in contradiction to Postulate 4 w hich states th a t “ There exists one and only one line through two points.” H ence, the bisector AD can not intersect side AC. By the same reasoning, it can no t intersect side AB. But by Pasch’s Axiom, it must intersect some side of the triangle! This will have to be side BC since i t is the only remaining side.

A proof such as the one ju st presented is called an indirect proof. W e will examine this type of proof m uch m ore carefully a t a later point in our work. Now, however, there is still one further question that has to be clarified

t R ecall th a t on page 54 it was s ta ted th a t the term s “ axiom ” ar.d " p o s tu la te” a re used in terchangeab ly .

and that is, "How do we know that the bisector of an angle even exists?”

i This we will have to assume.

Postulate 16: Every angle has a bisector.Now we have the tools necessary to prove the theorem about the isos

celes triangle.TH EO R EM 9: I f two sides of a trian g le a re co n g ru en t, th en th e angles

opposite those sides a re c o n g r u e n t .


Given: AB £= AC Concl.:

1. A B & A C { s )

2. L et AD be the bisector of LBAC.

\ . Given

2. Every angle has a bisector. (Postulate)

3. Pasch’s Axiom~—> .3. A D must intersect BC

a t some point E.4. L B A E ~ L C A E (a)

5. A E A E6. A A B E ^ A A C E7. L B ~ L C

C ' ■' Now is a convenient time to prove the reverse of Theorem 9, for we

can show the applications of both to the proofs of problems.

THEOREM 10: If two angles o f a triangle are congruent, then the sides opposite those angles are congruent.

4. Def, of the bisector of an angle

5. Reflexive property of congruence

6. S. A.S. (Postulate)7. Def. of congruent polygons

G iven : L B — L'.C Concl.: A B = AC

146

A n a l y s is : A s before, we will need the assumption on the bisector of angle of a triangle. This time, however, it will be the base angles th a t


anare

bisected ra th e r than the vertex angle. T o prove A S ~ AC, it is necessary to show that A A B E =£ A ACD. Exam ination of these two triangles brings to light the fact that there are only two pairs of elements th a t are congruen t: L A — L A , and LACD ~ L A B E (why?). T h e included sides between these angles a re A B and AC. But this is ju st the conclusion we are try ing to draw on the basis o f the fact that the triangles are congruent! Hence, another m ethod of a ttack will be needed to prove A 'A B E ~ A A C D .

By proving A D BC = A ECB we can obtain enough information to prove A A B E — A A C D . On the basis of the analysis, try to prove this theorem yourself before reading the proof below.

P R O O F ) STATEMENTS

1. L B S = L C ( a )

2. Let B X be the bisector of

L A B C and CY be the bisector of LAC B.

_____3. B X m ust intersect AC at

some point E, while C y

m ust intersect A B a t some point D.

4. L E B C — LD C B (a)

5. B 5 ^ B C ( s )6. A D B C S iA E C B

7. S E ^ C B (s)8. L B D C Z L L C E B9. LA D C is supp. to LBDC.

10, L A E B is supp. to LCEB.

R EA S O N S

1. Given

2. Every angle has a bisector. (Postulate)

3. Pasch’s Axiom

4. Halves of congruent angles are congruent. (Postulate)

5. Reflexive property o f congruence6. A.S.A. (Postulate)

7. Def. of congruent polygons .8 . Same as 7

9. Reverse of def. of supp. angles 10. Same as 9

j

THE ISOSCELES. TRIANGLE 147

11. LADC 9* LA E B (a)

12. L A B E 9Z LACD (a)13. A A B E — AAC D

14. AB ^ AC

11. If two angles are supplem entary to two congruent angles, then they are congruent. (Theorem)

12. Same as 413. /4.S./4. (Postulate)

14. Same as 7

It is interesting to note that although we seem to have come a long way from our original definitions and postulates, all but one of the reasons used in the preceding proof is either a definition or a postulate. We must never lose sight of the fact that every one of our proofs can always be traced back to the postulates and definitions upon which'we previously agreed.

In the proof of Theorem 10 we did not use nor did we need the fact that the bisectors BE and CD intersected. Can you apply Pasch’s Axiom to prove that these rays intersect?

Before we proceed further, there are three line segments whose definitions we need. These are

D efinition 33: An altitude of a triangle is a line segment draw n from any vertex perpendicular to the opposite side (extended, if necessary) of the triangle.

D efinition 34: A median of a triangle is a line segment draw n from any vertex to the midpoint of the opposite side of the triangle.

D efinition 35; An angle bisector of a triangle is a line segm ent that bisects any angle of a triangle and terminates in the opposite side.

In Figure 5-36 an altitude was drawn from vertex A to side BC. In Figure 5-37 the median was drawn from B to AC, while in Figure 5-38 the

F ig u re 5-36. F ig u re 5-37. F ig u re 5-38.

angle bisector of LA C B was drawn. How m any altitudes will a triangle have? How m any medians? How m any angle bisectors? D raw an obtuse triangle and draw in the three altitudes, Do all three of them lie w ithin the interior of the triangle? Draw a right triangle with its three altitudes. W hat did two of the altitudes turn out to be? W hat is true about all three altitudes of the right triangle? If in the obtuse triangle th a t you drew you extend the altitudes far enough, will the altitudes also meet a t a common point?

Draw two of the medians of a triangle. Can you prove on the basis of Pasch’s Axiom that they must m eet in the interior of the triangle? Can you prove that two angle bisectors of a triangle must also intersect in the in-

148

terior of a triangle? M ust two altitudes of a triangle meet in the interior of the triangle? Give an illustration to justify your answer.

Illustration 1 of the Theorems on the Isosceles Triangle:


Given: A ABC is isoscelcs with A b £=• AC.

CD is the m edian to AB. B E is the m edian to AC.

Concl.: CD S B E

A n a l y s . s : By p r o v in g A D B C a A £ C i? ) C7 ) can be shown to be congruent

P R O O F I STATEMENTS

1. A ABC is isosceles with

2. ZA B C £= ZA C B (a)

3. CD is the median to AB.4. D is the m idpoint of AB.

5. B E is m edian to AC.

6. E is the m idpoint of AC.

7. BD £ CE (s)

8. BC = BC (s)9. A D B C ~ A E C B

10. M S * CD

Illustration 2:

REASONS

1. Given

2. I f two sides o f a triangle a re congruent, the angles opposite those sides are congruent. (Theorem)

3. Given

4. Def. of a m edian

5. Given

6. Same as 4

7. Halves of congruent line segments are congruent. (Postulate)

8. Reflexive property o f congruence9. S.A.S. (Postulate)

10. Def. of congruent polygons

Given: Z l S Z2Concl.: A ABC is an isosceles triangle.

A n a l y s is : A A BC will be isosceles if we can show two sides to be congruent. Two sides will be congruent if two angles are congruent. Since Z l ~ Z2, it is possible to prove that /B A C — /.B C A . Hence,, two angles of the tri

angle are congruent and the proof can be completed.



1. Z l £ Z22. /.B A C is supp. to Z l.3. Z BCA is supp. to Z2.

' 4. Z BCA £ Z BAC

5. BA £ BC

6. A ABC is isosceles.

EXERCISES

1. Given2. Reverse of def. of supp. angles

3. Sam e as 24. If two angles are supplementary to

two congruent angles, then they are

0 congruent. (Theorem)If two angles of a triangle are congruent, then the sides opposite those angles are congruent. (Theorem)

6. Rev. of def. of an isosceles triangle

1. Given : A A B C is isosceles with AB £ AC.

AD is the median

to BC.

Conch: AD bisects I B AC.

3 . G iven: AB = AC B E = CD

Concl.: AD = AE

G iven: A ABC is isosceles w ith A B £= AC.

Concl.: Z l == Z2

Given: E B = EC/ A E C S i /D E B

Concl.: A B = CD


5. Given: AC S AD

EC = BD Cone].: LE A D & ZB A C

\ j ' 7. G iven: A B = - A E*—■4 —*/4C and AD trisect ZB A E .

Concl.: A A C D is isosceles.

•-J 9. Given: D is the m idpointof BC.

I B S* F 5 ~Z E D C Z P D B

Concl.: A ABC is isosceles.

Given: E B & E C 6.Z D E B S Z A E C

Concl.: A E A D is isosceles.

Given: A C ^ A D 8.C and D are trisection points of BE.

Concl:: A A B E is isosceles.

A

Given: A B ^ A C 10.AD bisects ZBAC .

Concl.: I D is the m edian

to SG.


/ 11. Given: AD is the m edian to BC.

AD is the altitude

to BC.Concl.: A ^ B C is isosceles.

A

1 3 i G iven: D E and FC ± BC Z \ ^ Z 2

D E = FG Concl.: A A B C is isosceles.

M , 2> f

/ 15. Given: A B & .D CZ B A D £ ZCD A

Concl.: A E A D is isosceles.

■

Given: BA S BC 12.D , E, and F are the

midpoints of BA, AC,

and CB respectively.

Concl,: D E = FE

Given: AD and FC trisect

BE.Z B £* Z E A B = FE

Concl.: A G C D is isosceles.

A F

14.

Given: A B and DC _L BC 16.A B S DC

C oncl.: A E BC is isosceles.

152

BE bisects LABC . Concl.: D C ^ E B

17. Given: A B 9 AC

CD bisects LACB .

19. G iven: L B ~ L CL I S L 2 B E

Concl. ■. AD AG

Given: A B '^ A C •JgCD bisects LA C B .

BD bisects LA B C .Concl.: A D SC is isosceles.


Given: AB Concl.: ZA

= AC = = L B i

: BC i L C

G iven: Isosceles triangles

A BC and D B C on

the same base BC Concl.: Z A B D S LA C D

A

20.t

22.

" oftcn consid" cd - a what would the 5!atemcnt of th;s

THE S.S.S. THEOREM 153

I The S.S.S. Theorem

There are four general methods for proving triangles to be congruent. Prior to now we have examined and assumed two of these methods. T he third we shall prove in this section, while the proof of the fourth will have to be delayed until m ore information has been established th a t will m ake th a t proof possible.

In order to develop this third statem ent of congruence, we shall need yet another theorem and another postulate.

THEOREM 11: If two triangles are congruent to the same triangle, then they are congruent to each other.

Given: A A B C ~ A D E F A X Y Z S A D E F

Concl.: A A B C S A X Y Z

F ig u re 5-41.

A n a l y s is : Triangles ABC and X Y Z can readily be shown to be congruent by the S.A.S. postulate. From the congruency correspondence ABC ^ D EF we can conclude th a t AB — DE. Similarly, X Y = D E and, therefore, A B = X Y . In the same way, BC can be shown congruent to YZ, and L B congruent to L Y .


\ . A A B C ^ A D E F 1. Given

2. A B ~ DE, L B != LE , 2. Def. of congruent polygons

~BC ~ E F3. A X Y Z A D EF 3. Given

4. X Y ^ M , L Y g * LE , 4. Same as 2

T z & e f

5. A B £ X Y (.r) 5. Transitive property of congruenceL B ^ L Y {a ) (Postulate)

B C ^ Y Z ( s )

6. A A R C S A X Y Z 6. S.A.S. (Postulate)

The postulate we need deals with angles and is comparable to one • already established for line segments. T he very first postulate m ade was


th a t a line can be extended as far as desired in either direction. O n e in terpretation of this statem ent is

I f we sta rt w ith a given line and a point P of th a t line, it is possible tofind a second point Q of this line such th a t PQ will be congruent to any line segment that is given to us.

p o a______ a----------- . ... ... ,,g»

F ig u re 5-42.

T hus, by starting with the point P it was possible to find Q such th a tI Q s z a E.

Now, we w ant to establish a com parable postulate for angles:

P o stula te 1 7 : At a given p o in t of a given line th e re exists an angle w hose vertex is th e given po in t a n d one of w hose sides is a ra y o f the g iven lin e such th a t th is an g le is co n g ru en t to a n y g iven angle.

T o illustrate, this postulate implies th a t by starting with some po in t P of th e line PQ it is possible to find an angle such as /.R P Q th a t will be congruent to th e given ZABC.

A

F ig u re 5-43.

We now have the weapons necessary to prove the following theorem .

T H E O R E M 12: T w o trian g les a re c o n g ru en t i f th e re exists a co rre spondence betw een the vertices >n w h ich th ree sides o f o n e a re co n g ru en t to those co rre sp o n d in g sides <>f th e o th er. (The symbols for this statem ent | r e S.S.S.) J

A. D. Given: i B & ' M ------

U S s z W 7 D s * W

~rf-------- E-------------------H Concl.: A A B C = A D E FFigure 5-44,

A na ly sis : O u r m ethod of attack will be to set up a th ird triangle. A A B C an d A D E F w ill bo th be proved congruent to this th ird triangle. Hence, by th e theorem just established, it will follow that A ABC S A DEF,

THE S.S.S. THEOREM155

^ I P

t t-

/ 1. A t point B .of line BC there exists an angle congruent to /D E F . Let this angle be

/S B C . (a)<-»

/ 2. Extend B S so that T iB & 'D E (s )

3. L et RC be the line through points R and C.

H4. L et RA be the line through

points R and A.

■ 5. W s * E F { s )' 6. A D E F S i A R B C

Now, it is necessary to show

7.8. A C ^ E F

✓ 9 , , : . A C ^ W ( s )10. Z CAR £ ZC R A

11. J B & W12. W ^ . ’D E

13. .■■ I S s M ( i )14. / B A R ~ /-B R A15. /B A C & /.B R C [a)16. A A B C S . A R B C

17. A A B C & A D EF

1. Postulate 17

I. A line can be extended as far as

desired. (Postulate)

3. T here exists one and only one line through two points. (Postulate)

4. Sam e as 3

5. Given6 . S.A.S. (Assumption)

that A A B C S * A R B C .7 . Def. of congruent polygons

8. Given9. Transitive property of congruence

10. If two sides of a triangle (A C A R )are congruent, the angles opposite those sides are congruent. (Theorem)

I I . Given12. Statem ent 2 vecopied

13. Same as 914. Same as 1015. Addition postulate of congruence .16. S.A.S. (Postulate)17. Theorem 1)

15.6

j Frequent application of T heorem 12 is made to figures that involve1 circles. Hence, it would be well to establish some of the properties of a circle.

■! D e f in i t io n 36: A circle is a set of points such that line segments draw n from cach of these points to a fixed point are congrucnt..

T he fixed point is called th e center. Although in Figure 5-46 there area great m any line segments th a t are congruent, such as AB, AC, and AD,


F igure 5-46,

a ll a re not congruent, for A E is no t congruent to the others. Hence, this c u rv e w ould no t be a circle. . ..

DEFiNmoN 37: A ^ adiuj o f a circle is a. line segment drawn from any point of th e circle to the center.

As an imm ediate consequence of the definitions of a circle and of the ra d iu s of the circle,'we have the following theorem :

T H E O R E M 13: All ra d ii o f a c irc le a re c o n g ru e n ^ 'i

T h e symbol for the word circle is a small cirqle, O , and to cam e a c irc le , we use the letter at its center. Thus, to nam e the circle in< Figure 5-47, w e would call it O O ; in addition, from the theorem lh a t all radii cf a c irc le are congruent, it would follow th a t OA £= OB.

Figure 5-47.

THE S.S.S. THEOREM157

Illustration:

Given: Q O and Q M intersect a t A and B,

Concl.: OM bisects ZAOB.

Figure 5-48.A n a l y s i s : In order to prove that OM bisects ZA O B , it will be necessary to prove that Z A O M S Z B O M . These angles can be shown to be congruent by proving them to be corresponding angles of congruent triangles. T here are, however, no triangles in the drawing. W hat lines would you suggest draw ing to obtain the triangles needed?

Figure 5-49.

P R O O F I s ta te m e n ts '

1. G O and O M intersect at A and B.

4—>2. L et M A be the line through

points M and A.

3. L et M B be the line through points M and B.

5. ~OA S WB U)

d. O M O M (s)7. A A O M & A B O M8. Z A 0 M 2 Z Z B 0 M

-->9. O M bisects ZA O B .

REASONS

1. Given

2. T here exists one and only one line 1 through two points. (Postulate)

3. Same as 2

4. All radii of a circle are congruent. (Theorem)

5. Sam e as 4

6. Reflexive property of congruence7. S.S.S. (Theorem)8. Def. of congruent polygons

9. Reverse of def. of bisector of an angle

\


EXERCISES

1. Given: A B — 5 U AD = BC

Coucl.: / .A ~ ZC

J 3. G iven: A B ~ ACAD is the median to BC.

Concl.: AD bisects /CBAC.

5. Given: W S ^ X T

W T ^ X S -

Concl.: A R S T is isosceles.

Given: A B AC D B = ~ D C

Cone!.: DA bisects /B D C .

Given: D B 9 Z E C DC = E B

Concl.: /B D C — /C E B

G iven: Point 0 is center

of 0 0 .

OC is m edian to A B.

Concl.: OC bisects /A O B .

V 6 .

^ 7 . Given: Q O .A B ^ C D C oncl.: Z l == Z2

THE S.S.S. THEOREM

t - 9. Given: © 4 and B intersect a t C and D

C oncl.: /CACB S / .A D B

^ 1 1 . Given: G O , A B ‘— CB \

Concl.: BO bisects /A B C . B

Given(^ ©A ahd B intersect 8 .•at.O and D.

Concl.: /C B A S /D B A

159

Given: A B £ AC 10. ^D B ^ D C .

Concl.: / B S Z Z C

Given: AB AC 12.Z l S Z2

Concl.: AD bisects /B A C .


The Hypotenuse-Leg Method of CongruenceYou may have wondered why the (‘‘side, side, angle” )

m ethod of congruence was not introduced. A glance a t tlVe Siagra'm T^e^ low m ight help clarify this. Notice that whereas Figure 5-51 is congruent to F igure 5-50, Figure 5-52 was so drawn that it is not congruent to Figure 5-50.

F ig u re 5-50. F igure 5-51.

And yet, the sam e corresponding parts in the three triangles are congruent. T hus, although triangles are sometimes congruent when two sides and an angle opposite one of them in one triangle are congruent to j ing parts in the other, they are not always congruent. T he “ side, side, angle’ . s ta tem ent can not be a theorem. There is, however, a special situation under which they are congruent with these conditions. T h a t case is when the triangles are right triangles.

Before developing the theorem on the congruence of right triangles, names will be given to the sides of a right triangle.

legF ig u re 5-53.

j l h e sides th a t form the right angle are called the legs-J-or occasionally, th /a w u ^ - w h i le the side opposite the right angle is called7th^ hypotenuse.)

T H E O R E M 14: T w o r ig h t triang les a re co n g ru en t i f tKere exists a correspondence be tw een th e vertices in w hich th e h y po tenuse a n d leg of one a re co n g ru en t to those corre- sp o n d ip g 'p a rts in the o th er. (The symbols for this statem ent a(re H.L.}.

THE HYPOTENUSE-LEG METHOD OF CONGRUENCE161

Given: Z B and Z E are right angles.

I c s K f

Z b s S I Concl.: A A BC 9= A D EF

Figure 5-54.

A n a l y s i s : T h e proof of this theorem follows identically the same pattern

as th a t used, for the proof of the S.S.S. theorem,;

P R O O F (The reason for each statem ent will be left for you to supply.)1 . ii_n r............ ...

1. A t point A of line AC there I exists an angle congruent to | Z F D E . L et this angle be

/.C A R . (a)

2. Extend A R so that

~AP ~ D E. (s)

3. L e t PC be the line through

P and C.

4. L et P B be the line through

P and B.

5. lC-^DF(s)6 . A D E F S A APC Hence, now it is necessary to

show that A A B C — AAPC.

7. Z A P C & Z E8. Z B and Z E are right angles.

/ 9. Z B ^ Z E10. .\ ZA P C & . Z B (a)

11. A B S + D E12. ~AP D E (See step 2.)

13. A B S Z A T ^14. Z A B P ££ ZA P B15. Z C B P < ~ Z C P B

16 . W ^ C P { s )17. A A B C £5 A A P C 1.8. /. A A B C ~ A D E F

162

Illustration:


Given: GO with OC X AB

Concl.: OC bisects Z.AOB.


1. 002. O A ^ O B

3. OC J_ A B4. Z.QCA and L.OCB are right

angles.

5. OC ~ OC6. A O C A ^ A O C B7. Z A O C ^ L B O C

8. OC bisects Z.AOB,

EXERCISES

1. Given: A B J . AD

CD X A D

E Fa e & E f

Concl.: Z A B F S * LD C E

1. Given

2. All radii of a circle are congruent. (Theorem)


5. Reflexive property of congruence6. H.L. (Theorem)7. Def. of congruent polygons

8. Reverse of def. of the bisector of an angle

G iven: BD ~ BC

BD X A D

BC X AC-4

Concl.: A B bisects Z D AC.

THE HVPOTENUSE-LEG M E T H O D O F CONGRUENCE 163

3. Given: 0 0 Given: GO with OD X AB 4.t~* *71 Concl.: D is the m idpointOA X AP __( - ► « - » °f AB.

OB X BP

Concl.: P A S P B

5 . Given: AE and CF 1 BD b F s S E a S & C d

Concl.: Z l S Z2

Given: GOBC = FD

AB and EF X CD

C oncl.: T B = E F

7 . Given: E is the m idpoint of S D .A E = CE

4—* ^AB and CD _L BD

6 *

Given: A ABC is isosceles with AB = AC.AD is altitude to BC.

Concl. : AD is m edian to BC. A

8 .


9. Given: D is the m idpoint of BC.D E == OF

D E X AC

Concl.

D F 1 AB

A B ^ T C

11. Given: CD is the altitude to AB.

M is the altitude to AC.

UB ^ B E C oncl,: A ABC is isosceles.

A.

G iven: DB S FC BC ^ CE *-* <->

DE x BC

Concl.FG X BC A A B C is isosceles.

4 G e t ^ £ j > 6 c o -jU i

Problems In of Triangles

Given: DG X AC

EF X AB

EF = UG

S B ^ E C Concl.: A A B C is isosceles.

10.

12.

0 B

■ Problems Involving the Congruence of More Than One Pair

In the proofs of three theorems—10, 12, and 14—it was necessary to prove two sets of triangles congruent in order to reach the conclusion we sought. In this section you will be given an opportunity to apply m ethods sim ilar to those employed to prove these three theorems.

T h e m ethod of approach for all the problems in this set of exercises is outlined below.

(1) D eterm ine the pa ir of triangles th a t contain the sides or angles th a t you are. trying to prove congruent.(2) A ttem pt to prove this pa ir of triangles congruent.

CONGRUENCE OF M O RE T H A N O N E PAIR OF TRIANGLES 165

(3) In trying to prove this pair of triangles congruent you will eventually discover that a pair of sides or a pair of angles is lacking.(4) Find a second pair of triangles containing the parts that are lacking.(5) Prove this second pair of triangles congruent.(6) T he rest of the proof will follow in the usual manner.

Given: A B = AD C B = *C D

Concl.: EC bisects /.B E D .

Figure 5-57.

A n a l y s i s : E C will be the bisector of / B E D if /B E C == /D E C . Hence, for step 1, the parts th a t we are trying to prove congruent are A BEC and DEC, while the triangles that contain them are ixB E C and A DEC. In attem pting to prove them congruent, following step 2, we find that CB = CD and CE — CE but that information is lacking concerning the included A BCE and DCE. Following the suggestion in step 4, we note that these angles may be corresponding angles of A ABC and ADC. These last two triangles can readily be shown congruent by S.S.S. From this it follows that A BCE and DCE are congruent. This, in turn, will make the A BEC and

DEC congruent, from which the conclusion th a t EC bisects / B E D is ap parent.

PRO O F I STATEMENTS REASONS |

1. C B S = W ( i ) 1. Given

2. CE = ~CE (r) 2. Reflexive property of congruence

At this point we discover that there are parts lacking and proceed to prove another pair of triangles congruent.

3. A B S AD (s)

4. AC S AC {s)5. A A B C ^ A A D C6. / B C E S /D C E (a)7. A B E C = A D EC8. /B E C /D E C

9. EC bisects /B E D .

3. Given

4. Same as 25. S.S.S, (Theorem)6. Def. of congruent polygons7. S .A S . (Postulate)8. Same as 6.

9. Reverse of def. of the bisector of ani angle

1


EXERCISES

1. Given: A B S CB AD ^ CD

Concl.; Z A E D ^ Z C E D

3. Given: A B != CD AD = BC Z l S Z2

Concl.: AF = CEA * --------------------------------- , D

5. Given: A B S AC

D B = DC

Concl.: AD bisects BC.

A

Given: A B = DC 2 .Z B ^ b UC is the m idpoint

of AC.C oncl.: G is the m idpoint

of EF.

Given: A B ^ D C 4 .A D ^ B C

C oncl.: A E ££ EC

Given: AC and BD bisect 6 .each other a t E.

C oncl.: E is the midpoint

of US.

CONGRUENCE OF MORE THAN ONE PAIR OF TRIANGLES 167

7 . Given: A B — DC AD = BC

A E bisects ZB A D .

CF bisects ZBC D .

Concl.: B E ^ F D

A ----------------------------_ o

9 . Given: d)/4 and B intersect a t C and D.*-* __

Concl.: A B bisects CD.

11. Given: 0 0 with A B — AC

Concl.: OA bisects BC.

A

Given: AB ~ AC B D = C E

D F S EF Concl.: Z l S Z2

A

Given: AB = AC E B ^ E C4—+ ____

Concl.: AD bisects BC.

8 .

10.

Given: A B JL BC

DC 1 BC Z l S Z 2 E is the m idpoint

of BC.

B F S iC G Concl.: Z A = Z D

12.


1 3 . * Given: A ^ f iC is isosceles

with A B ^ AC. A A D E is isosceles with AD ^ AE.

Ccncl.: A F B C is isosceles.

A

1 5 . * Given: AB ^ AC

A D ^ A E C oncl.: WF S CF

A

■ Test and Review

14.*

Given: AB « AC *A B ^ . A E

Concl.: AF bisects L B AC.

A

The diagram below has been used to prove the theorem that if two angles of a triangle are congruent, the sides opposite them are con-

I gruent. See if you can prove the theorem . by using this diagram.

Given: Z \ SS Z2

B D ^ C E Concl.: A B ^ A U

A^

A1, (a ) Express the following matchings as a single correspondence:

M ary *-* 1 Bill <-» 2 Fred «-> 3

(b ) L ist the m atchings th a t exist in the correspondence

1 2 3 4« -v l 10 11 100

2. T h e correspondence A BC D <-> R S 1 W is a congruence correspondence

TEST AND REVIEW 169

between the vertices of the polygon ABCD and the polygon R ST W . W hat does this statement imply?

3. (a) W rite a correspondence between the vertices of the two polygons below so that the corresponding parts will be congruent.

s

(b) W hat further information would be needed before it can be said that these two polygons are congruent?

4. W hat conditions would have to exist before it would be possible to have two triangles congruent under two different congruence correspondences?

5. T he theorem on the base angles of an isosceles triangle is sometimes proved by the following method. Justify each of the steps in this proof.

Given: A A B C is isosceles with A B 2 i AC.

Concl.: L B = Z C


A B S i A C ■ 1.2 . AC £ * A B 2.3. Z A & Z A 3.4. A W C S A CAB 4.5. Z B £ ZC 5.

6. If there exists a correspondence between the vertices of two right triangles such that a leg and an acute angle whose vertex is an endpoint of this leg in one right triangle are congruent to those corresponding parts in the sccond right triangle, will the triangles be congruent? Justify your answer.

7. If the definition of congruent polygons was applied to prove two triangles to be congruent, what would have to be shown to be true?

8. W hat conclusion can be drawn if the median and the altitude to a side of a triangle were shown to be the same line segment?

9. If A ABC== A EFG , then Z A = ZE . Does this imply that if A A B Cis not congruent to A E F C , Z A is net congruent to Z E i

10, Using a m ethod sim ilar to the one given in Problem 5, prove ihe theoremth at if two angles of a triangle are congruent, then the sides opposite these angles are congruent.


I

I

Prove each of the following:

1, Given: AC ^ BD

CE _L B E <-> <-+

D F ± A F C oncl.: Z A = Z B

3 . Given: Z \ & Z 3 Z 2 & Z4

C oncl.: CD bisects ZAC B.

G iven: A A B C ^ A EFG

AD bisects ZBAC .

E H bisects ZFEG.

Concl.: E D S z F H

G iven: M is the midpoint of A B in QO.D is the m idpoint of OB.C is the midpoint

of OA.

Concl.: M C & M B

TRY THIS FOR FUN 171

5. Given: A B ^ A C A D S I S

C oncl.: A FBC is isosceles.

Given: AB = ED BC = E C

AC 9 iE C

Concl.: AD = EB

H Try This (or FunD uring the M iddle Ages the theorem on the base angles

of an isosceles triangle was referred to a s ‘‘Pons Asinorum ,’’ or th e “ Bridge of Asses.” H e who passed over could proceed safely and enjoy th e vistas th a t lay exposed before him. H e who could not was lost forever in a mire o f Greek and Euclid, a castoff on the road to “success.”

I t seems likely th a t some M edieval wit fell upon the nam e for this theorem after examining the diagram that was used a t th a t time in th e proof o f this theorem. Notice below that it appears to resemble the cross sectionof the structure of a bridge.

Can you prove that in A ABC with A B = AC that Z A B C = Z A C B ?

Given: Isosceles A A B C with A B S Z A C 3 D & C E

Concl.: ZA B C S i ZA C B

Perpendicularity

AN IM P O R T A N T SEG M EN T O F T H E STU D Y O F geom etry concerns itself with the properties of lines th a t a re perpendicular to each other. Such questions as “ U nder what conditions are lines perpendicular?” and “ W hat conclusions can be drawn if lines are perpendicular?” will be the subject of our work in this chapter. !>v'v '■

T he definition of perpendicular lines informs us that should lines be perpendicular, then right angles will be formed at their point of intersection. Using the reverse of the definition, however, to prove that lines are perpendicular is quite a cumbersome process, for it is not a simple task to show angles to be right angles. Proving angles to be congruent is another m atter. W e . have m any ways of doing this, the most im portant of which is through the congruence of triangles. Hence, we are led inevitably to the need for proving th a t lines can be shown to be perpendicular through the congruence of angles. T he proof, though, depends upon a postulate and a definition th a t have not been established as yet.

T h e postulate was used m any times by you in your study of algebra.If you were asked to determine the value of a2 + 2a when a = 3, you would replace the a by 1, square 3, and add it to the double of 3, giving you a sum of 1 S. Should you be pressed to give cause as to why it is possible to replace a w iili 3, you would very iikely say, “ a and 3 a r t equal, therefore 1 can

172

PERPENDICULARITY 173

remove a and put 3 in its place.” I 11 fact, you may have expressed yourself more precisely by saying that since a and 3 are equal, a substitution of one for the other is permissible. Although you may not have considered th is as a postulate in your study of algebra, we must do so now,P o s t u l a t e 18:' If two numbers are equal, a substitution of one for the

other is permissible. (Substitution, or Replacem ent, Property)T he definition we need is based on one developed earlier in o u r work

concerning the concept of a ray being between two rays. This, as you recall,

was given as,D e f i n i t i o n 38: T he ray PB being between two rays PA and PC m eans that,

m /.A P B + m ZBPC = m ZA P C (See Definition 19)1 T.so '

Figure 6-1.

TV'-' L - will enable us to define a pair of angles known as adjacent angles. And that definition, in turn, will lead us to a relatively sim ple way for proving lines to be perpendicular.D e f i n i t i o n 39: Adjacent angles, ZA B C and ZD BC , are two angles such

th a t they have a common vertex, 8 , and a common side, BC, between

BA and BD.In Figures 6-2 and 6-3 the angles ABC and DBC have a comm on vertex

— r ----r

B and a common side BC that is between BA and BD . Notice th a t in Figure 6-2 the angles ABC and DBC are both acute while in Figure 6-3, one angle is acute while the other is obtuse. Could two angles be adjacent i f they both were right angles? I f both had been obtuse angles, could they have been adjaccnt? Before answering, refer to Figure 3-10 oh page 60. W hat condition m ust necessarily exist concerning the sum of the measures of the two angles in order, that they be adjacent? Similarly, in Figure 6-4, Z A B C an d Z D B C are not adjacent for although they have a comm on

174 PERPENDICULARITY

vertex B, their com m on side, BC, is no t between the o ther two rays, BA

and 3 D . T h e two angles in Figure 6-5 a re no t adjacent for they neither have a common vertex—in one it is B, in the o ther it is D —nor do they have

a comm on side— the side of one is BC while in the other it is DC.R eturn ing to the problem of proving lines perpendicular, we know th a t

F ig u re 6-6.

if A B _L CD, then Z l and Z2 must be right angles and, in turn, we can conclude that they are congruent. W hat we are searching for is the reverse

4—* 4—>of this; that is, if Z l = Z2, we would like to say that A B _L C,D. N otice that the angles 1 and 2 are adjacent angles. Hence, w hat we would like to prove is

T H E O R E M 15: I f two lines in te rsect to form co n g ru en t a d jacen t angles, th e n th e lin es a re p e rp en d icu la r .

|A

B

Figure 6-7.

Given: A B and CD intersect so th a t Z 2 = Z l .

Concl.: A B ± CD

A n a l y s i s : Since we have only one way of proving lines to be perpendicular, through the reverse of the definition of perpendicular lines, we have no recourse b u t to show th a t Z l (or Z 2) is a right angle. This we can do b y proving th a t m Z \ = 90 (or n / 2 = 90).



1. CD is a line. 1. Given

2. ZC B D is a straight angle. 2. F.ev. of def. of a straight an g le

3. Z l is supp. to Z2. 3. Rev. of def. of supp. angles

4. J7i Z l w Z2 = 180 4. Def. of supp. angles

5. Z 2 = Z \ 5. Given

6. m Z l + m Z \ = 180 or 6. Substitution postulate

2m Z l = 180Halves of equals are equal. (Postulate)

7. m Z l = 90 7.

8. Hence, Z l is a right angle. 8. Rev. of def. of a right angle

9. A B ± CD 9. Rev. of def. of perpendicular lines

Illustration:

Given: OO with OC the bisector of ZA O B

Concl.: OC _L AB

Figure 6-8.i—►

A n a l y s i s : Applying Theorem 15 to show that OC ± A 3 simply involves the need to show that ZOCA S ZOCB. This can be done by proving A A O C ^ & B O C .

PRO O F 1 STATEMENTS REASONS

1. OC bisects ZAOB. 1. Why?

2. Z A O C S i ZBO C (a) 2. Why?

3. O is the center of the O- 3. Why?

4. OA ~ OB (j) 4. Why?

5 . O C ^ U C ( s ) 5. Why?

6. A A O C & .L B O C 6. Why?

7. ZOCA S ZOCB 7. Why?<-> <->

8. OC 1 AB 8. If two lines intersect to form congruent adjacent angles, then the lines are perpendicular.


Thus, through the use of Theorem 15 it is possible to show that lines are perpendicular by proving th a t they intersect to form congruent adjacent angles. As it was noted earlier, justifying that angles arc congruent is far easier to do than justifying th a t they are right angles. T h e illustration ju st presented points up the fact that, to a large extent, the proof leading to perpendicular lines is bu t a repetition of the earlier proofs or. congruence of triangles with the added step inferring the perpendicularity of lines based on Theorem 15.

EXERCISES

1, Given: A B z= A CD is the m idpoint of

W .

C oncl.: AD J_ BC

3. Given: C A ^ C BCD is m edian to AB.

Concl.: CD _L A B

G iven: O 0 with C the m idpoint of A B

Concl.: OC _L AB

Given: CA = <52? D A S Z D B

Concl.: CD ± A B

2.

4.


5 . Given: A B ^ A C D B ^ D C

Concl.: A E 1 B C

7 . Given: B E \ bisects Zs A EC and ADC.

Concl.: B E J . AC

9. Given: © 0 with .<4C = £C

Conci.: OC 1. AB

Given: AC] bisects Z s B A D and BCD.

Concl.: AC 1 BD

Given: P A ^ P B0 is the center of the O .

Concl.: OP _L AB

8.

Given: Q O w ith AB = AC 10. Concl.: AD 1 BC

f T h e w ay this picce of G iven D a ta is w ritten is not m athem atica lly sound. H ow should it b e w ritten?


11. Given: © A and B

CDCone!.: AB

Given: D B _L BA 12.D C 1 C A

D B £* DC

Concl.: A D X BC

A

In each of the following problem s you are to draw your own diagram , then write the Given D ata and the Conclusion in term s of th e letters of your diagram. Complete the proof.

1. G iven: An isosccles triangle w ith th e angle bisector of the vertex augle. C oncl.: T he angle bisector of the vertex angle is perpendicular to the

base.

2. Given: A triangle with two congruent angles and a median to the sideth a t these two angles have in common.

C oncl,: This m edian will be perpendicular to the side.3. G iven: A four-sided polygon in which a ll the sides are congruent.

C oncl.: T he lines joining opposite vertices are perpendicular to eachother,

4. Given: An isosceles triangle w ith the bisectors of each of the base angles. C oncl.: T he line draw n from the vertex of th e vertex angle to the po in t

of intersection of the two bisectors will be perpendicular to the base.

5,* Given: An isosceles triangle with the two m edians to the legs.C oncl.: T he line drawn from the vertex of the vertex angle to the point

of intersection of the two medians will be perpendicular to the base.

DISTANCE AND RELATIONSHIP TO PERPENDICULAR LINES 179

B Meaning of Distance and Its Relation to Perpendicular LinesIf you were asked to find the distance from p o in t A to

point B, you would very likely place the edge of your ruler along th e two

A .• B *

points and read off the number ul incites between them. But, why didn’t you measure this distance along either of the paths shown in F igu re 6-9?

F ig u re 6-9.

Before answering this, let us examine another question th a t m igh t help clarify matters. How would you measure the distance from New York to Los Angeles? T h a t is, would you follow a route th a t passed th rough Dallas, Texas; or the South Pole; or perhaps one that went through London, England? “ No,” you say, “not any of these! I ’d try to find the shortest route between New York and Los Angeles.” And this is precisely w hat we m ean by distance-, it is the measure oj the shortest path between the objects involved.

D efin itio n 40: T h e d istance betw een two geom etric figures is th e m easure o f the shortest p a th betw een ’.h em .

T he shortest path between New York and Los Angeles, assum ing we stay ou the earth ’s surface, would be a section of a curve. At this point we are not prepared to say which curve it will be.

M athem aticians frequently use the lerm geodesic when referring to the shortest path. T he question of determining the distance from point P to circle 0 resolves itself to knowing what the geodesic, or shortest path , is from

a po in t to a circle. For it is the measure of the geodesic that vviil be the distance from P to circle 0 .

L et us re tu rn to the original question: “ W hat is the distance from point

180 PERPENDICULARITYA to point B?” The answer, apparently, is “ The m easure of the shortest p a th .” W e 'will assum»' that the path taken along th e edge of your ruler happens to be the shortest path. W hat is the name for the path along the edge of your ruier? W hat is the nam e for that p a r t of the p a th between points A and 5?

? o s t u i . a t e , 1 9 : T h e shortest p a th between two points is th e line segm entjoining the two points.

T hus, since distance means the measure of the shortest path , this postulate implies th a t the distance between two points is the m easure of the line segm ent joining the two points. Therefore, m AB, in addition to being the m easure of the line segment between A and B, can now be considered as the

distance between A and B. To say that AB £= AC implies not only that the line segments AB and AC are congruent but also th a t the distance from A to B is equal to the distance from A to C. Similarly, the statem ent th a t A is th e sam e distance from B as it is from C will be in terpreted as A B ~ AC.

T h e sentence th a t A is the same distance from B as it is from C is lar too long. I t is usually reworded as, A is equidistant from B and C, or, A is equally distant from B and C. E ither of these is interpreted as before, A B = AC.

.Pe

« Qfigure 6-12. figure 6-13.

In both Figure 6-12 and Figure 6-13 how do you in terpret the statem ent th a t P is equidistant from A and £? T h at Q is equidistant from A and B ? D raw figures similar to those above. Draw the line through the points P

an d Q and extend it, if necessary, to intersect AB. In w hat m anner does PQ


appear to be related to AB? In statem ent form, w hat theorem seems to be implied by the conclusions you have ju st noted?

T H E O R E M IS: I f two points a re each e q u id is tan t from th e en d p o in ts o f a lin e segm ent, th en the l in e jo in in g them w ill be th e p e rp en d icu la r b isecto r o f th e l in e segm ent.

_ AG iven: PA = PB

Q A 9= Q B4—►

Concl.: PQ bisects .■iB .

PQ X AB

Figure 6-14,

A n a l y s i s : Notice that rather than deal with the expression th a t P is equi

d istant from A and B, we have interpreted this as PA ~ ,F B . S im ilarly, Q being equidistant from A and B was restated as QA = QB. T he congruencies are m uch preferred, for these are the forms with which we have d ea lt in the past. T he problem thus becomes one ol showing that A R = B R and Z Q R B = ZQ RA, both of which can be done by proving A QRB ££ A Q R A T ry to do this before reading the proof below.

PR O O F ! (The reason for each statem ent will be left for you to supply.)

1. 7. A Q RB A Q R A

2. QR £=; QR (s) 8. A R = BR

3. P A 9 Z P B 9. R is the midpoint of AB.

4. P Q & P Q 10. PQ bisects AB.

5. .’. AQPA S A QPB W . Z Q R A ^ Z Q R B

6. Z P Q B ZPQA (a) 12. PQ X A B

Illustration: --------- --------'

G iven: CA — CB0 is the center of the O . i—► ____

Concl.: OC is the 1 bisector of AB,

CFigure 6-15.


A n a l y s i s : Applying Theorem 16 to this problem makes it necessary to find two points somewhere on f the line OC such th a t each of .these points is equidistant from points A and B. A pparently, the points to select are those about which we m ight know something. T o illustrate, one of the points to investigate would be 0 , since it is the center of the circle. Similarly, another point that bears investigation is C, for the Given Data: contains inform ation about C .^ n d , lastly, it often pays to examine the point of intersection of the two lines. “ ''

In this problem , 0 is equidistant from A and B by the theorem on the radii of a circle, while C is equidistant from A and B from the inform ation in the Given D ata that CA ~ CB. Therefore, the conclusion follows.

Figure 6-16.

1. W = CB (This statem ent implies that C is equidistant from points A and B.)

2. L et OA and OB be the lines through points 0 and A and points 0 and B. .

3. OA £= OB (This statem ent implies that 0 is equidistant from points A and B.)

4. OC is the _L bisector of AB.

1.

2 .

Given

O ne and only one line exists th rough two points.

3. All radii of a circle are congruent.

4. If two points (0 and C) are each equidistant from the endpoints (A

and B) of a line segment (AB), then the line joining thpse two points (0 and C) is the perpendicular bisector of the line segment.


t W henever we speak of a po in t as being “o n a line,” i t w ill im ply th a t th a t p o in t is an elem en t of the set of poin ts of w hich th a t line consists.

Theorem 16 enables us to prove a line to be the perpendicular bisector of a line segment. Now we would like to investigate those conclusions that can be drawn if a line is known to be the perpendicular bisector o f a line

i—V —

segment. To illustrate, let us say that PQ is the perpendicular b isector of AB,


Figure 6-17.

4—>while R is any point of PQ selected at random. W hat represents the distance from R to A? From R to B? W hat appears to be true about these line segments RA and RB> By expressing yourself in statem ent form, w hat theorem do you believe will be true in terms of the conclusion you have ju s t drawn?

T H E O R E M 17: I f a p o in t is on th e p e rp en d icu la r b isector o f a lin e segm ent, th tn it is equ id istan t (rom th e e n d p o in ts of the lin e segm ent.

Given: QP J . AB_

QP bisects AB. C oncl.: RA = RB

Figure 6-18.

A n a l y s i s : You will notice that, again, we have chosen to write HA = RB ra ther than R is equidistant from A and B. By doing this, it is imm ediately evident that to prove these line segments congruent, we need merely show th a t the two triangles are congruent. The proof will be left for you to complete.

There are a few interesting features about this proof that should be called to your attention. W hat did R represent on line PQ? Hence, considering what has been proved for R, what can be said about all o ther points

on PQ? M ight R be the point S ? If so, what would happen to triangles RAS

184 PERPENDICULARITYand RBS7 Since there are no triangles if R coincides w ith S, does this imply th a t S can not be shown to be equidistant from A and B?

T h e reverse of Theorem 17 is also a true statem ent. Its proof, however, involves a difficulty similar to the one encountered when proving that two angles of a triangle will be congruent if two sides are congruent. T here was a need a t that time for the assumption that every angle has a biscctor. Now, we are faced with the need for a comparable assumption w ith reference to a line segment.

P o s t u l a t e 2 0 : E v e r y l i n e s e g m e n t h a s a m i d p o i n t .

' W ith this postulate, we can prove that

THEOREM 18: If a point is equidistant from the endpoints o f a lin e segm ent, then it lies on the perpendicular bisector of the lin e segment.

G iven: PA ~ PBConcl.: P lies on the _L bisector of A B .

Figure 6-19.

f

- I

A n a ly sis : Perhaps the fastest way to show that a iine is the perpendicular biscctor of 2 line segment is to apply Theorem 1 6 .'From the Given D ata, P is already equidistant from A and B. Hence we need but find another point th a t is also equidistant from points A and B. Several suggestions w ere m ade in the analysis of the proof of Theorem 16 that will help us find this second

point. By selecting M , the m idpoint of AB, as the second point, the line th rough P and M will be the perpendicular bisector of A B. This condition, in tu rn , implies that P lies on the perpendicular bisector of AB ; and this, of course, is w hat we had set out to prove.



1. PA ^ PB2. Let M be the midpoint of

\ . Given2. Every line segment has a m idpoint.

AB.

3. M A ^ M B 3. Definition of the m idpoint of a line segment

*—¥ —-a—4. PM is the 1 bisector of AB. 4. If two points (P and M ) a re each

equidistant from the endpoint3 of a

line segment (AB), then the lin e join-

ing them (PM ) is the perpendicu lar bisector of the line segment.

Illustration:

Given: © 0 with CA ■= CBD is the m idpoint of A B .

Concl.: CD passes through 0.

Figure 6-21.

A n a ly sis : We have only one method. Theorem 18, for proving th a t a point falls on a line, and that method is dependent, upon two features:

(.1) If the point is equidistant from the endpoints of a line segment.(2) If the line it is to fall on happens to be the perpendicular bisector of that line segment.

Hence, our present problem becomes, a m atte r of showing th a t 0 is equi-

distant from points A and B and, secondly, that CD is the perpendicular

bisector of AB. I f OA and OB are drawn in the diagram , then from the theorem that the rad ii of a circle are congruent it would follow that OA ~ UB. This fulfills the first requirement that 0 m ust be equidistant from A and B. D being the m idpoint of ~AB implies that IDA S 255. In addition, knowing that Ua — US will give us two points, D and C, th a t are equidistant from

H -the points A and B. Hence, CD must be the perpendicular bisector of AB. This inform ation fulfills the second requirement.


Figure 6-22.

P R O O F | s ta te m e n ts REASONS

i . S J s c I 1. Given2, D is the m idpoint of AB. 2. Given

3. R A S D B 3. Def. of the m idpoint of a line segment_ _

4. CD is the X bisector of AB. 4. If two points are equidistant from theendpoints of a line segment, then theline joining them is the perpendicularbisector of the line segment.

i—►5. Let OA be the line that 5. There exists one and only one line

passes through 0 and A. through two points.

Similarly for OB.

6. O A ^ T T B 6. All the radii of a circle are congruent.

7. CD passes through 0 . 7. If a point is equidistant from the.endpoints of a iine segment, then it lieson the perpendicular bisector of theline segment.

In the d iagram of the preceding problem an arrowhead was p laced on4—CD to indicate th a t we are not certain that the line will pass through 0 but

will try to prove this. H ad 0 been placed on CD, it would imply th a t the point fell on the line w ithout need for proof. B ut this was exactly w hat we were try ing to show ! Two other ways of signifying that the position of 0 is in doub t are

Figure 6-23. Figure 6-24.


In Figure 6-23 a gap was left in the line CD, while in Figure 6-24 CD was deliberately made to appear as if it bypassed point 0 .

EXERCISES

A

A num ber of these problems are identically the same as those on pages 176-178. These problems, however, are to be proved by using Theorems 16, 17, and 18 only, not Theorem 15.

*|, Given: 0 0 with M the midpoint of AB

C oncl.: O M JL AB

3 ', Given: Z l = L I Z 3 S Z 4

Concl.: D B is the X bisector of AC.

Given: Isosceles A A BC and D BC on the same

base BU

Concl.: AD is the 1 bisector

of BC. (H int: Re-

2 .

Given: OO with AB = AC s 4 .

Concl.: AD is the Xbisector of BC.

A

188 PERPENDICULARITY5 . G iven: © A and B

Concl.: AB is the 1

bisector of CD.

7. Given: GO with CD J_ bisector of A B

C oncl.: CD passes through 0 .

G iven: D E is the _L bisector of BC, A A B C is isosceles

on base BC,

Concl,H

A lies on DE. (H int: Rewrite second piece of Given Data.)

G iven:

Concl,

0 0 with /.ABC, ~ /A C B

OA is the J_

bisector of BC.A

G iven: O 0 with A B S AC

AD is the m edian

to BC.

Goncl.: 0 Jies on AD,A

Given: A ABC and FBC are isosceles on base BC.

A E is the m edian

to BC.

Concl.: F lies on AE.

10.

Given: © C and D 12.CE bisects /A C B .


11. G iven: A A B C is isosceles on base BC.E is the midpoint

rJB C .Z3 S Z4

Concl.: D E passes through A.

13. Given: © C and D

CE X .A B

Concl.: D lies on CE.

C oncl.: CE passes through D.

Given: A B = AC

BD bisects /A B C .

CD bisects /A C B .<—

Concl.: A E is the X bisector

of BC.

14.


15. Given: A B =~AC

B E is the m edian

to AC.UE is the m edian

to AB.

C oncl.: AC is the _L

bisector of "BC.

17.* Given: DA = D B = DC4—>

i 'F i s the _L bisector of AB.

GH is the _L bisector

of AC.

Concl.: E F and GH intersect a t D.

Given: A A B C is isosceles 16.* with A B = AC.

BD bisects LABC .

CD bisects ZA C B

A E is m edian to BC.

Concl.: AE passes through D.

G iven: FD is the X bisector 1 0 .of 75.

F E is the X bisector of IC .

VPQ is the X bisector

of AB.<—i

C oncl.: PQ passes through F. (H int: Prove

FA = FC and

Fb SS FC.)

BIn each of the following problems you are to draw your own

diagram , th en w rite the Given D ata and the Conclusion in term s of the letters of your diagram . Com plete the proof.

CONDITIONAL AND CATEGORICAL STATEMENTS 191

An isosceles triangle with a median to the base.The median to the base is the perpendicular bisector of the base. Two isosceles triangles with the vertices of their vertex angles lying on the same side of their common base,The line joining the two vertices of their vertex angles is the perpendicular bisector of their common base.Two isosceles triangles on the same base and the bisector of the vei lex angle of one of them.The line of this bisector passes through the vertex of the vertex angle of the other.Two intersecting circles, a line segment joining their points of intersection, and a line drawn from the center of one to the m idpoint of this line segment.This line passes through the center of the o ther circle.The altitudes to two sides of an equilateral triangle.T he line joining the third vertex to the point of intersection of the two altitudes is the perpendicular bisector of the th ird side.

| Conditional and Categorical StatementsM athematical statements, or propositions, fall into one of

two forms. They are either

(1) Conditional Statements, or, as they are frequently called, Hypothetical Statements

or(2) Categorical Statements

The first of these, the conditional statement, we investigated a t great length earlier in this course, f Now, however, we would like not only to review th a t discussion but also to take a fresh look at this statem ent from, another point of view.

(1) T he conditional brobosilion is a statement containing two clauses, one beginninti with the word “ if.” while the other begins with the word “ then .” We are asked to accept the tru th of the antecedent, which, we leairned, is the information contained in the clause beginning with the word “ if.” T hen with the aid of our prior knowledge— that is, our definitions, our postulates, and our theorems—we are required to show in some m anner why the consequent must follow. And, as you recall, the consequent is that p a rt'o f the conditional statem ent that follows the word “ then.”

W e can express this concept somewhat differently: the Given D ata is contained in the antecedent, or “ if-clause,” while the Conclusion we are asked to justify is contained in the consequent, or “ then-ciause.” Hence,

t See pages 85 to 88.

1 . G iven :

C on cl.:

2. G iven :

C o n c l.:

^ 3 . G iv e n :

C o n c l.:

4. G iv e n :

Concl.: 5.* Given:

Concl.:

192p e r p e n d ic u l a r it y

when we are presented with a conditional statement whose consequent we are asked to. verify, our m ethod of attack will be as follows:

(a ) Draw the diagram containing ali the necessary parts.(b ) Express the Given D ata and the Conclusion in terms of the letters ,th a t appear in the diagram.(c) And, finally, prove th a t the Conclusion follows from the Given B a t a.

T o illustrate, consider the proposition

“ If a ray is the bisector of the vertex angle of an isosceles triangle, then it will bisect the base.”

We know that the diagram should contain an isosceles triangle and thebisector of the vertex angle. This we now draw.

Given: A A B C is isosceles with A B & .A C .

AD bisects /B A C .

c . C oncl.: AD bisects BC-

H aving labeled the diagram, wc proceed not only to indicate th a t the triangle is isosceles but also to point out which sides are the congruent legs. In addition, the “ if-clause” contained information that a ray was the bisector

o f th e vertex angle; this relationship we have shown by stating th a t AD bisects /B A C . I t is not necessary to include in the Given D ata th a t /B A C

is th e vertex angle, for the fact th a t A B is given as congruent to /1C will im ply this. Referring now to the “ then-clause,” we note th a t the ray th a t is th e bisector of the vertex angle will have to be proved to be the bisector

o f the base. Since AD was given as the bisector of the vertex angle, A D will h av e to be shown to be the bisector of the base.

T here are times w hen-the position of the “ if-clause” and the “ then- c lause” are interchanged in a proposition. T h at is, the “ if-clause” m ay a p p e a r as the second clause, while the ‘‘then-clause” appears as the first clause. T h e proposition analyzed in the preceding paragraph could have b e en ■written as:

“ A ray will bisect the base if it bisects the vertex angle of an isosceles triangle.”

T h is statem ent was made to appear even more obscure by the omission of t h e word “ then.” I t would still, however, be recognized as a conditional proposition by the fact that it contains two clauses, one of which begins with th e w ord “ if.” Should you prefer, it is always possible to interchange the tw o clauses to have them appear in type form: “ if-clause” first .followed, by th e “ then-clause.” O r as it is frequently called, an “ if-then” proposition.

Figure 6-25.

CONDITIONAL AND CATEGORICAL STATEMENTS 193

(2) A categorical proposition is a statem ent whose verb is some form of the verb “ to be.” T hat is, the verb may be any of the words is, u-as, are, were, will be, etc. To illustrate, the conditional proposition ju st analyzed m ight have been written in the categorical form :

“The bisector of the vertex angle of an isosceles triangle is the bisector of the base.”

W hat form of the verb “ to be” was used in this statement? In the conditional form what represented the “ Given D a ta” ? Relative to the position of the verb, where does the “ Given D ata” appear in the categorical form?In the conditional form what represented the “ Conclusion” ? Relative to the position of the verb, where does this same “ Conclusion” appear in the categorical form? Thus, it can be said that in a categorical statem ent

(a) T hat part of the sentence that is prior to the verb is the “ Given D ata. I(b ) T hat part of the sentence that follows the verb is the “ Conclusion.”l ‘

There are occasions when categorical statem ents are. disguised through the use of verbs other than the verb “ to be.” Thus, the statem ent th a t

“ The bisector of the vertex angle of an isosceles triangle bisects the base”

is the same as the one given earlier. Now, however, the verb is bisects ra ther • than is. Be this as it may, the. method for determ ining the. “ Given D a ta” and the “ Conclusion” remain the same. W ith a little effort every statem ent of this form can be rewritten to contain the verb “ to be.”

EXERCISES

A

For each of the following propositions draw the diagram , write the “ Given D ata” and the “ Conclusion,” but do not w rite the “ Proof.” You are not prepared to prove most of these propositions.

1. If a line segment joins the midpoints of two sides of a triangle, then its measure is equal to one-half the m easure o f the third side.

2. Two isosceles triangles are congruent if a leg and a base angle of one are congruent to those corresponding parts of the other.-

3. If the bisectors of two angles of a triangle are congruent, the triangle is

isosceles.4. I f the m edians to two sides of a triangle are congruent, then the. triangle

is isosceles.5. The line segments joining the midpoints ol the sides of an equilateral

triangle form another equilateral triangle.

194 PERPENDICULARITY6. T he opposite sides of a four-sided polygon are congruent if the opposite

angles are congruent.

7. I f perpendiculars arc draw n from any point of the bisector of an angie to the sides of the angle, then these perpendiculars are congruent.

8. If perpendiculars are drawn from the m idpoints of the legs of an isosceles triangle to the base, then these perpendiculars are congruent.

9. The line, segments drawn from the m idpoint of the base of an isosceles triangle perpendicular to the legs are congruent.

10, T he acute angles of a right triangle are complementary.11. A triangle is equilateral if its altitudes are congruent.

[Bj]>>■ ' 1----- 1

For each of the following propositions, draw the diagramand w rite the “ Given D ata ,” the ' “Conclusion,” and the “ Proof.”

1. If a line bisects the vertex angle of an isosceles triangle, then it will be perpendicular to the base.

2. T he m edian to the base of an isosceles triangle bisects the vertex angle.3. T he line segments joining the vertex of the vertex angle of an isosceles

triangle to the trisection points of the base are congruent.4. If the opposite sides of a four-sided polygon are congruent, then the

line join ing a pair of opposite vertices divides the polygon into two congruent triangles.

5. Line segments drawn from the m idpoint of the base of an isosceles • triangle to the midpoints of the legs are congruent.

6. If perpendicular segments drawn from the-m idpoint of one side of a triangle to the other two are congruent, then the triangle is isosceles.

7. If two altitudes of a triangle are congruent, then the triangle is isosceles.8 . If two triangles are congruent^ the angle bisectors of a pair of cor

responding angles are congruent.9. If two triangles are congruent, the medians to a pair of corresponding

sides are congruent.10. I f the line join ing a pair of opposite vertices of a four-sided polygon

bisects these angles, then the remaining two angles are congruent.11. T h e m edian to the base of an isosceles triangle is perpendicular to the

base. '12. If there exists a correspondence between the vertices of two isosceles

triangles in which a leg and a vertex angle of one are congruent to those corresponding parts in the second, then the two triangles are congruent.

13. T h e angle bisectors of the base angles of an isosceles triangle form another isosceles triangle with the baise.

CONDITIONAL AND CATEGORICAL STATEMENTS

14. The medians to the legs of an isosceles triangle are congruent,15. A line perpendicular to the bisector of an angle forms with the sides

of the angle an isosceles triangle.16. If the perpendicular bisector of a side of a triangle passes through the

opposite vertex, then the triangle is isosceles.17. If the opposite sides of a four-sided polygon are congruent, then the

opposite angles are congruent.18. If the line segments joining the two pairs of opposite vertices of a four

sided polygon bisect each other, then the opposite sides of the polygon

are congruent.19. If the line segment joining a pair of opposite vertices of a four-sided

polygon bisects these angles, then this line is the perpendicular bisector of the line segment joining the other two vertices.

20. If two line segments are drawn as the perpendicular bisectors of the legs of an isosceles triangle and they term inate in the base, then thfese line segments are congruent.

21. If a radius is drawn bisecting the line segment th a t joins two points of a circle, then the radius is perpendicular to the line segment.

22. The perpendicular bisector of a line segment whose endpoints lie on a circle passes through the center of the circlc.

23. If two circles intersect, then the line joining their centers is the perpendicular bisector of the line segment joining their points of intersection.

24. If the median to the side of a triangle is also the .a ltitude to the side,, then the triangle is isosceles.

25. If two isosceles triangles have the same base, then the line joining the vertices of their vertex angles is the perpendicular bisector of the base.

26. T he perpendicular bisector of the base of an isosceles triangle passes through the point of intersection of the bisectors of the base angles.

27. If a point on the base of an isosceles triangle is equidistant from the midpoints of the legs, then that point is the m idpoint of the base.

28. T he perpendicular bisector of the base of an isosceles triangle passes through the point of intersection of the medians to the legs of the tr i

angle. ,29. If all four sides of a four-sided polygon are congruent, then the two

line segments joining pairs of opposite vertices are, the perpendicular bisectors of each other.

30.* If a point is equidistant from the vertices of the base angles nf an isosceles triangle, then it will lie on the bisectoi of the vertex angle of the triangle. (H int: Show th.it this line is the perpendicular bisector

of the base.)

196PERPENDICULARITY

31.* T he bisector of the vertex angle of an isosceles triangle passes through the point of intersection of the bisectors of the base angles.

32.* T he bisector of the vertex angle of an isosccles triangle passes through the point of intersection of the m edians to the legs.

O Test and Review

tIi


1 . Given: Circle 0 with OC the bisector of ZA O B

Concl.: OC J. A B

3 . Given: AB = = £ C S C Z ) SS e ^ a e

Z D & . Z E F h the midpoint of KE.

Concl.: C F is the 1 bisector of

'4 (H int: Prove Fequidistant from B and D.)

G iven: Circle 0 withZC A B £= ZC BA

Concl.: OC 1 AB

G iven: Z C S i Z B

C E S I B B

M E ± AC

M D X AB

C oncl.: A M _L BC

TEST AND REVIEW 197

5 . Given: AD bisects Z.BAC. AB = AC Z \ S Z2

C oncl.: E lies on AD.

Given: Circle 0 withZC AD S ZC BD Z l S Z2

Concl.: CD passes through 0 ,

BProve each of the following statem ents:

The altitude to the base of an isosceles triangle is the m edian to the base. T he angle bisectors of the base angles of an isosceles triangle are congruent. ■If two triangles are congruent, then the line segment joining the m idpoints of two sides of one triangle is congruent to the line segment jo in ing the midpoints of the pair of corresponding sides in the other triangle.

4. If the bisector of an angle whose vertex lies on a circle passes through the center of the circle, then it will be the perpendicular bisector of the line segment joining the points of intersection of the sides of the angle and the circle.

Perpendicularity in Space Geometry

J U N T IL N O W WE HAVE ONLY C O N SID ER ED ' those figures that can be drawn on a "fiat surface” but have m ade no

a ttem pt to either nam e or define Jhis^surface. We have reached the point w here w e w ould like to ex len S w m e of M r( concepts to the geometry of “ space” ra th e r th a n keeping them c’on^neQ to a “ flat surfM ^’’ TWs par-*" ticuiar time has been selected since there is a great deal of s¥ra|j^n|£petw een the ideas about perpendicularity that we have just estaBflshea and those th a t we can now establish in “space” geometry.

Space and Surface

■ ■ ■ ■ ■ ■ ■ ■ B ut how can v/e define these terms, space and surface? As w ith several of th e term s we encountered earlier, it is not possible to <^e$ne these words. A lthough they can be classified, we are 'im able to. d istingu ish^ them from the o ther members of their class.

Both space and surface can be classified as a set of points’,

In fact, surface is a subset of those points that are members of the set of points called space. C an you illustrate this by pointing to objects in your classroom? Unfortunately, it is not possible to show how the points in each of these

I

sets differs from other sets of points. Hence, space and surface m ust be considered the first of our undefined terms in space geometry.

The word plane, though, presents a different picture; this we can define. Surfaces such as the blackboard at school, the desk on which you write, . floor on which you stand, the ceiling overhead'—these are all callfedplanes. All of the surfaces selected have certain common properties. Select any two points on the blackboard and draw the line th a t exists through them . Are all the points of the line also points of the pland? Select any other two points and draw the line through them. Are there any points of this line th a t are no t points of the plane? Can you find any two points on the blackboard through which a line can be drawn so that there are some points of the linethat do no t lie on the blackboard?

Now let us tak i a look a t this same situation when the surface is either a ball or a tin can having no top or bottom . I f the line is draw n th a t exists through the points A and B, how m any points of this line would also be

PERPENDICULARITY IN SPACE GEOMETRY 199

Figure 7-1. . •

points of the ball? Are there any points of the line that are not points of the ball? W here are these points? If 3 line were draw n through points C and D on the tin can, how many of the points of this line would also be points of the tin surface? Are there any points of the line th a t are not points of the tin surface? W here are these points? Can you find two points on the ball such that all the points of the line through them are also points of the ball? Can you find two points of the tin can such that all points of the line through them are also points of the tin surface? ' r, > ^ . '

These illustrations should have helped to m ake clear th e properties of aplane as given by its definition.

■ x u W «\ft D efin itio n 41 •/ A plane is a surface such th a t if a line is draw n through

any two points of this surface, then the points of the line Will also be points of the surface.

In the case of the ball the only points of the line that were also points of the ball were A and B. If, in the case of ths tin can, the point C was “ directly above” the point D on the tin surface, then all points of the line CD would also be points of the tin surface. This, however, was tru e only if one point was “ above the other” on this surface. T h e definition of the plane

insists that the points of the line be points of the plane no m atter where the two points be located on the surface.

Naming a Plane

■ ■ ■ ■ H H T o give the appearance of a table top, a plane is often d raw n as shown in Figure 7-2. You arc to picture this plane as lying horizontally

200 PERPENDICULARITY IN SPACE GEOMETRY

before you, not vertically. Although there are numerous ways of nam ing a plane, the two. most comm on are using either^orce small letter ox two capital letters placed at opposite vertices. This plane can be called either plane A B or p lane m. ' ~

I The Meaning of Determine

Before examining any of the other properties of a plane, we will have to backtrack to take a close look at the m athem atical in terpretation of the word determine. Early in the study of geometry we assumed that

“T here exists one and only one line through two points.”

R ather than use the words "one and only one,” it is felt that equally as expressive as this phrase is the use of the word determine. Thus, whenever the term determine is used in the m athematical sense, it implies that “ one and_gply..Qne’’ of these creatures exists under the conditions described. Based on this, the assumption just stated m ight have been worded as

“Two points determ ine a line.”

How would you interpret the statem ent “ A fixed point as center and a given line segment as a radius determ ine a circle” ?

W e would like, now, to examine those conditions under which a plane is determined. Boys will recall from their shop work th a t they constantly had trouble m atching the four legs of a table to keep it from wobbling. Yet, .tables with three legs seemed always to be firmly planted on the ground no m atter how careless we were in cutting the lengths of the legs. T he principle behind this is the fact that three points will always lie in the same plane, Should we have four points, however, any combination of three of them w ill. lie in one plane, bu t it’s only through chance that the fourth will lie in the same plane with the remaining three. Points that do lie in the same plane are called coplanar points; similarly, points that lie in the same line are collincar.

D e f in i t io n 42; Coplanar points are points that lie in the sam e plane. D e fin it io n 43'. Collinear points are points that lie in the sam e line.

The surveyor uses the principle stated above when he places his transit on a tripod, for he knows th a t the three legs of the tripod m ust rest firmly on the ground. T he photographer uses a tripod, also, each tim e he takes a

“ tim e” exposure. This principle is stated as:P o s t u l a t e 21: Three noncollinear points determ ine a plane.

W hy was it necessary to say th a t the three points had to be noncollinear? I f the three points were on the same line, then there would be m any planes th a t would contain them, as pictured in Figure 7-3. T hese planes would

THE MEANING OF DETERMINE 201

Figure 7-3.

resemble the pages of a book where the line of which the points A, B , and

C are members is the binding of th e book.

EXERCISES

1 ,-T h e diagram) above resembles an Egyptian pyram id and in fact is called a Pyramid. T tie questions below are to be answered in terms of this figure,

(a ) How. m any planes are there in this figure?


(b ) How m any planes contain the line AB? Using three letters, two on <—>A B and the third a t some other point in the plane, name the planes

that contain AB.(c) Name two sets of three coilinear points.(d ) Name two seb cf five coplanar points.(e) Name three noncollinear points.(f) Nam e four noncoplanar points.(g) How m any planes contain all three points, A, C, D ? Is this plane in

the diagram? How m any planes contain the points E, F, G? Is this plane in the diagram?

2. T h e diagram below is to be .used for the questions th a t follow it.

(a) W hy were some of the lines drawn as dotted lines in this diagram ?(b ) How many planes are there in this diagram? Using either two or

three letters, where necessary, name these planes. ,(c) How many of the planes contain four coplanar points that are nam ed

in the diagram?(d ) Name a plane that contains the points G and D. Will all the points

of the line GD lie in this plane? <x(e) Name a second plane that contains the points G and D. Will all the

points of the line GD lie in this second plane? W hat conclusion can you draw concerning the points of line GD with reference to these two planes? D a you think there are any points other, than those of

w " - ' t f o ;line GD that are common to these two planes? GD is called the in te rsection of planes H D and BD.

( f) • Nam e the intersection of planes H B and CA. O f planes H A F and FEH.(g ) Nam e the intersection of planes AG, GC, and AC. Pj>

3. (a ) How m any lines exist through one-fixed point?(b) How m any planes exist through a fixed point?(c) How m any lines exist through two fixed points?(d ) How m any planes exist through two fixed points?

CO N DITIO N S UNDER W H IC H A PLA N E IS DETERMINED 203

(e) How many lines exist through three fixed points?(f) How many planes exist through three fixed points?

4. (a) Given three noncollinear points, how m any lines can be draw n suchthat each line will contain a t least two of the points? Can any of these lines contain more than two of the points?

(b ) Given three noncollinear points, how m any planes can be draw n such that each plane will contain all three points?

5. Given four noncoplanar points, how many planes can be draw n such that each plane will contain a t least three points?

6. Given points A, B, and C. U nder w hat conditions will these points determ ine a plane?

7. In terms of your answer to Problem 6, are the sides and the vertices of a triangle always coplanar?

8. In terms of your answer to Problem 6, will it always be possible to balance a book that has been placed on the points of three thum b tacks? Assume that the thum b tacks are lying on a table pointing upward.

H Further Conditions Under Which a Plane Is DeterminedConsider a bicycle positioned so th a t it is resting on its

handlebar and seat. Can you explain why it should be balanced in this position? Now spin the front wheel about its axle and fix your a ttention on a single spoke as it revolves. Should we want to stop the wheel in any position, we would place a small stick between the spokes to force the wheel to come

to rest a t th a t spot. In some ways this can be considered an application of th e assumption that there exists one and only one line through two points. In this situation, the spoke passed through the point A (the axle), and as it revolved it took on many, many, different positions. However,’as soon as we insisted, .that the spoke pass also through point F, then it was stopped dead in its\trac'A T here could be bu t one such iine, or spoke, passing through A and

T he revolving door of a departm ent store can be exam ined in very

m uch the same way. T he door can take on many, m any different positions as it revolves about its inner edge. But let someone place his foot in the path


___ I...

1

p.1111

11

J

r\ ]

Figure 7-5,

of the door and its m ovement will be stopped immediately. So, too, there are infinitely m any positions that a plane can take as it revolves about line A B . Insist, however, th a t this plane also contain the point P , and there is only one position for which this will be true. This is w hat we shall now prove.

THEOREM 19: \ ijne an(j a point not on that lin e determ ine a plane.

Given: Point P and line /C oncl: Point P and line I determine a

plane.

Figure 7-6.

A n a l y s is : T hus far there is only one m ethod available to us for proving th a t a plane is determ ined: three noncollinear points determine a plane. Hence, there is no alternative but to attack this proof by using this postulate. Select points A and B as any two points on line I . f Since P, A, and B a re three noncollinear points, they determ ine a plane. Since from the definition of a plane two points of /, A and B, lie on this plane, then all points of I lie on this plane. Hence, the conclusion follows.

In the same way, it is possible for us to prove a second theorem for determ ining a plane,

t A single le tte r is often used to nam e a line.

METHODS OF DETERMINING A PLANE 205

TH EO R EM 20: T w o in tersecting lines d e te rm in e a p lane .

Given: Lines / and m intersect at P. C oncl.: Lines / and m determ ine a plane.

F ig u re 7-7.

A n a l y s is : The analysis is much the same as in the preceding theorem. Now, however, we must be careful about the selection of one of the three points that determine the plane. Let one of these points be P, the point of intersection of / and m. T he other two points, A and B, m ust lie on each of the respective lines. W hy should all points of line I be on the plane determ ined by B, P, and A? Why should all points of line m lie on this plane?

Jgj Methods of Determining a Plane

T hree noncollinear points

A line and a point not on the line

Figure 7-8.

Two intersecting lines

EXERCISES

1. If three lines pass through a common point; will the three lines lie in the same plane? V c i ulw#t\S'

2. If three lines that are not coplanar intersect at a common point, howmany planes are determined that contain at least two of these lines? ^ ^

3. If A B and AC are two lines, what can be said about the points of lineSC?

4. If P is a point th a t is not on line / and Q is a point that is on line /, then what can be said about line PQ?

5. Points R, S, and T are in plane m. Points R, S , and T are also in plane p. Under what conditions will plane m be different from piane p?

6. Lines / and m intersect. P is a point on neither I nor m.(a) How m any plants will contain at least two of these elements?

206 PERPENDICULARITY IN SPACE GEOMETRY(b ) If P fell on I, how m any planes would contain a t least two of the

elements?

(c) I f P was the intersection of I and m, how m any pianes would contain a t least two of the elements?

7. If points A, B, C, D ail lie in the same plane, prove that the lines AB, AC, and AD are coplanar.

8 . Points A, B , and C are the points of intersection of the three lines. Prove that the three lines lie in the same plane.

9. A B and CD intersect at E. Prove that all p o in t!o f ACBD lie in the same plane.

r—f ^—,

10. If D lies on A B and E lies on AC, prove th a t the five lines of this d iagram are coplanar.

A

11. T he drawing below is a “ space” polygon. A, B , C, and D are non- .

PERPENDICULARITY BETWEEN A LINE AND A PLANE 207

<—> «—►coplanar points. In any m anner whatsoever, justify that AC and BD do not intersect.

■ Perpendicularity Between a Line and a Plane 'You may at some time have observed a m an trying to drive

a stake “ vertically” into the ground, perhaps in the process of setting up a post for his mailbox. After having driven the post several feet into the ground, he will stand off a few yards to observe if it appears “ stra igh t.” Having satisfied himself that all is well from that position, he then moves to the right and again examines the position.of the post relative to the ground. Although he may again be pleased with w hat he observes, he may 'still have doutas as to whether the post is actually “ vertical.” Hence, he repeats the process several times more by moving to the right aijDund the post. By doing this, he is applying the definition of a line perpendicular to a plane.

D efinition 44: A line p e rp e n d ic u la r 'to a p lan e is a line th a t is p e rp en d ic u la r to .every line in th e p lane th p t passes th ro u g h its foot.

And, of course, it should be pointed out that the point of intersection . of a line and a plane is called the foot of the line,

Ju s t as the definition of perpendicular lines was a cumbersom e tool to use to prove two lines to be perpendicular, so, too, is the above definition a bit unwieldy. Proving a line perpendicular to every line passing through its foot may become somewhat wearisome. Hence, effort is m ade to cut the num ber down. From our analysis of the man and his post, we know that a line will not be perpendicular to a plane if it is perpendicular to only one line in the plane. If we can show, however, that it is perpendicular to at least two lines in the plane, then it will be perpendicular to the plane.

THEOREM 21: A lin e is perpendicular to.a plane i f it is perpendicular to at least two lines in the plane that pass through its foot.

Given: BC and BD in plane m

T b 1 BC

A B 1 BD

Concl.: A B X m

Figure 7-9.


A n a l y s is : In order to prove th a t A B is perpendicular to plane m, it will be necessary to prove that it is perpendicular to every line in m passing through B. T his would not be possible. W e get around this difficulty by selecting any .line at randofn that lies in m ar>d passes through B. W hat we can prove to be true for this line must then be true for all lines in m passing through B.

T he additional lines are draw:'; in the figure so as to enable us to appiy the theorem concerning two points each equidistant from the endpoints of

a line segment. By extending AB so that RB ~ VB , we immediately obtain4'—>

one point of B Y equidistant from R and V. Point P will be the other point i—'f

of B Y that will be proved equidistant from R and V.


1. Let R he any point of A B

and extend A B so that

BR= * W .

2. L et S be any point on BD

and T b e any point on BC; then draw line ST .

3. W ithin the in terior o f

Z S B T d r a w BY. It wilJ in

tersect S T in some point P. ■ {

4-. A B X -B C

A line can be extended as far as desired in either direction.

2. T here exists one and only one Jine through two points.

3. Pasch’s Axiom

4. Given

PERPENDICULARITY BETWEEN A LINE A N D A PLANE 209

5. /. TR = TV (s)

6 . AB X BD

7.8. S T ^ S T i s )

9. A f i s r s i w r10. Z R T P — Z V T P (a)

11. 7 7 £= 7 712. A R T P = A V T P

13. M s F F

14. PB X RV

15. AB X m

5. If a point is on the perpendicular bisector o f a line segment, then it is equidistant from the endpoints of the

i—► ,

line segment. (T B is the X bisector

of /?]/.) See Statem ent 1.

6. Given

7. Same as 5. (SB is X bisector of RV.)

8. Reflexive property of congruence9. Why?

10. Why?

11. Why?12. Why?

13. Why?

14. If two points are each equidistant from the endpoints of a line segment, then the line joining them is the perpendicular bisector of the line segment. (W hat are the two points and what is the line segment?)

15. Reverse of the definition of a line perpendicular to a plane.

Illustration:If a line is perpendicular to a plane of a circle at its center, then any.

point on the line is equidistant .from all points of the circle.A| £<.Ukr>fJ is f ' 0 n (lh>-|

HaJE U n i j lu c?J

Given: A B X mB is center of circle lying in m.

Concl.: 7 6 9 * 7 5

Figure 7-11.

A n a l y s is : Let P be any point on AB. If it can be shown that P is equidistant from two points selected a t random on circle B , then it will be equidistant from all points on circle B. The-two points selected on circle B are C and £>.

Hence, the problem simplifies to one where it is merely necessary to prove that P C ^ P D .


PR O O F | STATEMENTS REASONS

1. A B jL m 1. Given<-» <-> <->

2. A B _L BC, A B _L BD 2. Def. of a iine perpendicular to aplane.

3. Z P B C and ZP B D are right 3. Def. of perpendicular linesangles.

4. Z P B C -1— Z P B D (a) 4. If two angles are right angles, thenthey are congruent.

5. B is the center of the circle. 5. Given

6. ~BC ~ BD (s) 6. Why?

7. P B S * P B (s ) 7. Why?8. & P B C 9 * A P B D 8. Why?9. P U ^ P D 9. Why?

Some of the problems in the exercises that follow depend upon an understanding of the m eaning of an oblique line to a plane.

D e fin itio n 45: An oblique line to a plane is any line th a t is not perpendicu lar to the plane but has one and only one point in com m on w ith the plane.

<■) <—>In the illustration above, PC and PD are oblique lines to p lane m.

EXERCISES

1. If a line is perpendicular to a line in a plane, will it be perpendicular to th e plane also? Explain.

2. How m any lines can be draw n perpendicular to a given line a t a given point of the line? W hat do you believe will be tru e about all the lines th a t are perpendicular to a given line at a given point of that line?

3. Lines a, b, and c are such that a ± b, a ± c, and b j . c. Explain why each line is perpendicular to the plane determined by the o ther two lines.

4. In the diagram for the proof of Theorem 21, page 208, can you show that the points R, S, P, and T must be coplaner?

5. I f a line passes through a vertex of a triangle and is perpendicular to

two sides at this vertex, is it perpendicular to the plane of the triangle?

Justify your answer.

>ERPENDICULAR1TY BETWEEN A LINE AND A PLANE 211

B

"I, Given: BC, BD, and BE in plane m <-> «->

AB 1 BC

AB 1 BD

Concl.: AB ± BE

3 , Given: A C L mm bisects AC a t B.

Concl.: PB bisects ZAPC.

Given: AB L m

CB and D B in plane m Z C A B ■= Z D A B

Concl.: ZA C B S Z A D B

i—>Given: A B _L m

AB 1 n I d ^ b S

Concl.: AC — ~BD

4 .


G iven: A B and CD are two line segments in m that bisect each o th er at E.

P A ^ P B

PC = PD

C oncl.: P E X m

BC S BD Concl.: Z A C D ^ Z A D C

Given: A B X m

CD X mE is the midpoint

of ID .

A B S Z D C Concl.: £ is equidistant

from B and C.

Given: QO lies in plane m. 6 . A U and BD intersect at 0 .P is equidistant from A, B, C, and D

4—►Concl.: PO X m

Given: AD X m E A S Z E D

Concl.: A A B C ^ Z A D B C

A

BD L m

IC S * BD P is the m idpoint

of CD.Concl.: P is equidistant

from A and B.

P E R P E N D IC U L A R IT Y BETW EEN A L IN E A N D A PLANE

11 . G iven : A R X m

AR X n

BD ~ BE

Concl.: C F ^ C G

B E is the X bisector o f5 C .

Concl.: A E X DC (H int: Use information from Problem 12.)

G iven: A B L m <—*BE is the X bisector

of DC..

C oncl.: AD ~ AC

213

12.

G iven: EA ^ EC

BA S BC Z A E D £ ZCED

Concl.: D B X AC

15.* Given: A A B C is in plane m.P is not in plane m. A P B C — A ABC

Concl.: ZA P D = ZP A D(H int: Prove A D P A to be isosceles.)

14.


01. If from a given point not on a plane two oblique line segments and a

perpendicular are drawn to a plane such that the loot of the perpendicular is equidistant from the feet of the oblique line segments, then the oblique line segments are congruent,

2. If a plane is the perpendicular bisector of a line segment, then any point on the plane is equidistant from the endpoints of the line segment.

3. If from a point on a line perpendicular to 'a plane congruent oblique line segments are draw n to the plane, then the foot of th e perpendicular will be equidistant from the feet of the oblique line segments.

4. I f from any point on a line perpendicular to a p lane oblique lines are drawn to the plane so that they form congruent angles with the perpendicular, then the foot of the perpendicular is equidistant from the feet of the oblique lines.

5. I f a line is perpendicular to the plane of a circle a t its center, then all oblique lines draw n from any point on the perpendicular to points of the circle make congruent angles with the perpendicular.

6. Two planes are perpendicular to the same line segment at its endpoints. O blique line segments a r t drawn through the m idpoint of this segment term inating in the two planes. Prove that this point is also the m idpoint of the oblique line segments.

7. If the foot of a perpendicular to a plane of a triangle is equidistant from the vertices of the triangle, then every point on the perpendicular is equidistant from the vertices of the triangle.

■ Test and Review

[A|1. T he diagram below is to be used in answering the questions that follow it.

TEST AND REVIEW 215

(a) Name three sets of four coplanar points.(b) How m any planes are there in this figure?(c) Name two points that are common to planes HF and BG.(d) Name the planes that have as two of its elements the points A and E.(e) Name all the planes that have the point H in common.

If CG is perpendicular to plane HF, what conclusion can be drawn? W hat plane is determ ined by the point E and the line AB?W hat plane is determ ined by the points £ , G,. and H?

/(i)j W hat plane is determ ined by DA and HD?2. If the four vertices of a four-sided polygon are coplanar, will the lines

joining the pairs of opposite vertices also lie in that plane? Justify your

answer.3. Will the m edian of a triangle lie in the plane of that triangle? Justify

your answer.4. I f a line is not perpendicular to a plane, what is the m axim um num ber

of lines in the plane to which it can be perpendicular?5. If a line is perpendicular to each of two intersecting lines at their point

of intersection, what can be said of the plane determined by these two.

intei'secting lines?6 . O f four points, three are collinear. Are the four coplanar? Justify your

answer.7 . -If the lines A B and CD have the point P in common, then what can be

said of the five points A, B, C, D, and P?8. In the diagram at the right, point D

is not in the plane ABC. Can you justify in any way why point A is the

only point that DA has in common with plane ABC?

g

*QA n A b C ’ U \


B

1. Given: PlaneProve ench of the following:

m and G O I Given: Plane m bisects AB.

PO _L m Concl.: ZAPO S ZBPO

3 . Given: PA A. mZ A B C ~ ZAC B

C oncl.: Z P B C SS ZPC B

5 . Given: Plane m w ith <2 the

m idpoint of A B and

CDP is equidistant from A, B, C, and D.

Concl.: PO X m

A B X m L-oncl.: A PA B is isosceles.

AD S AC E is the m idpoint of .DC.

Concl.: B E X CD

G iven: Circles A and B in plane m

P Q X m Concl.: A P R S is isosceles.

\

TEST A N D REVIEW 217

Prove each of the following statem ents: ~J Or F

1. If a line is perpendicular to the plane of a circle a t its center, then all oblique lines drawn from any point on the perpendicular to points of the circle will make' congruent angles with the radii drawn to these points.

2. A line is perpendicular to the plane of an isosceles triangle at the point of intersection of the angle bisectors of the base angles. If from any point on this perpendicular line segments are drawn to the vertices of the base angles, an isosceles triangle will be formed.

3. If from any point on the perpendicular bisector of a line segment a line is drawn perpendicular to the plane determ ined by the line segment and the perpendicular bisector, then any point of the perpendicular will be equidistant from the endpoints of the line segment.

4. Two planes are perpendicular to a line segment at its endpoints. Two oblique line segments are drawn, one from each foot and term inating in the other plane. If the oblique segments make congruent angles with the perpendicular segment, then the oblique segments are congruent.

A

The Indirect Proof and Parallelism

A PR IN C IPL E T H A T HAS G R E A T IM P O R T A N C E in geometry concerns itself with the relationships that exist between various angles of a triangle. No m atter w hat the triangle may be—w hether it be acute as in Figure 8-1, obtuse as in Figure 8-2, or even right— Z \ will always bear the same relation to each of the other angles of the triangle.

T he apparent relation, and one with which we are familiar, is the fact th a t is both supplem entary and adjacent to Z2. In addition, moreover, we will prove that its measure is greater than the measures of either Z i or Z4. T h e m ere reference to the words greatsr than implies th a t we will have to create new assumptions, for nowhere, thus far, have we established the tools for showing when the measure of one angle will be larger than another

. or when the measure of one line segment will be larger than another.

218

while the expression

is read as

In the event that you may have forgotten some of your work in algebra, the symbol > is used to replace the words “ is greater th an ,” while the symbol < replaces the words “ is less than .” Thus, the expression

“ 5 > 2”is read as

“ 5 is greater than 2”

“ 4 < 7”

“ 4 is less than 7”

Each of these relations, or inequalities, can be read in either direction; that is, either from left to right, as

“ 5 is greater than 2”or from right to left, as

“ 2 is less than 5”

Students who are just learning these symbols frequently confuse one with the other. A simple device for keeping them clear in your m ind is to remember that the arrowhead always points to the smaller number.

W ith these symbols a t our disposal, we are prepared to create a postulate establishing when one num ber will be larger than another.

P ostulate 22: If a, b, and c are positive numbers

where a = b + cthen ' a > b and a > c

Actually, this postulate points out no more than to say that

t if 5 is equal to 2 plus 3; then 5 must be greater than 2

and 5 must be greater than 3

| From our point of view, also, there was no need to state th a t a, b, and cwere positive numbers, for we deal with no others in our work.

W hen, though, have we encountered a situation in geometry wherein a single num ber represented the sum of two other numbers? Actually, this has occurred but twice in our work. T he first time was when we defined the sum of two line segments and the second wfyen we defined the sum of two

angles. Thus, in Figure 8-3 the m AC was defined as the sum of the m A B

THE INDIRECT PROOF AND PARALLELISM 219

Figure 8-3.

V)

220 THE INDIRECT PROOF AND PARALLELISM

and the m BC, while in Figure 8-4 m Z A B C was defined as the sum of the measures of Z A B D and ZD BC. By using symbols these relations can be expressed as

(1) T't AC ~ m A B 4 ■ m BC(2) m Z A B C = m Z A B D + m Z D B C

Thus, by using Postulate 22, it imm ediately follows that

( t ) m A C > m A B and m AC > m BC(2) m Z A B C > m Z A B D and m Z A B C > m ZD BC

Q uite often A B and BC are referred to as the parts of AC, while Z A B D and Z D B C are the parts of ZABC. In view of this and the relations shown above, it is ap paren t why Postulate 22 is frequently quoted as

P o s t u l a t e 2 2 a : T he whole is greater than any of its parts.

Let us re tu rn to the figures on page 218. An angle such as Z \ is called an exterior angle of a polygon. In this case, of course, it would be an exterior angle of a triangle.

D e fin it io n 46: An exterior angle of a polygon is an angle that is adjacent to and*supplem entary to an angle of the polygon.

How m any exterior angles does the triangle have a t vertex C? From the definition of an exterior angle, why is Z b not an exterior angle? How is Z 6 related to / 2? How .m any exterior angles will a triangle have? How are the two exterior angles at any vertex related?

Occasionally, for emphasis, the angles of a triangle— A 2, 3, and 4 in the d iagram above— are called interior angles of the triangle. W ith reference to either exterior Z \ or Z i , the A 3 and 4 are spoken of as the remote interior angles. T hey are the interior angles th a t do not have a vertex in common with the exterior angle being considered. W hat are the remote interior angles with reference to exterior Z \ \ ? W ith reference to exterior Z&? T o which exterior angles are Z 2 and /£4 the rem ote interior .angles?

T H E O R E M 22; T h e m easu re of an ex te rio r ang le of a t r ia n gle is gre a te r th a n the m easure ol e ith e r oftKe~remcte in te rio r angles.


Given: A ABC with exterior ZAC D Concl.: m ZAC D > m ZA

m ZA C D > m Z B

A n a l y s is : By creating 'an angle th a t is part o f ZAC D , its measure will be less than the measure of ZACD. Furtherm ore, we will show th a t this angle is congruent to ZA . From this it will follow that the measure of ZA C D is greater than the measure of ZA. W hich angle in Figure 8-7 is part of ZACD1 W hat else will have to be shown to be. true about this angle?

T he proof th a t m ZACD > m Z B will be given as an exercise for you

to do. ->> V .

P R O O F STATEMENTS REASONS 1

1. Let M be the midpoint of 1. Every line segment has a midpoint.

AC.

2. Let B M be the line through 2. Why?

points B and M .

3. Extend B M to point P 3. A line can be extended as far as de

such th a t M F ■= BM . (s)sired in either direction.

4 . L et PC be the line through 4. Why?

P and C.

5, f f i s C f f ( s ) ' 5. Def. of a m idpoint

6. Z A M B ^ ZC M P (a) ‘ 6. Why?

7. & .AM B — A C M P 7. Why?

8. Z A & Z \ 8. Why?

9. m Z A — m Z \ 9. Def. of congruent angles

10. But, fn ZA C D > m Z \ 10, T he whole is greater than any of itparts.

11. ,\ m ZA C D > m Z A 11. Substitution postulate

222 THE INDIRECT PROOF AN D PARALLELISM

EXERCISES1. By using the diagram at the right,

justify the drawing of each oi the red lines and prove that m ZACD > m Z b by following the pattern of proof for Theorem 22.

2. (a ) In the drawing a t the right, angles D and C are the remote interior angles with reference to which exterior angles?

(b ) How is Z A B D related to ZD?(c) How is Z E B C related to Z D ?

« ( d ) How is Z A B E related to ZDBC?'

■ 3 .)(a) How does the measure of Z \ com-pare with that of Z2? m O >

(b ) If the m Z 2 = 80, what can besaid of Z t? „ g o

(c) If PR is made to ro tate about P in a counterclockwise direction, what will happen to the measure of Z \? To the measure of Z2? Hence, eventually, how should the measure of Z \ compare with the measure of Z 2 ?

4. (a) W hat is the measure ofZ A B D ?

(b ) W hat can be said of the measureof z a

(c) W hat can be said of the measure of ZD?

5. (a) W hat is the measure of

Z A C m(b ) W hat is the measure of ZACD?(c) Can you give any justification for

the fact that m Z A B C coult not be 110 as noted in the Given Data?

Given: m Z D B C = 110

Given: m Z A B C = 1 1 0

Z S s z a U


6. Can you justify the statement that one of the base angles of an isosceles triangle can not be ah obtuse angle?

7. Can you justify the statement that one of the base angles of an isosceles triangle can not be a- right angle?

8. (a) Can you justify why Given: m ZACD — 40

m ZACD can not be 40 as stated in the Given Data?

(b) Is it possible for the m ZACD to be 90? Why or why not?

(c) W hat is the least value that the measure of ZACD may have? Why?

9. (a) W hat is the value of m Z \ + m Z2?(b ) How is m ZC related to « Z l?(c) By using the information from (a)

and (b), w hat can be said of m Z C -I- tn Z 2 i

(d) W hat can be said of the value of m Z D + m Z l?

10. (a) W hat is the m ZB D C ? Given:(b ) W hat is the m ZBCD?(c) W hat is the m ZBCA?(d ) W hat is the m ZACD?'(e) How does m Z B D E compare with

m ZACD ?(£) ABDC is a four-sided polygon.

Will the measure of an exterior angle of any polygon be greater than the measure of any of the remote interior angles? Justify your answer in terms of polygon ABDC.

A B ^ A C

A A B C and A CBD are congruent isosceles triangles as m arked in the diagram . m Z B D E = 1 0 0

11. In isosceles A ABC, AB = AC. Base BC is extended to point D. T he bisectors of AABC and ACD intersect a t E. Can you justify in any manner whatsoever that ZE C D can not be an angle of 30°?


M Nonintersectins Lines and the Indirect ProofW e would like to puisue further the analysis begun in

Problem 3 b f the preceding exercises. W hen we examine Figure 8-8, we note

Figure 8-8,

Vthat m Z l > m Z 2 on the basis of the exterior angle theorem. Now, should we allow I to ro tate counterclockwise about point P, although nothing h a p pens to the m easure of Z2, the m easure of Z l grows smaller and sm aller. Eventually, of course, I will coincide with line PQ, and when this occurs, the m easure o f Z l will be zero. However, at some position between its present m easure and that of zero, the measure of Z l will approach and be equal to the m easure of Z2. But, w hat happens to other points in the d iagram as I rotates about P? Points P and Q will rem ain fixed, but point R will move off to the right.

Notice, in Figure' 8-9, that although the am ount of rotation from position

R P to R'iP is' approxim ately the same as from R^P to RiP, the line segment

__ ^RtRs is far greater than RRt. And should I continue to be rotated to a fourth position so th a t another angle is formed whose measure is again the same as m Z R P R i, the po in t of intersection of I with m will be thrown far, far to the right. Thus, as the measure of Z l approaches the measure of Z2, the point of intersection of / with m appears to recede so rapidly to the right that it seems to disappear completely. W ere this situation expressed formally as a problem , it w ould be

NONINTERSECTING LINES A N D THE INDIRECT PROOF 225

G iven: Z l = Z2Concl.: I does not intersect m.

Figure 8-10.

A nalysis: This problem, obviously, involves a difficulty that we have not encountered before. Until now we have examined situations in which angles or line segments had to be proved congruent, lines had t ^ b e proved p e rpendicular, triangles had to be proved congruent, and the like.'' But at no tim e had it been necessary to justify that lines do not intersect, as we are confronted with in this problem. T o show how to overcome this difficulty, we will have to take a ra ther lengthy detour.

Let us assume that you and your family are sitting a t the dinner table one evening when the lights suddenly go out. T he group of you decide to seek the cause of this event without having to leave the table. And so you m ake a list of possible causes:

(1) T he electric power in this section of the city is out.(2) T he light bulb has become .defective.(3) T he fuse for this room has blown out,(4) T he switch has become defective.(5) T he wires leading to the light have separated.(6) T he electric power on this street is out.(7) T he m ain fuse for the entire house has blown out.

You then proceed to examine ways of elim inating some of these possible causes. By looking through the window you discover that the lights in the house behind you are still on. This would elim inate cause num ber 1. T h e fact that the lam p in the livingroom is still on does away with causes 6 and 7. T h e sound of the refrigerator motor, which operates on the same fuse as the light overhead, makes short work of cause 3.

/ 2 X 4 5 X /

And so you move from possibility to possibility, hoping to elim inate all but one of these possible causes. If this can be done, then w hat conclusion can be draw n concerning the possibility that remains? We are overlooking an im portant factor though. W ho set up these possibilities? Could there, perhaps, have been some causes that you may not have considered? If this was so, then were you to eliminate all but cause #5, would it then follow th a t this cause le<| to the event? W hat is the m ajor weakness in using the m ethod of elim ination to’ justify a conclusion?

The m echanic who examines yo;ir father’s car to determ ine why it is


not functioning properly, your family physician who examines you to learn why you are not functioning properly, and even your m other who peers helplessly down the kitchen drain to see why it is not functioning properly— all of them are making use of the ‘‘proof by elimination.” Each knows m any causes that might have led to the event that is troubling him, and each, in his ov/n way, is trying to eliminate all but one of these. Should he be successful in discarding all but one possibility, he heaves a sigh of relief and lays blame on this rem aining cause. There remains always, however, a shadow of a doubt: “ H ave I been careful to list all the possibilities? Are there any, perhaps, that I m ay have overlooked?” This question of doubt is something that the m athem atician will not and dare not contend with. Should- he use the proof by elimination, he must be certain that he has listed all the possibilities.

T o understand how this difficulty is overcome, we resort to the Venn diagram . Consider a universal set of elements where set A is a subset of the

• 0 u

*v / ^ \.r V .r J .M

Figure 8-11.

universal set. Quite apparently, every mem ber of U will be either a m em ber of A or not a 'm em ber of A. Using symbols, we express the set of elements that are members of U but not of A by saying that they are members of the set ~ A , this being read as “ not A.” Thus, in the diagram above either an element is a member of A or it is a mem ber of ~ / l , and no other possibility exists. Specifically, one of these two statements m ust be true:

<2 is a m em ber of A.or

Q is a member of ~yf.

Furtherm ore, both of these statements can not be true a t the same time. T h a t is, the element Q can not be both a member of the set A and not a m em ber of the set A a t the same time.

Let us bring this discussion closer to home. I t would seem th a t no o ther possibility exists concerning an angle A but

(1) Angle A is a right angle,

and

(2). I t is not the case that angle A is a right angle.

Furtherm ore, it seems that one and only one of these two possibilities can be true a t the same time. T he tru th of (1) would immediately imply th a t (2)

was false, while the tru th of (2) assures the falsity of (1). Statem ents such as (1) and (2) are called contradictory statements, and the symbols usually applied to the two are

p andFor the statements above

p is “ Angle A is a right angle”

while is “ It is not the case that angle A is a right angle.”

W e usually find it more convenient to express

“ It is not the case that angle A is a right angle”

in the simpler form of“ Angle A is not a right angle.”

In summary, the m athem atician has established the following two postulates so that he might apply the proof by elimination.

P ostulate 23: Either p or is true. No other possibilities exist. (Law of the Excluded Middle)

P o s t u l a t e 24: Both p and can not be true a t the same time. (Law of Contradiction)___ ,

The existence of both p and in a proof is called a logical inconsistency,for by the law of contradiction both can not be true a t the same time.Hence, in applying the proof by elimination, possibilities th a t are listed' can be elim inated by showing in some m anner that they lead to statem ents th a t are contradictory to either the given data, a postulate or a definition we have agreed upon, or a theorem we have proved. Since we accept the tru th of these latter statements, it will imply that the possibility we selected must be false, for by the law of contradiction, both p and can not be true at the same time.

Basically, the mechanics of the proof by elimination are as follows:

(1) Examine the conclusion that you are asked to prove.(2) Set up the statem ent that is contradictory to this conclusion.(3) Show that the acceptance of this possibility leads to a logical inconsistency: that is, it leads to a statem ent that ts contradictory to one of the following:

(a) T he given data(b ) An assumption(c) A definition(d ) A theorem

(4) T he conclusion then follows, for having eliminated one of the two possibilities that existed, the remaining one will have to be true.

This m ethod of proof is called the indirect proof, for we do no t attem pt to move from the given da ta directly to the conclusion, but ra ther we s ta r t w ith a statem ent that is contradictory to the conclusion and somehow justify

NONINTERSECTING LINES AND THE INDIRECT PROOF 221


that this statem ent leads but to a logical inconsistency. Experience has shown that it is far easier to w rite the indirect proof in paragraph form than resort to the two-column form th a t has been used until now. Several examples will be given to illustrate this m ethod of proof.

Illustration 1:

Given: A B S Z D E

B € = £ E F A C 3 iD F \

Concl.: Z E g Z Z B

Figure 8-12.

P R O O F

By the law o f the excluded m iddle one of these statem ents must be true and no o ther possibilities exist:

or Z E S i Z B

L et us accept the possibility th a t Z E ~ Z B . From the Given D ata we note

th a t A B ^ D E and BC = £ F ; hence, if Z E ^ Z B , it follows th a t

A CAB £= A F D E . If this is so, then AC ~ DF, according to the definition

of congruent polygons. T he Given D ata, however, indicates that 4 C 3 - DF. Therefore, accepting the possibility th a t Z E ~ Z B led to the logical in

consistency of AC S i D F and A C $£ DF. According to the law of contradic

tion, both can not be true at the same time. Since AC DF must be true,

for it is p a rt of the Given D ata, then AC ~ D F must be false and, so too, must Z E ~ Z B be false. Hence, Z E 3= Z B m ust be true, for it is the only rem aining possibility.

Illustration 2:

A triangle cannot have m ore than one obtuse angle.

t W henever a slash is d raw n th rough a sym bol, it signifies th a t the word “ n o t” should p rcccdc th e sym bol. In th is case it im plies “ n o t co ng ruen t."

NONINTERSECTING LINES AND THE INDIRECT PROOF 229

Given: A ABCConcl.: Z \ and Z2 are not both obtuse

angles.

PRO OF

By the law of the excluded middle one of these statements must be true and no other possibilities exist:

Z l and Z2 are not both obtuse angles.or

Z \ and Z2 are both obtuse angles.

Let us accept the possibility that Z l and Z 2 are both obtuse angles. Z l and Z3 are supplementary angles since their sum is a straight angle. However, from the fact that Z l is obtuse, Z3 must be acute, for the sum of their measures is 180—the measure of one being greater th in 90, the measure of the other must be less than 90. This would make the m Z3 not greater than the m Z2, since Z3 is acute but Z2 is obtuse.' But this is contradictory to the theorem that the measure of the exterior angle of a triangle is greater than the measure of either of the remote interior angles.

Therefore, accepting the possibility that Z l and Z2 are both obtuse led to the logical inconsistency of the tru th of both m Z 3 m Z2 and m Z 3 > m Z2. By the law of contradiction both can not be true a t the same time. Since m Z3 > m Z2 must be true, for it is the result of a theorem, then m Z3 £ m Z 2 must be false. Therefore, the statem ent th a t Z l and Z2' are both obtuse angles is also false. Hence, the statem ent th a t Z l and Z 2 are not both obtuse must be true, for it is the only rem aining possibility.

EXERCISES

A1. W rite two contradictory statements th a t m ight occur in everyday ex

perience.2. W rite two contradictory statements that m ight occur in geometry.3. W rite up a situation in everyday experience wherein the proof by elim ina

tion is used.

230 THE INDIRECT PROOF A N D PARALLELISM

Bin each of the following problems you are to do three things:

(a) Decide what the extraordinary occurrence is for which a causa! explanation has been formulated.

(b ) State the cause that has been given that led to th a t event.(c) Give at least tone other possible cause that m ight have led to th a t

event.

1. D uring a debate a child psychologist made the following sta tem en t: “ T he amazing increase in the sale of comic books with their sordid in te rp reta tion of crime has been accompanied by the equally am azing increase in the rate of crime committed by teenagers.”

2. “ Because we did not live up to our opportunity of join ing the League of Nations after W orld W ar I, the League failed to prevent the most tragic war in history—the Second W orld W ar."

3; “ Another advantage of establishing standards is that, inevitably, a rise in teaching standards must follow as the night the day. We are graduating only 40,000 scientists annually. Every day we postpone im proving and upgrading education is a day lost in our scientific advancem ent.”

4, “ Since achieving independence, the gross volume of business in C entralia has increased by 65 percent, employment has risen, and exports have grown from 25 million to 125 million dollars annually .”

Use the indirect proof in the proof of each of the followingproblems;

1 , G iven: AC ~ DF..~BC ~ EF Z C g Z F

Concl.: A B 3 Z D E

Given: AB ~ AC 2 .D is no t the midpoint of 'BC.

Concl.: AD does not bisect ZBAC .

A

NONINTERSECTING LINES AND THE INDIRECT PROOF

3 . Given: A B £ AC

AD / BC Concl.: D is not the m idpoint

o i W .

G iven: B is the m idpoint

of AC.D A & D C

Concl.: D B / . AC

5 . Given: GO where OD does not bisect Z A 0 B .

Concl.: OD is not the median

to AB.

231

4 .

Given: AB = AC D B g ~ D C

Concl.: AD does no t bisect ZBAC .

G iven: BA = BD c d g C A

Conci.: BC does not biscct

AD.

Given: 0 0 with AB 3^ AC

Concl.: A 0 does not bisect

BC.

232

10.THE INDIRECT PROOF AND PARALLELISM

9 . G iven: A ABC I Given: AC J_ mConcl.: Z2 and Z3 are not AB AD

both right angles. I Concl.: BC g DC(Sec Illustration 2.) j a

A

11, Given: A B JL m

DC X m

A B S Z D C E is not equidistant from A and D.

C oncl.: E is not the mid

point of BC.

13. Given: m is the _L bisector of AC. -

D A & D C C oncl.: D does not lie in m.

G iven: O B in plane m A C m A D

C oncl.: A B £ m

12.

Given: A B _L m

B E ± CD A e m Z B

Concl.: C E m B E

14.

PARALLELISM—SECTION I 233

D1. If two angles of a triangle are not congruent, then the sides opposite

them are not congruent.2. If two sides of a triangle are not congruent, then the angles opposite

them are not congruent.3. If a triangle has no two sides congruent, then the perpendicular bisector

■ of one of the sides does not pass through the opposite vertex.4. Two lines perpendicular to the same line do not intersect. (H int: See

Problem 9, Group C.)5. A base angle of an isosceles triangle can not be an obtuse angle.6. A base angle of an isosceles triangle can not be a right angle.7. In a right triangle neither of the two angles th a t are not the right angle

can be obtuse.8. Any point that is not equidistant from the endpoints of a line segment

is not on the perpendicular bisector of the line segment.9. A line segment can have only one midpoint. (H int: Use the postulate

th a t the whole is greater than any of its parts.)10. An angle can have only one bisector.11. At a given point o n a given line there can be only one line perpendicular

to the given line.^ 12. From a given point not on a given line there can be only one line per

pendicular to the given line.13. Tw o lines perpendicular to the same plane can not intersect.14. If from any point'on a perpendicular to a plane line segments of unequal

measures are drawn to the plane, they will make angles of unequal measures with the perpendicular.

15. From a given point not on a given line there can be only one plane th a t is perpendicular to the given line.

16. From a given point in the interior of an angle perpendicular segments are draw n to the sides of the angle. If the ray whose endpoint is the vertex of this angle and which passes through the given point is no t the bisector of the angle, then the perpendicular segments are not congruent,

E Parallelism— Section IWe have detoured so far from our original path th a t you

m ay have forgotten why the need for the indirect proof arose. W e wei'e trying to prove the problem below when we realized that we had no way to prove that two lines cannot intersect. A new and different approach was necessary in order to draw our conclusion. Now we are ready to take a fresh look a t this problem.

;

A n a ly sis : By using th e indirect proof, the solution to this problem is very m uch the same as that used in the preceding exercises.

P R O O F

By th e law of the excluded m iddle one of the following statem ents m ust be tru e and no other possibilities exist:

/ does not intersect m or / does intersect m

L et us accept the possibility that I intersects m. This implies th a t I will m eet

m in some point R. T he lines I, m, and PQ will form the triangle PQR. Hence, the measure of Z l will be greater than the measure of Z 2. T he Given D ata, however, states th a t m Z l = m Z2. Therefore, accepting the possibility th a t / intersects m leads to the logical inconsistency of the tru th of both m Z l = m Z2 and m Z l ^ m Z2. By the law of contradiction bo th can not be true a t the same time. Since “ m Z l = m /LI" m ust be true, for it is p a rt of the Given D ata, then “ m Z l ^ m Z 2” must be false and, therefore, the statem ent th a t “ i intersects m" is also false. Hence, "I does not intersect m” m ust be true for it is the only remaining possibility.

In the same m anner it can be shown that I cannot intersect m to the

left of Q.

Tw o coplanar lines, such as I and m, th a t do not intersect are called parallel lines.

D efin itio n 4 7 : Parallel lines are two coplanar lines th a t do no t intersect.

Before it is possible to express the statement of the theorem that has ju st been proved, it will be necessary to define a num ber of o ther terms.

PAR ALLEUSM—SECTION I 235

In the diagram above PQ is called a transversal with reference to I and m. ■<rs puVCftjj.

D efinition 48: A transversal is a line that intersects two other lines in two

distinct points.

In Figure 8-16, n is a transversal with reference to I and m, for it inter

sects these lines in the two distinct points P and Q. In Figure 8-17 it is not a transversal, for it intersects the two lines in bu t one point, P.

A point A may be selected on the transversal, and the three points Q, P, and A can be read in the order in which they appear on th a t transversal, I f this order is either QP/I or AQP (see Figures 8-18 and 8-19), then point A

is said to be in the exterior region with reference to the lines I and m. In either event, A is not between P and Q. Should A be between P and Q, then the point A falls in the interior region w itnreierence to Pan'S m. jThil occurs in Figure 8-20. «. • ■■ 1 c f ' - ' i ' t ' t ,,

For simplicity the regions are pictured in Figure 8-21. •


W hen a transversal intersects two lines, eight angles are formed. Every pair of angles is given a special name. Several of these names are fam iliar to you. T hus, w hat is the nam e for the pair of angles 6 and 8? Angles 1 and 3? Angles 5 and 6? Angles 1 and 4?

T he pairs of angles such as 2 and 8 or 1 and 7 that are in the interior region bu t on opposite sides of the transversal are called alternate interior angles. These angles can be recognized by the fact that they form the letter Z, either

in norm al form or backwards or sideways

Figure 8-23.~ ' t o l a - ■D efin itio n 49: A lternate interior angles are two angles formed by a trans

versal intersecting two lines, both angles being in the interior region, on opposite sides of the transversal, and a t different vertices.

In Figure 8-22 the pairs of angles such as 4 and 8, 3 and 7, 5 and 1, or 2 and 6 that are on the same side of the transversal so as one is interior and the o ther is exterior are called corresponding angles. These angles can be recognized by the fact th a t they form th e le tte r F in m any ways.

T h e pairs of angles 3 and 5, or 4 and 6, of Figure 8-22, being on opposite sides of the transversal, are known also as alternate angles. Now, however, since they are in the exterior region, they bear the full name of alternate exterior angles, ....

PARALLELISM—SECTION I 237

D efinition 50: Alternate exterior angles are two angles formed by a transversal intersecting two lines, both angles being in the exterior region, on opposite sides of the transversal, and at different vertices.

D efinition 51: Corresponding angles are two angles formed by a transversal intersecting two lines', the angles are a t different vertices, one being interior and the other exterior, but they are on the same side of the transversal.

EXERCISES

1. (a) Nam e two pairs of alternate interior angles. ^ $(b ) Nam e two pairs of alternate exterior angles. < V 8 > S ^ 1(c) Name four pairs of corresponding angles. < J } i i ^ | , M l i ^(d ) Name a pair of interior angles on the same side of the transversal.(e) Name a pair of exterior angles on the same side of the transversal.

2. (a) Using / as a transversal with reference to m and n, nam e two pairs of alternate interior angles. Name a pair of alternate exterior angles.

(b ) By using m as a transversal, what nam e does the pa ir of angles 3 and 8 have? W hat name does the pair of angles 2 and 5 have?

(c) Using n as a transversal, name a pair of interior angles on the same side of the transversal. Name a pair of corresponding angles.

T“f TT3. (a) By using BD as a transversal w ith reference to AD and BC, w h a t is

the name of the pair of angles A D B and D EC ?

Bflwr

r-

238 THE INDIRECT. PROOF AN D PARALLELISM<—V

(b ) List a pair of angles formed by the transversal BD and the lines A B and DC. W hat is the nam e of this pair of angles?

A,------------- -------------. 0

4. (a ) W hat is the only line in the diagram that, w ithout being extended,

can act as a transversal with reference to A B and EF?

(b ) W hat is the name of the pair of angles that this line forms with A B

and EF}

(c) By using B E as a transversal with reference to the lines AC and D F, w hat is the nam e of the pair of angles ACB and FDE? O f the pa ir of angles ACD and FDC?

5. E F is used as a transversal for the questions that follow.(a ) W hat nam e can be given to the pair of angles E N H and E M K ? Angles

C N M and B M N ?

(b ) Nam e a pa ir of a lte rna te exterior angles w ith reference to th e lines J K and CH.

(c) Is there any special nam e for the pair of angles B M K and D N H 7 Justify your answer.

(d ) Nam e a pair of angles in the diagram where the transversal is no t a side of either angle. Does this pair of angles have a special name? Justify your answer.

PARALLELISM—SECTION II 239

■ Parallelism — S ectio n II

In terms of the words that we have recently defined, the problem that was proved on page 234 can now be stated as

TH EO REM 23: I f two lines are cut by a transversa l such th a t th e a lte rn a te in te rio r angles a re co n g ru en t, th en th e lines a re paralle l, f

As an exercise, without referring to the proof on page 234, see if you can prove the above theorem.

Theorem 23 is the principal m ethod we have for proving lines parallel. The next two theorems that we are about to present are developed on the basis of Theorem 23. There is no necessity to apply the indirect proof when we attem pt to justify these theorems, for now we need but show th a t the alternate interior angles are congruent to prove lines parallel. T h e symbol || represents the word parallel. ^

TH EO R EM 24: I f two lines a re cu t by a transversa l such th a t th e corresponding angles a re co n g ru en t, th e n th e lines a re p a ra lle l.

2

Given: Z l = Z2 Concl.: / || m

Figure 8-25.

Analysis: Simply show that Z t = Z3, then by the theorem on the alternate interior angles the lines will be parallel.


1. Z l S Z2 1. Given2. Z3 £ Z2 2. Why?3. Z l £ Z3 3. Why?4. / 1| m 4. If two lines are cut by a transversal

such that the alternate interior angles■r~ are congruent, then the lines are

, parallel.

t IlencefcVth, unless otherwise sta ted , all points and 'lines in th e s ta tem e n t of any theorem o r probler i should be considered to be coplanar.


T H E O R E M 25: I f two lines a rc cut by a transversal such th a t th e a lte rn a te ex terio r angles a re co n g ru en t, th en th e lines a re p a ra lle l.

G iven: Z l 3S Z2 Concl.: / || m

Figure 8-26.

A n a l y s i s : By s h o w in g t h a t Z 3 = Z4 i t w ill fo llo w t h a t I || m f r o m t h e t h e o r e m o n t h e a l t e r n a t e i n t e r i o r a n g le s .

P R O O F

T he proof will be left for you to do.

T H E O R E M 26: If two lines a re p e rp en d icu la r to th e sam e lin e , th en th ey a re para lle l.

A c

! r . .. _ . r

Given: A B _L I

CD 1 I

Concl.: AB II CD

B DFigure 8-27.

A n a l y s i s : By showing that a pair of corresponding angles are congruent

it will,follow th a t A B || CD.

Although there are a num ber of other methods for proving lines parallel, the three given in theorems 23, 24, and 25 have the widest application.' Should you be required to prove that lines are parallel, try to prove th a t one of the following relations is true:

(1) T h e alternate interior angles formed by the lines and a transversal are congruent,(2) T h e alternate exterior angles formed by the lines and a transversal are congruent.(3) T he corresponding angles formed by the lines and a transversal are congruent.

PARALLELISM—SECTION II 241

Illustration:

F ig u r e 8-28.

G iven: AC ^ EF

B E S D F Z l S Z2

f C oncl.: A B || CD

G

A nalysis: Bv using AC as a transversal with reference to A B and CD, we note that a pair of corresponding angles is formed. These are A A and DCF. By proving A A B E — A C D F these angles will be congruent, and hence,

► 4-+A B will be parallel to CD.

PRO O F | (The reasons will he left for you to supply.)

1. A C ^ E F 7. But, Z l m Z2

2. C E SZU E 8. Z A E B S /-CFD [a)

3. AE — CF (s) 9. A A B E S A CDF

4. B E = D F (s)10. Z A S ZD C F

*-> «->5. Z AEB is supp. to Z l.6. ZCFD is supp. to Z2.

11. .-. A B || CD

EXERCISES

1. G iven: Z l = Z2 A B S Z D C

Concl.: l o l l BC

Given: A B and CD bisect each other a t E.

Concl.: A D l| CB


3 . Given: Z l — Z2

B D ^ C E

AC S DF

C oncl.: AB II EF

5 . G iven: CE = B f A E ^ D F / .A EC S Z D F B

Concl.: A B II CD

7 , Given: 0 0 with 0 the m idpoint of E F

Concl.: A B \ \ c b

Given: A B ^ E F 4 .BD S CE

AC ^ D F

Concl.: AC\\ D F

Given: A B ££ FC ■ 6 .B C S Z m *-* <->

A B 1 B E<-► «-vFC JL B E

C oncl.: AD || E F

By using the diagram below 8 > prove Theorem 24 again.Do not label any other an gles in the diagram .

9 . By using the diagram below prove Theorem 25 again. Do not label any other angles in the diagram.

PARALLELISM—SECTION II

“11. Given: Z l is supp. to Z2. Concl.: || m

13. Given: Z l = Z4

Z2 S Z3

Concl.: A 5 | | ^ £

Given: Z l is supp. to Z2.

Concl. ■ T b \\ CD

10.

243

Given: Z l S Z2Z 2 is supp. to Z3.

12.

Given: CD X FD

E F 1 FD /.C D A S Z E F B

C oncl.: AD || FB

14.


15. Given: /A G E SZ Z D EG

GH bisects /A G E .

EF bisects /D E C .

C or.cl.: G H \\<EF

17. Given: A ABC is isosceles with AB S AC. / \ ^ / C

Concl.: I d || BC

19.* Given: A B ^ A C

a d ^ a S

A F bisects /D A G .

Concl.: S C | |5 G ( H in t : Prove that A AEC and AFG are right angles.)

Given: / J E B S /E G D 16 .

E F bisects / J E B .

GH bisects /E G D .

Given: & A B C is isosceles 1 8 . with A B ^ A C .

CB bisects /A C D .

G iven: O 0 with E F bi- 20 .* secting A B and CD

Concl.: AB || CD

E

PARALLELISM— SECTION III 245

B1. If two lines are cut by a transversal such that the interior angles on the

same side of the transversal are supplementary, then the lines are parallel.2. If the opposite sides of a four-sided polygon are congruent, then they are

also parallel.3. If a pair of corresponding sides of two congruent triangles fall on the same

line but do not coincide, then either the other pairs of corresponding sides will be parallel or they will intersect to form isosceles triangles when extended.

4. Use the indirect proof to prove the theorem that if two lines are cut by a transversal such that the corresponding angles are congruent, then the lines are parallel.

■ Parallelism— Section IIIW hen developing the theorem th a t two lines will be parallel

when the alternate interior angles are congruent, we examined the relative positions of the lines I and m as the line I ro tated about the point P. In its original position we knew that the measure of / \ had to be greater than that of / 2 . But as I was rotated about the point P, although the measure of

/ 2 did not change, m / \ decreased toward zero. By using the indirect proof it was possible for. us to show that when m / \ reached the m Z 2, the lines I and m could not intersect. W hat we did not investigate, however, was the possibility that I m ight not intersect m for some value of m / \ where m / \ was ro t equal to m Z 2 !

We did note in passing (see page 224) that as m / \ approached m / 2 in size, a small amount of rotation threw the point of intersection of I and m a great distance to the right. Assume that in rotating I counterclockwise the position was eventually reached wherein the point of intersection between I and m could not be found. W hat would that imply concerning these two. lines? Let us say also that when this position is reached, the m / \ is still


larger than m Z2. Since we already know that / does not intersect m when m Z l equals m Z2, we are faced with the situation th a t there m ust be at least two lines through P that do not intersect m. In fact, had we rotated line I about, point P in a clockwise direction, we m ight again come to a first position of ! on the left where / does not intersect m. B y’virtue of this pos-

mFigure 8-31.

sibility, it would seem th a t we have a first parallel on the right, P R ; a first

parallel on the left, PL\ and a whole mass of lines, such as PA, w ith in the angle formed by their intersection that also do not intersect line m. Can you

show by the indirect proof that if PR does not intersect m, PA can not intersect m?

For over 2,000 years m athematicians were disturbed by this analysis. I t was their feeling that the drawing above was merely an optical illusion and th a t the two parallels through P were in reality but a single line. And so, they, like we, were faced with two roads down which we m ight travel: either

(1) Through P there exists one and only one line that does not intersect m.or

(2) Through P there exist m any lines that do not intersect m.

W hich of these possibilities do you believe to be the more plausible?T h e collector and organizer of the subject of geometry, Euclid, lived

approxim ately 330 B.C. In the process of logically arranging the definitions, postulates, and theorems, lie, too, encountered the difficulty described above. After trying unsuccessfully ta_£rove that through P there can be only one line that does not intersect/m .flife finally resorted to listing this as his fifth

t T h e form of Euclid's s ta tem e n t was not the sam e as this, b u t equ ivalen t to this. T h is p a rtic u la r statem ent is a ttr ib u te d (o Playfair, a m athem atic ian w ho lived d u rin g the e igh teen th century. ~--------

PARALLEUSM-SECTION III 247

postulate. M athem aticians that followed him, however, were not satisfied that this statement could not be proved in terms of E uclid’s preceding four postulates. Hcnce, for over 2,000 years, “ proofs” containing subtle errors in reasoning were presented for the fifth postulate.

Around the year 1815 A.D., two men, living thousands of miles apart, came out with separate studies showing what would happen to the subject m atter of geometry were the second road followed ra ther than the first. These men were .Johann Bolyai and Nicolai Lobachevsky.

This, though, was not the end. About the year 1855 a th ird m athem atician, Bernhard R iem ann, set forth the possibility that: •

(3) Through P there exists no line that can be draw n th a t does not intersect m.

Your immediate reaction will be to say, “ This is foolish. H ave not we proved that when the alternate interior angles are congruent, the lines can not intersect!” Yes, but if you examine the proof of this statem ent, you will see th a t it depends on the theorem about the exterior angle of a triangle. If the proof of that theorem was carefully analyzed, it would tu rn up the fact that that proof depends upon the postulate th a t one and only one line exists through two points. Hence, in developing his geometry, R iem ann had to discard this postulate also.

Well, where does this lead us? Can we ask the question, “ W hich of the three is the ‘correct’ postulate?” No, for this would indicate th a t we have forgotten the foundation laid in the earl;? weeks of this course. At that time it was pointed out that it m attered not what our definitions and postulates were just so long as we agreed upon them and that they did not contradict one another. Each m athem atician has the right to start w ith his own postulates and pursue them as he will.

A more interesting question might have been “ W hich postulate leads to results that are in keeping with the world we observe about us?” This, too, has to be answered by hedging. R iem annian geom etry gives most accura te results when dealing with astronomical distances; Lobachevsky’s geometry leads to greatest accuracy when dealing With the m ovem ent of atoms within a molecule; all three geometries give com parable results when dealing with measurements on a small area of the earth’s surface. Euclidean geometry, however, is far easier to apply and far easier to learn. In view of the latter reason, this is the path we will pursue. At some la te r date we will take a short detour to browse about the fields we m ight have observed had we followed the roads taken by Lobachevsky and R iem ann.

P ostulate 25: Through a given point not on a given line there exists one and only one line th a t is parallel to the given line. (T he Parallel Postu- late, or Euclid’s Fifth Postulate)

W ith the aid of this postulate we are prepared to prove the converses, of the earlier theorems on parallel lines. This is the first tim e the term

248 THE INDIRECT PROOF AN D PARALLELISMconverse has been used, although there were earlier occasions when, perhaps, we should have used i t j If two statements are so related wherein the given da ta of the first is the conclusion of the second and also the conclusion of the first is the given data of the second, then the second statem ent is said to be the converse of the first.

First Statem ent:given conclusion

Second Statem ent, or Converse:

You may have noticed that were these related statements definitions, then the second would be the “ Reverse” of the first. All such related statements other than definitions, however, are called the converses of each other. L ater in our work we will examine converse statem ents m uch m ore thoroughly. * '

THEOREM 27: If two parallel lines are cut by a transversal, th*n the alternate interior~an gl es~arecnngriient.

Given: a || b Concl.: Z l S Z2

Figure 8-32.

A n a l y s is : T here is nothing we know that relates parallel lines to congruent angles. Thus, as with the first theorem on proving lines parallel, we will have to resort again to the indirect proof. Assuming th a t Z l Z 2 will enable us to draw a second line through P to m ake an angle w ith c that is congruent to Z2. Hence, this second line through P will also be parallel to b. W hy?

t See theorems 9 and 10, page 145.

This will lead to a statement contradictory to Postulate 25. W hat is this

postulate?

PRO O F

By the law of the excluded m iddle one of the following statements must be true and no other possibilities exist:

Z l S Z2 or Z l m C l

Let us accept the possibility that Z l C l. T hen by Postulate 17 concerning the existence of an angle at a particular point on a line, we can say that there exists an Z3 at point P that is congruent to Z2. From this it follows that 1 1| b, for if the alternate interior angles a re congruent, the lines are parallel. But the Given Data states that a is already parallel to 6. Therefore, there are two lines through P parallel to b. Postulate 25, however, limits io only one the num ber of lines that can exist through P parallel to b. T herefore, accepting the possibility that Z l Z2 led to a logical inconsistency of the truth of both statements. By the law of contradiction both cannot be true a t the same time. Since we have accepted Postulate 25 as true, the statem ent that both / and a are parallel to b m ust be false, and, therefore, the statem ent that Z l 3^ Z2 is also false. Hence, Z l == Z2, for it is the only

remaining possibility.T H E O R E M 28: If two p aralle l lines a r f m t ny a transversal, th en th e

corresponding angles a re congruen t.

PARALLELISM—SECTION III 249

A nalysis: Since the lines are parallel, Theorem 27 enables us to say that Z2 — Z3. But Z l is also congruent to Z3, hence Z l and Z 2 are congruent.


1. a | i i 1. Given

2. Z2 S Z3 2. If two parallel lines are cut by atransversal, then the alternate interior ang\r* are congruent.

3. Z l £* Z3 3. Why?

'A. Z l £* Z2 4. Transitive property of congruence.

250 THE INDIRECT PROOF A N D P A R A LLE LIS M

T H E O R E M 29: If two p a ra lle l lines are cut by a tran sv e rsa l,.th e n th e a lte rn a te ex te rio r angles are co n g ru en t.

----------------------- — -

Figure 8-35.

A n a l y s j s : From the fact that the lines are parallel Z3 ~ Z4. Since Z l and Z 2 are congruent respectively to these two angles, they will be congruent to each other.

P R O O F

The proof will be left for you to do.

The next two theorems can very easily be proved on the basis of the previous three theorems on parallel lines. T heir proofs will be left for you to do.

\J THEOREM 30: If two lines are parallel to the same lin e , then they are parallel to each other.

THEOREM 31: If a lin e is perpendicular to one of two parallel lines, then it is also perpendicular to the other.

Illustration:

■If a line intersects the legs of an isosceles triangle and is parallel to the base, it cuts off another isosceles triangle.

Given: A ABC is isosceles with J B ^ A C .

D E !| BC C oncl.: A AD E is isosceles.

Figure J-36.

A n a ly sis : Since Z B S ZC, it w'H be possible to show th a t / '.ADE is congruent to ZAED . Hence, it follows th a t AD £= A E and, therefore, th a t A A D E is isosceles.

PARALLELISM—SECTION III 251

PROOF I STATEMENTS REASONS

i . j B S A C 1. Why?

2 . / B ^ / C 2. Why?

3. Afe || BC 3. Why?

4. /A D E = Z B 4. If two' parallel lines are cut by atransversal, then the correspondingangles are congruent.

5. / A E D & / C 5. Why?

6. /A D E -S /A E D 6. Why?

7. : . A D ^ A E 7. If two angles of a triangle are congruent, then the sides opposite theseangles are congruent.

8 . A AD E is isosceles. 1 8. Reverse ofdef. of an isosceles triangle

A

252

5 . Given: Q O with A B and CD intersecting a t 0

6 .

THE INDIRECT PROOF AND PARALLELISM

*-* (-*AD || CB

Concl.: O D S iO C

7 . Given: a || bc || d

Concl.: Z l S Z2

9 . Given: A B \\D C I «-> <->

AD || BC

Concl.: A B SS DC

M - D ‘ - $ / i - z i - ,

. . . . v - ^

Given: a || b c || d

C onci.: Z l = Z2

Given: Z l ^ Z2 Concl.: Z3 — Z4

Given: AD and BC intersecting lines

A B || CD

EC ZZED Concl.: A E A B is isosceles.

8.

10.

PARALLELISM—SECTION III

11. Given: 0 0 with A B || CD Concl.: &OCD is isosceles.

13. Given: A B || DE

BC || E F .Concl.: Z B = Z E (H int:

4-4Extend D E until it

intersects BC.)

15. Given: A B \\ DC A B SH D C

~ B F ~ E D

A E 1 BD

C oncl.: CF X BD

Given: AB || CD Z D S Z B

Concl.: B C \\D E

Given: A B || DC

A D \\B C

J f ^ d e

Concl.: *AE || FC

Given: A B || DC

J B ^ W+-+ __

E F bisects AC.«-+ __

C oncl.: AC bisects EF.

£


17. Given: EC bisects /A E D .

A B || CD Cone!.: A D E C is isosceles.

19. Given: 0 0 with AD || BC Concl.: DC (H int:

Draw lines AO and DO.)

Given: AD bisects /.C A E. 18.AD || BC

Concl.: & A B C is isosceles.

Using the diagram below and 20. the indirect proof, prove the theorem that if two parallel lines are cut by a transversal, then the corresponding an gles are congruent.

B

1. If two sides of a four-sided polygon are congruent and parallel, then the other two sides are congruent.

2. If two sides of a four-sided polygon are congruent and parallel, then the other two sides are parallel.

3. If both pairs of opposite sides of a four-sided polygon are parallel, then the opposite angles are congruent.

4. If both pairs of opposite sides of a four-sided polygon a te parallel, then the line segments join ing opposite vertices bisect each other.

5. I f two parallel lines are cut by a transversal, the interior angles on the sam e side of the transversal are supplementary.

6. If a line is drawn parallel to a leg of an isosceles triangle and intersects the other two sides, then it cuts off another isosceles triangle,

7. If through any point on the bisector of an angle a line is drawn parallel

UNIQUENESS AND EXISTENCE 255

to one of the sides of the angle, then an isosceles triangle is formed.8. The bisectors of a pair of corresponding angles of parallel lines are

parallel.9. The bisectors of a pair of alternate extefior angles of parallel lines are

parallel.10. State and prove the converse of example 4.11. If the bisector of an exterior angle of a triangle is parallel to one of the

sides of the triangle, then the triangle is isosceles.12. If a triangle is isosceles, then the line through the vertex of the vertex

angle parallel.to the base bisects the exterior angle at th a t vertex.13. If both pairs of opposite sides o f a four-sided pclygon are parallel, then

a line joining opposite vertices that bisects one of the angles will bisect

the other also.14. If two lines are parallel, then perpendicular segments draw n from two

points of the first to the second are congruent.15. I f the sides of one angle are parallel to the sides of ano ther angle, then

the angles are either supplementary or congruent.16. By using the indirect proof, prove the theorem that if two lines are

parallel to the same line, then they are parallel to each other.17. If a line intersects one of two parallel lines, then it intersects the other

also. (Use the indirect m ethod of proof.)18. If a line intersects one side of a triangle and is parallel to the second •

side, then it must intersect the third side of the triangle, (Use the

indirect method of proof.)19.'' If two lines are parallel respectively to two intersecting lines, then the

first two lines must intersect each other. (Use the indirect m ethod of proof.)

20.* By using the indirect proof and Pasch’s Axiom prove that a line that is parallel to one side of a triangle and passes through the vertex formed by the other two, can not pass into the interior of the triangle.

■ Uniqueness and ExistenceT he Parallel Postulate and several of the problem s in the

set of exercises on page 233 afford us an excellent opportunity to call a ttention to an im portant concept in mathematics. Thus, Problem 11 stated, “ At a given point on a given line there can be only one line perpendicular to the given line.” By proving this statem ent you have emphasized the fact that no more than one line can be draw n perpendicular to a line a t a particular point on that line. W hat you have confirmed, from a m athem atical standpoint, is the uniqueness of this perpendicular. You, as an individual, stand out, for there is only one such as you in the entire w orld; th a t is, you are


unique. Uniqueness, from a m athem atical standpoint, implies, too, th a t there is only one of these creatures under the conditions stated.

T hus, through the proof uf Problem ) 1 you have confirmed the uniqueness of a perpendicular at a given point on a giver. line. Yet, is this enough? No, for we wouid also like to know if such a line does exist? To show existence we m ust verify that such a creature if possible ir. terms of prior definitions and assumptions that we have made. T o illustrate this procedure we wil! show th a t there.m ust exist a perpendicular at a given point on a given line.

G iven: Line I and point P C oncl.: At point P there exists a perpen

dicular to line I.

P R O O F | STATEMENTS REASONS

1. L et A be a point of line / o th er th an point P.

2. Extend / through point Pso

th a t A P S * BP.3. A t A let Z.QAB be any acute

angle.4. A t B let / R B A be the angle

th a t is congruent to /Q A B .

5. QA and R B must intersect a t some point C.

6. Let CP be the line through points C and P,

l . C A & C B

■ ■ CP 1 I

1. A property given to a line was th a t it was a set of points.

2. A line can be extended as far as desired.

3. Postulate on the existence of an an gle. (Postulate 17)

4. Sam e as 3

5. See if you can prove by the indirect

m ethod that QA and RB can not be parallel.

6 . Why?

7. If two angles of a triangle are congruent, the sides opposite these angles are congruent.

8. If two points (C and P) are each equidistant from the endpoints of a line

segment (A B ), then the; line (CP) jo ining them is the perpendicular bisector of the line segment.

T he proof above and your proof to Problem 11 on page 233 establishes the theorem that

UNIQUENESS A N D EXISTENCE 257

T H E O R E M 32: A t a g iven p o in t on a g iven l in e there , exists one a n d on ly one line th a t is p e rp e n d ic u la r to th e g iven line ,

T he word “ one” in the above statem ent implies the existence of this perpendicular. Its existence we established by the proof above. T he words “only one” imply the uniqueness of this perpendicular. This uniqueness you established in the proof of Problem 11 on page 233. From a similar point of view, the Parallel Postulate establishes the uniqueness of a line through a given point parallel to a given line, while the theorem that if the alternate interior angles are congruent, the lines are parallel establishes the existence of such a line.

T here are several other statem ents on. existence and uniqueness that you have exam ined as problems in the past th a t will be re-examined now. O ne of these is Problem 12 on page 233. T hrough this problem you proved the uniqueness of a perpendicular from a given point not on a given line to the given line. Now, we shall establish th e existence of this perpendicular.

G iven : Line I and point P C oncl.: T hrough P there exists a per

pendicular to I.

P R O O F (The reasons will be left for you to supply.)

6. CA ~ CA7. A PAC ^ A B A C8. /.PCA /B C A

9 . P B 1 1

1. Let point A be any point of line I.<—►

2. Let PA be the line through P and A.3. Let /R A Q be the angle a t A that is congruent

to /P A Q .

4. Extend AR so th a t BA = PA.

5. Let PB be the line through P and B.

T h e proof above and Problem 12, page 233, establish the theorem that

T H E O R E M 33: From a given p o in t n o t on a g iv en line th ere exists nnff and n n l y n n - l u g th at is p e rp e n d ic u la r to th e g iv en line .

By your proof of Problem 9, page 233, you confirmed the uniqueness of the m idpoint of a line segment, while Postulate 20 informed us. of the existence of this point. Can you combine these two principles into a single statem ent? Similarly, proving problem 10, page 233, implies the uniqueness of a biscctor of an angle, while Postulate 16 brings to our attention the existence of this ray. C an you combine the two principles into a single statement?


EXERCISES

t . Is Problem 15, page 233, a “ uniqueness” statem ent or an “ existence” statement? How would the statem ent have been worded had it been both?

2. Prove: At the endpoint of a given ray there is only one angle on one side of this ray that is congruent to a given angle.(a) Is this a “ uniqueness” statem ent or an “ existence” statem ent?(b ) W hen have we encountered the other of these two statements?

3. Prove: There exists a point that is equidistant from two given points. (H int: See the diagram for the proof of Theorem 32.)

Is this point unique? If not, how m any such points are there and where do they lie?

4. Prove: There exists a point on a circle that is equidistant from two other points on the circle.

Is this point unique? If not, where does the other, or others, lie?5. Prove: There exists an isosceles triangle whose vertex angle is a given

angle.Is this triangle unique? If not, w hat relation do you believe will exist

between the bases of these triangles?6 . Prove: A line segment has one and only one perpendicular bisector.

(H in t: First show how to get one perpendicular bisector and then use the indirect method of proof to prove that this is the only perpendicular .bisector of this line segment.)

■ The Parallelogram— Part IReferring to a four-sided polygon as we have been until

now is both unnecessary and a bit awkward. Not only are we going to name this polygon but also we will devote this section to a discussion of the properties of spccial four-sided polygons.

D e f in i t io n 52: A quadrilateral is a polygon that has four sides.

In order to obtain an over-all picture of the words that are about to be b rought into this language, a family tree of the quadrilateral and its offsprings are pictured in Figure 8-39.

From the diagram in Figure 8-39 we note that there are two special types of quadrilaterals: the parallelogram and the trapezoid. In turn, the general variety of parallelograms has two special varieties: the rectangle and the rhombus. We will work our way down one side of the tree a t a time.

D e f in i t io n 53: A parallelogram is a quadrilateral in w hich the opposite sides are parallel.

It is quite simple to recall the property of a parallelogram given to it by its definition: the word parallelogram contains the term parallel as part

THE PARALLELOGRAM-PART I 259

p o l y g o n

Iq u a d r i la te ra l

p a ra l le lo g ra m

r e c t a n g le r h o m b u s

tr a p e z o id

iso sce les I r a p e i o i d

sq u a reF ig u r e 8-39.

of its spelling, while the property given to this quad rila te ra jjiy its definition is that the opposite sides are parallel. The symbol representing .the word parallelogram is the drawing of a small parallelogram, O . W hat property about the sides of a parallelogram do you believe yoti will be abl£ to prove?

Figure 8-40.

W hat.property about the angles? Draw a parallelogram ; then draw the.line segments joining the two pairs of opposite vertices. These line segments are called the diagonals of a parallelogram. W hat do you believe will be true about the diagonals of a parallelogram? “£m e c

D e f in it i o n 54: A rectangle is a parallelogram with one right angle.

Figure 8-41.

You may have wondered why the rectangle is not defined as a parallelogram w ith four right angles. T he reasoning behind this is very m uch the same as the reason for not defining an isosceles triangle as a triangle

(1) with two congruent sides,(2) w ith two congruent angles,(3) whose bisector of the vertex angle is perpendicular to the base,(4) whose bisector of the exterior angle at the vertex is parallel to the base.

Certainly what we have said of the isosceles triangle is true, for we haveproved all but one of these a t some time in our work. In form ulating a


EXERCISES

1. Is Problem 15, page 233, a “ uniqueness” statem ent or an “ existence” statement? How would the statem ent have been worded liad it been both?

2. Prove: At the endpoint of a given ray there is only one angle on one side of this ray that is congruent to a given angle.(a ) Is this a “ uniqueness” statem ent or an “ existence” statem ent.'1(b ) W hen have we encountered the other of these two statements?

3. Prove: T here exists a point that is equidistant from two given points. (H int: See the diagram for the proof of T heorem 32.)

Is this point unique? If not, how m any such points are there and where do they lie?

4. Prove: There exists a point on a circle that is equidistant from two other points on the circle.

Is this point unique? If not, where does the other, or others, lie?5. Prove: There exists an isosceles triangle whose vertex angle is a given

angle.Is this triangle unique? If not, w hat relation do you believe will exist

between the bases of these triangles?6 . Prove: A line segment has one and only one perpendicular bisector.

(H in t: First show how to get one perpendicular bisector and then use the indirect method of proof to prove that this is the only perpendicular .bisector of this line segment.)

■ The Parallelogram— Part IReferring to a four-sided polygon as we have been until

now is both unnecessary and a bit awkward. Not only are we going to nam e this polygon but also we will devote this section to a discussion of the p roperties of special four-sided polygons.

D efin itio n 52: A q u a d rila te ra l is a polygon th a t has four sides.

In order to obtain an over-all picture of the words th a t are about to be brought into this language, a family tree of the quadrilateral and its offsprings are pictured in Figure 8-39.

From the diagram in Figure 8-39 we note that there are two special types of quadrilaterals: the parallelogram and the trapezoid. In turn, the general variety of parallelograms has two special varieties: the rectangle and the rhombus. We will work our way down one side of the tree a t a time.

D e f in it i o n 53: A parallelogram is a quadrilateral in which the opposite sides are parallel.

I t is quite simple to recall the property of a parallelogram given to it . by its definition: the word parallelogram contains the term parallel as p a rt


polygon

I ,q u a d r i la te ra l

p a ra lle lo g ra m

re c ta n g le rh o m b u s

tr a p e z o id

isosce les t r a p e z o id

sq u areF ig u r e 8-39.

of its spelling, while the property given to this quadrilateraJJjy its definition is that the opposite sides are parallel. T he symbol rppfresenting .the word parallelogram is the drawing of a small parallelogr/m , O . W hat property about the sides of a parallelogram do you believe yoi» will be a b lt to prove?

Figure 8-40.

W hatp roperty about the angles? Draw a parallelogram ; then draw the.line segments joining the two pairs of opposite vertices. These line segments are called the diagonals of a parallelogram. W hat do you believe will be true about the diagonals of a parallelogram? ^ti e tfor w k .

D e f in i t io n 54: A rectangle is a parallelogram with one righ t angle.

Figure 8-41.

You may have wondered why the rectangle is not defined as a parallelogram with four right angles. T h e reasoning behind this is very m uch the same as the reason for not defining an isosceles triangle as a triangle

(1) with two congruent sides,(2) with two congruent angles,(3) whose bisector of the vertex angle is perpendicular to the base,(4) whose bisector of the exterior angle a t the vertex is parallel to the base.

Certainly what we have said of the isosceles triangle is true, for we have,proved a ll bu t one of these a t some tim e in. o u r work. In form ulating a


definition, however, we try to list as Jew properties as possible that will distinguish th a t term from all other v/ords in its class. All other properties of the figure we prove. Thus, if we say that a rectangle is a parallelogram that simply lias one right angle, this will suffice to distinguish it from all other parallelogram s. Furthermore, on the basis of this definition we can prove th a t a rectangle has lour right angles.

D e f i n i t i o n 55: A square is a rectangle with two adjacent sides congruent.

/v . ^ ^

V-'"

F ig u r e 8-42.

In view of the discussion about the rectangle, can you justify why the square was not defined as a rectangle with four congruent sides? If we had defined the square by placing it in the category of polygons rather than in its nearest class, the rectangle, then w hat properties should we have included in its definition to distinguish the square from the other polygons?

D e f i n i t i o n 56: A rhom bus is a parallelogram with two adjacent sides congruent.

D

W hat property do you believe, from observation, that a rhom bys has th a t is no t given to it by its definition? W hat property does the square have th a t the rhom bus does not have?

Now we will move down the other branch of the quadrilateral tree to exam ine the second set of terms.

D e f i n i t i o n 57: A trapezoid is a quadrilateral that has one and/only one. pair of sides parallel.

JJEEor base

lower base

Figure 8-44,


T h e parallel sides of a trapezoid are called the bases. In w hat way does a trapezoid differ from a parallelogram?

D e f in it io n 58: An isosceles trapezoid is a trapezoid in which the nonparallel sides are congruent.

^ u p p e r b a s e 0

lo w e r b as e a n g le s

u p p e r b as e a n g le s

l o w e r b as e

Figure 8-45.

Notice that there is a similarity between nam ing the parts of an isosceles trapezoid and the parts of an isosceles triangle. W hat property does the isosceles trapezoid have that is not given to it by its definition?

THEOREM 34: T he opposite sides of a parallelogram are congruent.

^ '*------------------------------’ 0 G iven: ABCD is a O .Concl.: A D ^ B C

~a b s d €_^CFigure 8-46.

Analysis: By drawing a diagonal we can prove that the two triangles that are formed will be congruent and, hence, our conclusion will follow. .


1. ABCD is a parallelogram.

2. AD || BC, A B || DC

3. Let AC be the line through points A and C.

4 . Z D AC S ZB C A ( a )

5. Z D C A ^ Z B A C [ a )

6 . AC = 3 C ( j )

7. A B A C & A D C A

8 . AD S 5 C , A B S DG

1 . Given

2. Def. of a parallelogram

3. Why?

4 . Why?5. Why?

6 . Why?7. Why?

8. Why?

THEOREM 35: The opposite angles of a parallelogram are congruent.

A , --------------------------------------- , D G i v e n , A B C D k a E 3 .

C oncl.: / .A — Z C Z B & Z D

Figure 8-47.


A n a l y s is : A s in the preceding proof, drawing a diagonal will give us a pair of congruent triangles, and, hence, / B can be shown congruent to / D . By drawing the other diagonal the other pair of angles can be shown to be congruent.

PRO O FT he proof of Theorem 35 will be left for you to do.

T H E O R E M 36: T h e d iagonals of a para lle lo g ram bisect each e th e r.

® Given: ABCD is a O .—> ___

Concl.: AC bisects BD.__

BD bisects AC.Figure 8-48.

A n a l y s is : By proving A AED ~ A C E B , both conclusions will follow.

PR O O F | (The reasons will be left for you to supply.)

2. AD || BC3. /A D B ~ /C B D4. /.C AD — /B C A

5. A D = BC6. A AED = A C E B

1. ABCD is a □ . 7 . B E S iD E

8. E is the m idpoint of BD.

9. AC bisects WD.10. AE ^ CE

11. E is the m idpoint of AC.<-> __

12. BD bisects AC.

In summary we can say that should a quadrilateral be a parallelogram , there are immediately four conclusions that can be drawn:

(1) The opposite sides are parallel.(2) T he opposite sides are congruent.(3) T he opposite angles are congruent.(4) T he diagonals bisect each other.

It should be borne in m ind that these properties are true not only of the general parallelogram but also of each of the quadrilaterals that were classified as parallelograms. These are the rectangle, the square, and the rhombus.

T H E O R E M 37: All four sides of a square are congruent.

A —---------------- .0

G iven: ABCD is a square with

AB S DA.Concl.: J B ~ B C ~ C D ^ D A

Figure 8-49.

\


PROOF | STATEMENTS REASONS

1. ABCD is a square with 1. Given

A B S D A .

2. BC S DA 2. The opposite sides of a parallelogramare congruent.

3. CD = AB 3. Same as 2

4. 4. Transitive property of congruence

THEOREM 38: All four sides of a rhom bus are co n g ru en t,

The proof of T heorem 38 is left for you to do.

TH EO REM 39: T h e low er base angles of an isosceles trap ezo id are congruen t.

Given: ABCD is a trapezoid with

■ D C S A B .Concl.: / B ~ / C

Figure 8-50.


1. ABCD is a trapezoid. 1. Giveni—►

2. Let DE be the line through 2. T he Parallel Postulate

D that is parallel to AB.■*-> <-*

3. AD || BC 3. Def. of a trapezoid4. ABED is a parallelogram. 4, Reverse of def. of a parallelogram .

5. D E ^ A B 5. Why?

6. DC SZ IB 6. Given

7. D E — DC 7. Why?8. / C £ / D E C 8. Why?9. / B £ /D E C 9, If two lines are parallel, then the

corresponding angles are congruent.10. .-. / B £ / C 10. Why?

EXERCISES


A

1 . Given: ABCD is a □ .

E is the midpoint

of FG.Concl.: E is the midpoint

of AC.A 'f 0

j , G iven: ABCD is a O .

B E — D F4—> __________ I

C oncl.: E F bisects AC. ]

5 . Given: ABCD is a O .

B F S Z D E Concl.: Z l S / 2

Given: ABCD is a O . 2 .

Concl.: Z P s CE

G iven: ABCD is a O . 4 .F is the midpoint of AB.G is the midpoint

of GD.C oncl.:. £ is the midpoint

of I ? .

Given: ABCD is a O w ith 6. EF passing through G.<-♦ _EF bisects AD.<-> _

Concl.: EF bisects BC.


7 . Given: ABCD is a O .E is the midpoint

of AD.F is the midpoint of BC.

C oncl.: AG = HC

9 . Given: ABCD is a trapezoid

with AD || BC.

CA bisects /.BC D .

BD bisects /C B A . Concl.: ABCD is an isosceles

trapezoid (H int: Prove A, ADC and D AB are isosceles.)

A ______________ _ D

W , Given: ABCD is an isos. trap,

with AD || BC.

R S I b l W4—f

C oncl.: R S passes thru E.

A D

G iven: ABCD is an isosceles trapezoid with

AD II BC.C oncl.: A E B C is an isos

celes A .

Given: ABCD is a rectangle. 10. P , Q, R, and S are the midpoints of their respective sides.

Concl.: P Q ^ ^ R ^ R S . ^ S P (H int: Prove that all the angles of a rectangle are right angles.)

P

Given: ABCD is an isos. trap. 12. with AD || BC.

R S 1 bi. BU

C oncl.: RS passes thru E.


13. Given: ABCD is an isoscelestrapezoid with

* <—*AD ]| BC.E is the midpoint of AD.F is the m idpoint

of BC.

Concl.: EF JL AD (H int: Draw A F and DF, then prove them congruent.)

A £ o

Given: A A B C is isosceles 14.* with A B AC.

<r->PQ II AC

PR || AB Concl.: m A B + m AC —

perim eter of AQPR (H int: How would you define the perim eter of ;> polygon?)

B1. T he diagonals of an isosceles trapezoid are congruent.2. T h e diagonals of a rectangle are congruent.3. T h e JirjeJoining the m idpoints of two opposite sides of a parallelogram

bisects either diagonal of the parallelogram.4. A pa ir of consecutive angles of a parallelogram are supplem entary.5. T he upper base angles of an isosceles trapezoid are congruent.6. I f the nonparallel sides of an isosceles trapezoid a re extended un til they

intersect, two isosceles triangles will be formed.7. T h e diagonals of a rhom bus are perpendicular to each other.8. I f a diagonal of a parallelogram bisects one of th e two angles whoss

vertices it connects, then the parallelogram is a rhom bus.9. If the diagonals of a parallelogram are perpendicular to each other,

then the parallelogram is a rhombus.10.* If the bisectors of the upper base angles of an isosceles trapezoid in te r

sect a t the m idpoint of th e lower base, then two of th e three triangles formed will be congruent isosceles triangles.

11.* If the diagonals of a parallelogram are congruent, then th e parallelo-

THE PARALLELOGRAM-PART II 267

gram is a rectangle. (Hint: Extend the base through one of its vertices and prove that the adjacent angles at that vertex are congruent.)

12.* If the lines joining consecutive midpoints of the sides of a parallelogram form a rhombus, then the parallelogram is a rectangle. (H int: See the suggestion for Problem 11.)

13.* The perpendicular bisector of the lower base of an isosceles trapezoid passes through the midpoint of the upper base. (H int: Use the information from Problem 5.)

14.* The perpendicular bisector of the lower base of an isosceles trapezoid passes through the point of intersection of the bisectors of the lower base angles.

I The Parallelogram— Part I!

The theorems on the parallelogram in Part I enabled us to draw conclusions in the event the quadrilateral was a parallelogram. In this unit we are going to investigate those properties that will make a quadrilateral a parallelogram. Primarily, of course, we have at our disposal the reverse of the definition of a parallelogram. T h at is, by showing that the opposite sides of a quadrilateral are parallel, the quadrilateral will be a parallelogram. Hence, the first theorem developed to show that a quadrilateral is a parallelogram will, of necessity, be based on the reverse of the definition of a parallelogram.

TH EO R EM 40: I f the opposite sides of a q u a d rila te ra l a re congruen t,then th e q u a d rila te ra l is a paralle log ram .

Given: AD ~ BC AB ~ DC

Concl.: ABCD is a D .

Figure 8-51.

PRO O F (The reasons will be left for you to supply.)

1. AD S BC (s)

2. A B ^ 'S C (r)<—■>

3. Let BD be the line through points B and D.

4. BD BD (j)5. A A B D £* A CDB

6. Z A B D ZC D B

7. A B || CD8. Z A D B ZC BD

9. AD !| BC10. ABCD is a O . (Rev. of the def. of a

parallelogram.)


T H E O R E M 41: If th e d iagonals of a q u a d rila te ra l bisect each o th e r, then the q u a d rila te ra l is a p a ra lle lo g ram .

__Given: AC bisects bD .

BD bisects ~AC.C cncl.: ABCD is a CJ.

A n a l y s i s : By proving iw o pairs o f triangles congruent it is possible to show '

that AD == BC and A B — CD. T hen by Theorem 40, ABCD will have to be a parallelogram,

PR O O F

The proof of Theorem 41 is left for you to do.

You may have noticed that Theorems 40 and 41 are the converses of theorems that we had proved earlier. W hat were those theorems? T h e last of the basic methods for showing a quadrilateral to be a parallelogram is not the converse of a prior theorem. It does have wide application, though.

In each of the methods examined thus far, it has been necessary to show something to be true about both pairs of opposite sides of the quadrilateral before being able to conclude that it was a parallelogram. If, however, our information concerned itself w ith only one pair of sides, w hat properties would have to hold before we m ight conclude th a t the quadrilateral was a parallelogram? W ould it be sufficient to know th a t this pair of sides was ■ congruent? Can you draw a quadrilateral with one pair of sides congruent and yet with the quadrilateral not being a parallelogram? W ould it be sufficient to know th a t one pair of sides was parallel before concluding that the quadrilateral was a parallelogram? Can you draw a quadrilateral with one pair of sides parallel and yet the quadrilateral is not a parallelogram?

THEOREM 42: If a quadrilateral has one pair of sides congruent and parallel, then the quadrilateral is a parallelogram .

Given: AD || BC lD = * B C

Concl.: ABCD is a O .Figure 8-53.

A n a l y s is : By drawing in the diagonal, BD, it is possible to prove that the

two triangles formed are congruent. From this it will follow th a t AB S CD, and hence, by Theorem 40, ABCD will be a parallelogram.

Figure 8-52.

THE P A R A L L E L O G R A M -P A R T II 269

PR O O F

Sum m arized below is allgram .

Ways to Prove That a Q uadrilateral Is a

Parallelogram

(1) I f the opposite sides are parallel.

(2) I f the opposite sides are congruent.

(3) If the diagonals bisect each other.

(4) If two sides are congruen t and parallel.

T he proof of T heorem 42 is left for you to do.

the information we know about a parallelo-

Conclusions T hat Can BeDrawn If a Quadrilateral

Is a Parallelogram

(1) T he opposite sides are parallel.

(2) T he opposite sides are congruent.

(3) T he diagonals bisect each other.

(4) T he opposite angles are congruent.

Illustration:I f the bisectors of a pair of opposite angles of a parallelogram do not

coincide, then they form another parallelogram.

Given: ABCD is a O .

A F bisects / .B A D .

CE bisects /.B C D .Conc.l.: AFCE is a O .

A n aly sis : By proving A A B F to be congruent to A CDE i t will follow th a t TiF ~ ~UE. Since ABCD is a parallelogram, AD = and, therefore, Z S — FS. But A E and 7 C are also parallel by virtue of th e fact th a t ABCD is a parallelogram . Hence, AFCE is a parallelogram, for one pa ir of sides

are congruent and parallel.

Stating that A1i || FC implies the very same m eaning as AD || BC, since a line can be nam ed by the letters of any two points of th a t line. I t is usually considered best to nam e the parallel lines in terms of the letters th a t appear a t the vertices of the quadrilateral that is being shomn to be a parallelogram. In

this problem , this would imply saying that AF || FC rather- than AD j| BC:

Figure 8-54:.

\i


A E

I P R O O F (The reasons will be left for you to supply.) 1

1. ABCD is a parallelogram.

2. A B S i CD (s)3. / B ^ / D (a)4. Z BAD ^ /B C D

— ■)5. /4F bisects / .B A D .

6. CE bisects /B C D .7. /B A F 9 * /D C E (a)8 . A CDE

9. B F ~ D E

10. But, B C ^ J 5

11. ■■■FC ^ J E

12. However, AE || FC (Def. of a pa rallelogram)

13. /. AFCE is a parallelogram . (Theorem 42)

EXERCISES

1. Given: ABCD is a O . B E = zT )F

C oncl.: AECF is a O .

3 * G iven: AC || D F Z l £* Z2

C oncl.: ~BD = UE

Given: ABCD is an isosceles 2 . trapezoid with

Concl.

~AB^.7)G.Z l S Z2 ABED is a O .A____________D

Given: OO w ith AC and B D 4 , intersecting at 0

Concl.: ABCD is a O .


5 i Given: ABCD is a □ .*-* irX

A E and DF X BFConcl.: AEFD is a rectangle.

7 . G iven: ~AC and EF bisect each other at G.E is the m idpoint

of AD.F is the midpoint

oi~BC.Concl.: ABCD is a O .

9 . Given: ABCD is a □ .

mC oncl.: PQRS is a O .

Given: ABCD is a □ .

F A S f i C Concl.: PQRS is a O .

A P

6.

Given: ABCD is a O .A P -9 iC §

Concl.: PBQD is a CD,

G iven: M is the m idpoint o CBD.P is the m idpoint

oiA D .Q is the m idpoint

ofBC.Z l S Z2

Concl.: ABCD is a O .

10.


11. Given: BE is the m edian in A ABC./LDAC ~ ZFCA

Cocci.: ADCF is a O .

13. Given: All lines in this figure are coplaner.ABD E is a O .AEFC is a O .

Concl.: BDFC is a O . I

15. Given: ABCD is a O .~EE = & F

Concl.: AC bisects ~EF, (H int: Prove that AECF is | a O .) j

Given: ABCD is a O . 12.R, S, T , W a re the

midpoints of A M ,

’EM , UM, D M respectively.

Concl.: R S T W h a a .

Given: ABCD is a O . 14.E is the m idpoint ofZZ).F is the m idpoint

o ( W .Concl.: AFCE is a O .

G iven: D is the m idpoint 16. of A B .E is th e m idpointof 1C .E is the m idpoint

o fS F .Concl,: DBCF is a O .


17. Given:

Concl.

ABCD is not a O . P is the midpoint

of AT).Q is the midpoint

of W .AQCP is not a O . (H int: Use indirect proof.)

Given: D M — M BABCD is not a O .

Concl.: J M Q I M C

18.

LU1. If two parallel lines are cut by a transversal, then the bisectors of all

the interior angles will form a parallelogram.2. If a pair of opposite exterior angles of a parallelogram are bisected,

then another parallelogram will be formed.3. If the m edian to one side of a triangle is drawn and then extended its

own length, the line segments joining this endpoint with the endpoints of the side will form a parallelogram.

4. If consecutive midpoints of the sides of a parallelogram are joined in order, then another parallelogram will^be formed.

5. If consecutive pairs of angles of a quadrilateral are supplem entary, then the quadrilateral is a parallelogram.

6. If the diagonal AC of parallelogram ABCD is trisected a t points P and

Q where the order of the points on diagonal AC is A, P, Q, C, then PBQD is a parallelogram.

7. If each diagonal of a parallelogram is extended congruent segments in both directions and these new endpoints are joined in order, then the quadrilateral formed will be a parallelogram.

8. T he line segment joining the midpoints of a pair of opposite sides of a parallelogram will be congruent to either of the o ther two sides.


10 . ’

11.

T he bisectors of all four angles of a parallelogram form another pa rallelogram. (H int: Use information proved in the illustrative problem on page 269.)T he m edian to ihe hypotenuse of a right triangle is congrucnt to either of the segments of the hypotenuse. (H int: Extend the m edian its own length; then draw line segments from this endpoint to the endpoints of the hypotenuse.)If a line bisects one side of a triangle and is parallel to the second side, then it bisects the third side. (H int: See the diagram of Problem 16, G roup A.)

| Test and Review

Prove each of the following

1. Given: A B (| CD

GH || EF Concl.: Z l S Z2

AD || BC

A F = EC

Concl.:

G iven: A B and CD intersect a t center of 0 0 .

Concl.: AC || D B

Given: ABCD is a □ .A FED is a O ,

Concl.: 'BC — FE

\

TEST AND REVIEW 275

G iven: AF bisects exteriorangle DAB of A A B C . <-> ■*->

GB \\ AF C oncl.: &ABG is isosceles.

Given: ABCD is an isosceles 6 . trapezoid w ith

AD l| BC.R is the midpoint

of PB.S is the m idpoint

of PC.C oncl.: A P R S is isosceles.

(H in t: P r o v e A PBC iso s c e le s .)

7 . Given: ABCD is a quadrilateral.E is the midpoint

of AD.F is the midpoint

of "BC.

~AS and EF bisect each other.

C oncl.: ABCD is a O .

Given: A B = A C "BD S— C®

A F 1 DE

Concl.: BC || D E (H int:

IProve AF _L BC.)


9 . G iven:. A B 3= AC <-> <->

AD ± BC Gone!.: D is not the midpoint

of BC.

Given: A B ~ AC 1 0 .

AD does not bisect

~BC.

Cone!.: 'D B '^ .D C


1. A diagonal of a rhombus bisects two of the angles of the rhombus.2. I f the bisectors of the lower base angles of an isosceles trapezoid are

extended until they intersect, they will form an isosceles triangle with the lower base of the trapezoid.

3. I f a pair of opposite angles of a trapezoid are supplementary, then the trapezoid is isosceles.

4. If a pair of opposite angles of a quadrilateral are congruent while a pair of opposite sides are paraliel, then the quadrilateral is a parallelogram .

5. T he line joining the midpoints of a pair of opposite sides of a parallelogram is parallel to the remaining two sides of the parallelogram.

6. If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus.

7. If the lower base angles of a trapezoid a re congruent, then the trapezoid is isosceles.

8 . If a m edian is drawn to a side of a triangle and perpendicular segments are draw n from the endpoints of this side to the m edian (extended if necessary), then the perpendicular segments are congruent.

9. If the perpendicular bisector of the lower base of a trapezoid passes through the m idpoint of the upper base, then the trapezoid is isosceles.

10. I f a line is draw n through the endpoints of the medians to the legs of an isosceles triangle, i t will be parallel to the base of the triangle.

TRY THIS FOR FUN

Using the indirect proof, prove each of the following sta te

ments.1. If the lower base angles cf a trapezoid are not congruent, then the

trapezoid is not isosceles.2. If two line segments do not bisect each other, then the line segments

joining their endpoints do not form a parallelogram.3. If a line is perpendicular to one side of an angle, then it is not perpendicu

lar to the other side also.4. I f the diagonals of a parallelogram are not congruent, then the parallelo

gram is not a rectangle.5. If a line is not perpendicular to the bisector of an angle, it will not form

congruent angles with the sides of the angle.6. If the diagonals of a parallelogram are not perpendicular, then the ■

parallelogram is not a rhombus.7. If the perpendicular bisector of one side of a triangle does not pass through

the intersection of the other two sides, then these two sides are not c.on-V

gruent.8. If the bisectors of two angles of a given triangle do not form an isos

celes triangle with a side of the given triangle, then this triangle is not

isosceles.

■ Try This For FunIf the three medians of a triangle are drawn, they will

intersect a t a point that is one of the trisection points of each of the medians. W ere we to draw the triangle on a piece of cardboard, then cut the triangle away from the rest of the cardboard, it would be possible to balance this figure on the point of a pin th a t had been placed at the intersection of the medians. For this reason, this point of intersection is called the ccnter of

gravity of the triangle.Although we are not prepared to prove all th a t we, have ju st stated, we

can show that the point of intersection of two of the m edians is a trisection point of both of them.

Given: A ny■ A A B C with. A M and B N two of the medians

Concl.: m GM = \m AG

277*1


_ _ Suf f Sti0n: E x t e n d ^ s o that C M ~ M P . Ako extend -B N so that G N =S NQ. Draw BP, CP, AQ, CQ, and GC.

A

Parallelism in Space

AS YOU M IG H T SU SPECT, T H E R E A R E MANY principles in space geometry that will be similar to those on parallelism that we encountered in the geometry of the plane. In this unit we plan to call your attention to a few of these. It would be best, of course, to start w ith the definition of parallel planes.

D e f in it i o n 59: Parallel planes are two planes that do not have a point incommon.

Now that we made this definition, another look at an earlier definition is needed. Parallel lines were defined as “ two lines in the same plane that do not intersect.” The point can be raised as to why we insisted that the two lines lie in the .same plane. This was. done so as to elim inate certain lines in space that do not intersect but th a t we prefer not to think of as being parallel. Consider the ceiling and floor of your room ; any line drawn in the ceiling would never intersect a line that you had draw n on the floor, Yet, only some of these would you consider to be parallel to the line t h a t . was draw n on the floor. To illustrate, we would probably consider linesa. b, and t, but not x, to be parallel to W hat is there peculiar about the relation that exists between a an d ^ , or h and y, or c a n d ^ th a t does not exist between x and y? Notice that it appears as if a single plane could be draw n

279

280 PARALLELISM IN SPACE

Figure 9-1.

th a t m ight contain both a and y \ the same can be said of b and y and of c and y. T he pair of lines x and y, however, do not lie in any common plane. Thus, to have the definition of parallel lines agree with our idea of w hat we would like parallel lines to be, we insist that they must lie in the same plane.

■ Hence, the existence of parallel lines implies the existence of a plane that will contain them. This is m ade even a bit more em phatic by the agreem ent th a t this be the only plane that will contain them. Thus, we have created a fourth m ethod for determ ining a plane.

■ P o s t u l a t e 26: Two parallel lines determ ine a plane.

Lines such as x a n d y in the diagram above are called skew lines. Since . two intersecting lines determ ine a plane and so do two parallel lines, it

would seem alm ost natural that the definition of skew lines be

D e f i n i t i o n 60: Skew lines are two lines that are not coplanar.

In terms of the intersection of two lines how might you have defined skew lines? Look around your classroom and find several pairs of skew lines.

•In our earlier contact with space geometry there was no need to consider th e nature of the intersection of two planes. Henceforth, however, m any of the properties we develop will have their origin in knowing the set of points in which two planes intersect. T he drawing below is th a t of a room very m uch like your classroom.

Notice th a t the front wall and floor of the room have in common the points of the line EH. Similarly, the intersection of the rear wall and the floor are those points of the line FG for they are points of both the rear wall

PARALLELISM IN SPACE 281

and the floor. Where will the points lie that represent the intersection of the side wall AF and the rear w a l l? Where will the points lie that represent the intersection of the side wall CH and the front w a l l?

P o s t u l a t e 27: The intersection of two planes is a line.

W hat name would you give to the pair of planes that are the ceiling and floor of your classroom? In Figure 9-2 these are the parallel planes BD and FH. Notice that the ceiling intersects the front wall in the line AD, while the floor intersects the front wall in the line EH. It would appear as

if these two lines of intersection, AD and EH} are parallel. Similarly, the ceiling and floor intersect the side wall AF in the lines AB and EF, which also appear to be parallel: In what way are the two side walls related to each other? In what lines does the rear wall intersect the two side walls? W hat appears to be true about these lines? In view of this analysis, w hat proposition do you think can be proved?

THEOREM 43: If a plane intersects two parallel planes, the lines of intersection w ill be parallel.

z .Given: a || b

Plane c intersects a and b in / and m.

Z Concl.: / 1| m

A n a l y s is : M any, many of the theorems in space geometry— and particularly those on parallelism—are developed on the basis of the indirect proof. At this stage of the work this method of approach will probably lead to success more often than not.

PRO OF

By the law of the excluded m iddle one of the following statements must be true and no other possibility exists:

/ || m or / X m

Let us accept the possibility that I % m. Since the two lines lie in the same plane, c, then being “ not parallel” implies that they intersect a t some point* P. Since P lies in I, P must lie in a by definition of a plane Similarly, since P lies in m, it must lie in b. This means that a and b are not parallel, for they have the point P in common. The Given D ata, however, states th a t a |j b. Therefore, accepting the possibility that / % m • led to the logical inconsistency of the truth of both a || b and a X b. By the lav,' of contradic

» \ / T .Z /

Z l

Z . y iFigure 9-3.


tion both cannot be true at the same time. Since a l| b must be true, for it is part of the Given D ata, then a X b must be false and, therefore, I X m is also false. Hence, i || m must be true for it is the only rem aining possibility.

THEOREM 44: If two p lanes are p e rp en d icu la r to th e sam e lin e , th en th ey a re p a ra lle l.

Given: / _L a I X b

C oncl.: a |j b

PR O O F

By the law of the excluded m iddle one of the following statem ents must be true and no other possibility exists:

a || b or a X b

If we accept the possibility that a X b, it will imply that a and b have some point, P, in common. T he line / and the point P will determ ine a plane th a t

intersects plane a in the line PR and the plane b in the line PS. PR m ust be perpendicular to line I, for, by definition, if a line is perpendicular to a plane, it is perpendicular to every line in the plane th a t passes through its

foot. This will make Z l a right angle. In the same way, PS is perpendicular to line I, and therefore Z2 is also a right angle. Thus, m Z l = m Z2.

PARALLELISM IN SPACE 283

However, the theorem on the exterior angle of a triangle enables us to say that m Z l > m Z2. Therefore, accepting the possibility th a t a X b led to the logical inconsistency of the tru th of both m Z l > m Z 2 and m Z l > m Z2. By the law of contradiction both cannot be true a t the same time. Since m Z l > m Z2 must be true, for it is a theorem, then m Z l > m Z2 must be false and, therefore, the statem ent a X b is also false. Hence, a || b must be true, for it is the only remaining possibility.

Notice the similarity between the statem ent of Theorem 44 and that of Theorem 26.

TH EO R EM 45: If two planes a re p a ra lle l to th e sam e p lan e , th en they are p a ra lle l to each other.

/ H 7

r ~ z u

n m uFigure 9-6.

PRO O F


a || b or a X b

If we accept the possibility that a X b, it will imply that they have some point, P, in common. Let this point and any line, /, in plane c determ ine a plane. This plane will intersect plane a in line PR, which will be parallel

R

5

J_________________________________

Figure 9-7.

to line I, for if a plane intersects two parallel planes, the lines of intersection are parallel. In the same way, this plane will intersect plane b in line PS,

Given: a || c b || c

C oncl.: a || b


which will also be parallel to line I. This means that there are two lines through P parallel to I. This, however, contradicts the Parallel Postulate, which states that through P there can be only one line parallel to I. T h ere fore, accepting the possibilit) that « if led to the logical inconsistency of the truth of both statements. By the iaw of contradiction both cannot be true at the same time. Since wc have accepted the truth of the Parallel

Postulate, the statem ent that both PR and PS arc parallel to I must be false and, therefore, the statem ent that a Jf b is also false. Hence, n || b must be true, for it is the only remaining possibility.

Notice the similarity between the statem ent of Theorem 45 and th a t of Theorem 30. If, however, the statem ent of Theorem 30 were applied to space geometry, its proof would be quite difficult. We will assume, though, that it is true.

P o stula te 28: I f t wo hnes in space are paralle l to th e sam e line, they a re p a ra lle l to each o ther.

TH EO REM 46: ^ a ^ ne ** perpendicular to one of two parallel planes, it is perpendicular to the other also.

Given: a |[ b I ± a

Concl.: I X b

A n a l y s i s : In o rd e r th a t / be p e rp en d icu la r to b, it is necessary to p rove th a t it is p e rp en d icu la r to tw o lines passing th rough its foot; these will be

Q T a n d Q W .


1. In plane a select any point R<-4

and let PR be the line passing through these two points.

2. 1 and P R determ ine plane c.

1. T here exists one and only one line through two points.

2. Two intersecting lines determ ine a plane.

3. T he intersection of two planes is a line.

4. Given

5. If a plane intersects two parallel planes, the lines of intersection are parallel.

6 . Given

7. Def. of a line perpendicular to a plane

8. If a line is perpendicular to one of two parallel lines, it is also, perpendicular to the other.

i ^In the same way, by draw ing PS in plane a it is possible to show th a t I

is perpendicular to QfV.9. / X b '

285

EXERCISES

1. Given: a || b

A B || CD

C cncl.: A B ^ C D

9. A line is p e rp e n d ic u la r to a p lan e if it is p e rp e n d ic u la r to a t least two lines in th e p lan e passing th ro u g h its

foot.

A |

Given: PQ || RS 2 .

PQ || V T

oi l *Concl.: A R P V = & S Q T

PARALLELISM IN SPACE i—►

3. c intersects b in Q T.

4. a || b♦ <r-t

5. PR || Q r

6. I X a

1. I X PR

8. ! X QT

3 . G iven: a j| b

P R \\S Q C oncl.: a nd J r bisect

each other.

286p a r a l l e l is m i n s p a c e

4.Given: a || 4

P $ s F r Concl.: Q S ^ R T

5 . Given: a X I

b l l

PQ = RS

c contains PQ and RS.Concl.:

G iven: a X / b l l

P $ = z K s

c contains PQ and RS. C oncl.: PS S RQ

6 .

PARALLELISM IN SPACE

Concl.

p i L .a l l b X Ic and d contain I. ZPQR S Z S T W

(Hint: Make .PQ = S T

and RQ = TW, then prove A PQR = ASTW .)

9 . By using the diagram and Given D ata in Problem 8, prove that OABCD ^ OEFGH. (H int: See the definition of congruent polygons on page 118.)

Given: plane BD j| plane FH plane BE |] plane CH plane AH || plane BC

C oncl.: AC = EG (The figure in this problem ..

is called a parallelepiped.)

G iven: Each of the pairs of 10. opposite planes are parallel.Lines FC, EH, AD, and BC are perpendicular to planes AF ■ and DG.

Concl.: AG == FD (Assume that AG

and FD intersect.)


11, Given : a j| bc || b

> __ 'W S bisects PR at Q.

Cone!.: PR bisects IVS.

G iven: a || b 1 2 . *c II h

J T = T W

Concl.: = QR

(H int: Let UV be the line through Tthat is parallel to

.PR.)

BUse the indirect proof for each of the problems in this group.

1. Tw o lines can not be perpendicular to the same plane at the same po in t on th e plane, f

2. From a given point not on a given plane there can be no more than one line perpendicular to the given plane, t

3. Tw o lines that are both perpendicular to the same plane can not in tersect.

4. T hrough a given point not on a given plane two planes can not exist th a t are both parallel to the given plane,

5. A t a given point on a given line two planes can not exist that are both perpendicular to the given line.

6. T h rough a given point not on a given line two planes can not exist th a t are both perpendicular to the given line.

7. I f a given line is perpendicular to a given plane, then any line th a t is perpendicular to the given line a t its foot lies in the given plane.

t T h e se s ta tem en ts arc often considered as theorem s.

DIHEDRAL ANGLES 289

8. * If a given line is parallel to a given plane, then the intersection of anyplane containing this line with the given plane must be parallel to the

given line.t9.* If each of two intersecting lines is parallel to a given plane, then the

plane determined by these lines is parallel to the given plane.f

B Dihedral AnglesAt this time we want to examine the figure in space geom

etry that is comparable to the angle in plane geometry. T h is figure is called a dihedral angle. T o define it, however, will require a num ber of terms that we have not encountered as yet. T he first of these is the half*plane. Consider in Figure 9-9 the points in plane m with reference to the line A B . If the line

PS is drawn, it will intersect AB in Q. When this point of intersection, Q, is not between the two points P and S, then these two points are said to be

on the same side of AB. O n the other hand, the line P N intersects A B at some point R such that R is between P and M. In this event, P and N are

<—►said to be on opposite sides of AB. Those points that are on the same side of

AB are said to be in the half-plane with reference to A B where AB is the edge of that half-plane,

D efinition 6 1 : Points on the same side of a given line in a given plane are the set of points not containing the line such th a t if a line is drawn through any two points o f the set, it will intersect the given line a t a point th a t is no t between these two points.

D efin itio n 62: a ha lf-p lane is the set of poin ts on the sam e side o f a given line in a given p lane .

' T he given line may act as an edge for m any half-planes. To illustrate, the bound edge of your book can be considered as the edge of those halfplanes consisting of the pages of the book,

t These statem ents arc often considered as theorem s.


A B is the edge of half-planes m, n, and p.

D efin it io n 63: A d ihed ra l ang le is a set of poin ts consisting of th e u n io n of tw o half-p lanes an d th e edge com m on to them .

From this definition it is quite apparent th a t unless two half-planes have a common edge, there will be no dihedral angle. T h e edge of a d ihedral angle is comparable to the vertex of an angle, while the half-planes A F and AG are similar to the sides of an angle. T h e half-planes A F and AG are called the facts of the dihedral angle.

A dihedral angle is named by using a point in each half-plane and two points along the edge. If there is no possibility for confusion, a dihedral angle can also be named by the two points along its edge. In the figure above, the dihedral angle can be named as either / G -A B -F or IA B .

DIHEDRAL ANGLES 291

O ur next concern is with the measure of a dihedral angle. This problem is overcome rather easily by relating an angle whose measure we know to that of the dihedral angle. To do this, we create the angle called the plane angle.

D e f i n it io n 64: A plane angle of a dihedral angle is an angle whose vertex lies on the edge of the dihedral angle and whose sides are perpendicular to the edgs, each side lying in a different face of the dihedral angle.

F ig u r e 9-12. F ig u r e 9-13 .

W ith this definition at our disposal, the measure of a dihedral angle becomes

D e fin it io n 65: T he measure of a dihedral angle is the measure of its plane angle.

I t m ight appear that by this definition we m ay have neatly entrapped ourselves, for a dihedral angle.has m any plane angles as seen in Figure 9-13. If the measures of all these plane angles are not the same, then it is evident th a t Definition 65 is poorly designed. Q uite fortunately this is not so, for it can be proved that

THEOREM 47: T he plane angles of a d ihedral angle are congruent.

P R O O FT he proof of this theorem is identically the same as that of

Problem 7, page 287. I t will be left for you to do.

I t seems almost needless to point out the next definition.

D e f in it i o n 6 6 : Congruent dihedral angles are dihedral angles that have thesame measure. “ " ’1A num ber of problems we would iike to present at this time are depend

ent upon a principle that seems obvious enough, yet would require several additional theorems hefore it can be proved. R ather than become involved in the proofs of these theorems that have little value for us in our work, wc will postulate the principle we need.

P ostulate 29: i f one 0 f two paralic! lines is p erp en d icu la r to a p lan e , th e o th e r is also.

Illustration:

I f two parallel planes arc cut by a third one, then the alternate interior dihedral angles are congruent.

A n a l y s i s : vVe will try to introduce a plane into this figure that will give usplane angles at both P and Q. I t will then be merely a m atter of provingp la n e ang les to b e co n g ru en t to show th a t th e a lte rn a te in terio r d ih ed ra l ang les a re co n g ru en t.


G iven: E F || K J Concl.: Z E - A B - H S

Z J -D C -G


1. E F \ \ K J

2. A B I CD

3. Let plane a be perpe'ndicu-4—>

lar to A B a t point P.

4. /, a J_ CD

5. a intersects planes EF, K J , and GH in lines PR, PQ, and VQ respectively.

. «-+ «->«-» <->6 . A B ± PR, A B ± PQ

REASONS

1. Given

2. If a plane intersects two parallel planes, then their lines of intersection are parallel.

3. See Problem $, page 288.

4. If one of two parallel lines is perpendicular to a plane, the other is also,

5. T he intersection of two planes is a line.

6. Def. of a line perpendicular to a plane

DIHEDRAL ANGLES 293

7. ZRPQ is the plane angle of Z E -A B -H .

8. CD 1 QV, CD 1 PQ9. ZPQ V is a plane angle

of Z J-D C -G .

10. But, PR || VQ11, m ZRPQ = m ZP Q V

12. m Z E -A B -H = m ZRPQ

13. m Z J-D C -G = m ZP Q V14. m Z E -A B -H

Z J-D C -G

15. Z E -A B -H Z J-D C -G

7. Def. of a plane angle

8. Same as 69. Same as 7

10. Same as 211. If two parallel lines are cut by a

transversal, then the alternate interior angles are congruent.

12. Def. of the m easure of a dihedral angle

13. Same as 1214. If two num bers are equal to two

equal num bers, they are equal to each other.

15. Def. of congruent dihedral angles

Having defined a dihedral angle, we are now in a position to define perpendicular planes. By considering the definition of perpendicular lines, what do you believe the definition of perpendicular planes will be?

D efinition 67: Perpendicular planes are two planes th a t intersect to form right dihedral angles.

By virtue of this definition it will be possible to prove planes to be perpendicular if we can show that they intersect to form right dihedral angles. T o show that a dihedral angle is a right dihedral angle will necessitate showing that the plane angle of that d ihedral angle is a right angle. However, you m ay recall that proving an angle to be a right angle was no simple task. To avoid the difficulty of proving lines to be perpendicular by resorting to the reverse of the definition of perpendicular lines, we developed the theorem that if two lines meet to form congruent adjacent angles, the lines would be perpendicular. Similarly, in space geometry we will try to prove planes perpendicular not by referring to the reverse of Definition 67, bu t rather by proving that the planes intersect to form congruent adjacent dihedral angles.

. First, though, it will be necessary for us to prove this property.

T H E O R E M 48: If two p lanes in tersect to form ' co n g ru en t ad jacen t d ih e d ra l angles, th en th e p lan es a re p e rp en d icu la r .


G iven: Z E -A B -D =-------- -y Z E -A B -C

/ Cone!.: plane E F X plane CD

D /Figure 9-15.

A n a l y s i s : By showing plane angles GHK and G H J to be congruent it will—* 4—>

follow th a t GH X K J . Therefore, /.G H J is a right angle. This, in turn, will make Z E -A B -D a right dihedral angle. Hence, plane EF will be perpendicular to plane CD by the reverse of the definition of perpendicular planes.


1. Let plane m be perpendic- 1. See Problem 5, page 288.i—>

ular to AB at point H.2. m intersects EF in line GH 2. The intersection of two planes is a

and CD in line K J. line.<-> <-><-> <-*

3. A B X GH, AB X K J 3. Def. of a line perpendicular to aplane

4. Z G H J is the plane angle of 4. Def. of a plane angleZ E -A B -D .

5. ZG H K is the plane angle of 5. Same as 4Z E -A B -C .

6. m Z E -A B -D = 6. Givenm Z E -A B -C

7. m Z G H J = 7. Def. of the m easure of a d ihedralm Z E -A B -D angle

8. m ZG H K = 8. Same as 7m Z E -A B -C

9. m Z G H J = m ZG H K 9. If two num bers are equal to twoequal numbers, then they are equalto each other,

10. GH X K J 10. If two lines intersect to form congruent adjacent angles, the lines areperpendicular.

11. Z G H J is a right angle. 11. Def. of perpendicular lines12. Z E -A B -D is a right di 12. Why?

hedral angle.13. Plane EF X plane CD 13. Why?

DIHEDRAL ANGLES 295

EXERCISES

1. If two dihedral angles are right d ihedral angles, then they are con

gruent.2. If two dihedral angles are vertical dihedral angles, then they are

congruent.3. Using the problem proved on page 292 and Problem 2 above, prove

th a t if two parallel planes are cut by a third plane, the alternate exterior dihedral angles will be congruent.

4. If two dihedral angles are congruent, the supplements of these two dihedral angles will be congruent.

5. By using the problem proved on page 292 and Problem 4 above, prove that if two parallel planes a re cut by a th ird plane, the corresponding dihedral angles will be congruent.

By using the inform ation from Problems 3 and. 5 prove Problems 6

6. Given: J K || LM , R S |[ TW .Concl.: Z T - G H - K ^ Z R -A B -M

7. Given: J K || LM , RS || T W .C oncl.: Z S-C D -K = Z L -E F - T

8. Use Problem 5 to prove that if a plane is perpendicular to one of two ■ parallel planes, then it is also perpendicular to the other.

9. Prove th a t the edge of a dihedral angle is perpendicular' to the plane determined by one of its plane angles.

10. Use Problem 9 to prove th a t the planes determined by the plane angles • of a dihedral angle are parallel,

11.* If a line is perpendicular to a given plane, then any plane containing th a t line is perpendicular to the given plane, f

t T h is s ta tem en t is often considered as a theorem .


B Test and Review

Justify your answer to all questions except 4, 10, and 12.1. How would you account for the fact that the intersections of the side

wall with the ceiling and floor of your classroom are parallel lines?

2. If each of two lines is parallel to a plane, will the plane determ ined by these lines necessarily be parallel to the plane?

3. Four lines in space are parallel to each other. Will the plane determ ined by any two of these I'nes necessarily be parallel to the plane determ ined by the o ther two?

4. If two lines are parallel and one of them is perpendicular to a plane, m ust the other also be perpendicular to the plane?

5. (a ) Does a half-plane include its edge?(b ) How does a half-line differ from a ray?(c) Is there any figure in space geometry that is com parable to the

ray in plane geometry?

6. W hat is the m axim um measure that a dihedral angle can have?7. If a line is parallel to each of two planes, are the planes necessarily

parallel?

8. If a line is perpendicular to each of two planes, are the planes necessarily parallel?

9. (a) If a line is parallel to a plane, is it parallel to all lines in the plane? (b) If a line is parallel to a plane, will it intersect any line in the plane?

10. If two lines are perpendicular to the same plane, will the lines necessarily be parallel?

11. If two lines are parallel to the same plane, are they necessarily parallel to each other?

12. T h ere are six dihedral angles in the diagram below. Nam e them.

Da

A ir --------- \ -------- > C

B

TEST A N D REVIEW 297

8


p intersects m and n in

AB and CD.

~AB S CD

Concl.: A C ^~ B D

Given: m || np intersects m and n in

A B and CD.

T b ^ c d

Concl.: AD and ~BC bisect each other.

Given: m j| n~BC = ~DE

Concl.: A U ^ J B

G iven: PQ _L a

PQ 1 4M is the m idpoint

of PQ."RS is any line segment through M w ith endpoints in a and-A.

Concl.: M is the m idpoint

ofT tf.


5 . Given: A S I

p and q contain AB. p and q intersect m

4—r <—rin CD and EF.

C oncl.: CD \\E F

Given: a |j b

AD II CF

Prove each of the following statements:1. I f two dihedral angles are straight dihedral angles, then they are con

gruent.2. If two dihedral angles are com plem entary to the same dihedral angle,

then they are congruent.3. A line intersects two parallel planes. Any plane containing this line will

intersect these planes in two lines that form congruent angles with the given line.

■ Try This For FunFor years m athem aticians have been form ulating problems

in geometry wherein the reasoning that leads to the conclusion is perfectly valid, yet the conclusion is in contradiction to the postulates that have been established. In each, of course, there is some obscure error that the person who designed the problem has tried to hide. This error usually occurs in the m anner in which the diagram is drawn. One of the more interesting of these problems that has intrigued mathematics students for years appears here,

P r o b l e m : To prove that the measure of a right angle is equal to the measure of an obtuse angle.

TRy THIS FOR FUN 299

Given: ABCD is a rectangle.-

PM is the JL bisector of AB.—

P N is the 1 bisector of AE.

E C S ADConcl.: Z E C D S .Z A D C (Hence, the

measure of the obtuse angle ECD is equal to the measure of the right angle ADC.)

? '(1) Can you show this conclusion to be valid? (Suggestion: Prove A A D P = A £C P .)(2) Can you point out wher?. the error was made in the drawing of the diagram?

10The Angles of a Polygon

T H E ACCEPTAN CE O F T H E PARALLEL PO ST U - la te has m ade possible the proof of one of the most widely used theorems in plane geometry.

T H E O R E M 49; T h e sum of th e m easures of th e ang les of a t r ia n g le is equal to 180.

Given: A A BCConcl.: m /LA + m Z B + m ZC = 1 8 0

A n a l y s i s : T here is only one m ajor condition that we are aware of under which we would encounter 180; this occurs in the measure of a straight angle. Hence, an obvious move is to relate angles A, B, and C to a straight angle. In addition, since the proof of this theorem is based on the parallel postulate, the need for parallel lines is apparent. To bring this condition and the previous one into the picture, it would seem best to have one of the parallel lines pass through a vertex of the triangle.

THE ANGLES OF A POLVGON 301

F i g u r e 10-2.

PROOF STATEMENTS

4—V1. Let D E be the line through B that is

parallel to AC.2. m Z D B E - 1803. m Z D B E = m Z \ + m Z B + m Z24. m Z \ + m Z B + m Z 2 = 180

REASONS

1. Parallel postulate

2. Def. of a straight angle3. Def. of the sum of angles4. T ransitive property of

equality5. Why?6. Substitution postulate

5. But, ZA & Z \ and Z C 9=- Z26. -*• tn ZA -f- jn Z B -f- m Z C = 1 8 0

This theorem paves the way for two very im portant theorems, one of which is another and final m ethod for proving triangles congruent,

TH EO R EM 50: I f two angles o f one tr ia n g le a re c o n g ru e n t to two angles of a second trian g le , th e n th e th ir d angles a re congruen t.

Given: ZA ~ ZD Z B ~ Z E

F Concl.: ZC ~ Z F

Figure 10-3.

PROOF STATEMENTS REASONS ]

1. ZA £= ZD and Z B = . Z E 1. Why?

2. m ZA + m Z B = m Z D + m Z E 2. Why?

3. m ZA + m Z B + m ZC = 1 8 0 3. T he sum of the measures of the angl o of a triangle equals 180.

4. m ZC = 180 - (m. ZA + m ZB) 4. Subtraction postulate

5. m Z.D + m Z E + m Z F = 1 8 0 5. Why?

6. m Z F — 180 - (m Z D + m ZE) 6. Why?

7. m Z F = 180 - (m Z A + m ZB) 7. Substitution postulate

8. m ZC = m Z F 8. Transitive property of equality

9. Z C ^ Z F 9 W hy? '

302 THE ANGLES OF A POLYGON

T H E O R E M 51: Tw o triang les a re congruen t if th e re exists a co rre spondence b e tw een the vertices in w h ich tw o angles a n d a side opposite one of them in one tr ia n g le a re cong ru e n t to those co rresponding p a rts in th e secoSfli t r i angle . (The symbols for this statem ent are A.A.S.) }

Given: Z A =5 Z D Z B ^ Z E A C 2 £ D F

e c E Concl.: A A B C z = A D E F

Figure 10-4.

PR O O F | (The reasons will be left for you to supply.)

1. Z 4 S Z D (a) 4. AC = D F (1)2. Z B ~ Z E 5. A A B C ^ A D E F3. Z C = Z F (a)

G iven: D E X BC «-+ <-»

BA 1 AC Concl.: Z \ is supp. to Z B .

A n a l y s is '. Z l is already supplem entary to Z2, hence if Z 2 ~ Z B , the conclusion will follow. Notice that it will not be necessary to prove triangles to be congruent in order to show that Z l ~ Z B .

1 P R O O F (The reasons will be left for you to supply.) |«-+ ♦->

1. D E X BC 6. Z C S Z C2. Z D E C is a right angle. 7. Z B £= Z2 (Theorem 50)

3. BA X AC. 8. But, Z l is supp. to Z2.

4. Z B A C is a right angle. 9. Z l is supp. to Z B .

5. ZD E C S ZB A C

Illustration 2;

Illustration 1:

T h e perpendicular line segments from the m idpoints of the legs of an isosceles triangle to the base are congruent.

THE ANG LES OF A P O L Y G O N

Given

303

A B S i 'A C ___D is the m idpoint of AB. F is the m idpoint of AC.

D E and FC X BC

C oncl.: D E = FG

PRO OF (The reasons will be left for you to supply.) |

1. A B ^ A C 7. Z D E B is a right angle.

2. Z B ■= ZC [a) <-» <->8. FG X BC

3. D is the m idpoint of AB. 9. ZFGC is a right angle.4. F is the m idpoint of AC. 10. ZDE B = ZFGC (a)

5 . 5 7 ? f 5 (s) 11. A D B E ^ A F C G

12. ~DE = ~FG6. D E X BC

T h e following two theorems are an im m ediate consequence of the theorem on the sum of the angles of a triangle. T heir proofs will be left for you to do.

TH E O R E M 52: T h e m easure o f an ex te rio r a n g le o f a tr ia n g le is equal to th e sum of th e m easures o f th e rem ote in te r io r angles.

T H E O R E M 53: T h e acute angles of a r ig h t tr ia n g le a re com plem en- ta r y.

EXERCISES

1. In the diagram at the right, find the measure of(a) Z 3 , i f m Z l = 80, m Z 2 = 60(b ) Z4, if m Z l = 85, m Z3 = 35(c) Zl , if m Z2 = x, tn Z3 = y(d ) Z l , if m Z5 = 140, m Z2 = 50(e ) Z5, if m Z7 = 70, m Z2 = 60(f) Z6, if rr. Z5 = 150, m Z 4 = 120(g) Z3, if m Z l + m Z2 = x

2. In A X Y Z , J Y ^ J Z .

I


(a) If m / X = 80, / Y is an angle of how many degrees?(b ) If m / Y = 40, how many degrees are. there in / X ?

3, How m any degrees are there in each angle of an isosceles right triangle?4, How m any degrees ere there in each angle of an equilateral triangle?5, If the measures cl' two angles of a triangle are 80 and 60 respectively,

how m any degrees are there in the angic formed by the bisectors of these angles? ^

9. In the diagram at the right,m / A — 80 and m / C = 30. How m any degrees are there in /B D C

if B D bisects /A B C ?

7. In A ABC, AC £= BC, / A and / B are bisected, and the bisectors intersect a t D. I f m / C = 40, how m any degrees are there in /D ?

8. T h e side BC o f A ABC is extended to D. /A B C and /A C D are bisected and the bisectors m eet a t £ . If m /A B C = 80 and m /A C B — 60, how m any degrees are there in /E ?

9. Sides BA and BC of A B A C are extended through A and C to pointsE and D respectively. The bisectors of the exterior angles EAC and DCA m eet a t P. If m /B A C = 60 and m /B C A .= 40, how m any degrees a ie there in ZAPC? E

10, In the diagram at the right,

A B || CD, m / I = 70, and m / 4 =120. How m any degrees are there in each of the other angles?

11. In the diagram at the right, BE and

U S are altitudes, while m /A B C — 70 and m /A C B — 40. How m any degrees are there in /BFC ?

12. If, in A ABC, m / A = m / B + m / C , then / A is an angle of how

A—»

13.’

m any degrees?

In the triangle a t the right, BD and

CD are the bisectors of angles ABC and ACB. Can you show in any way th a t m / D = 90 + /A ?

THE ANGLES OF A POLYGON 305

1 4 . * In the diagram at the right, BD and

CD are the bisectors of the exterior angles EBC and FCB. C an you show in any way that m / D =90 - /A ?

15.* T he altitudes to sides A B and AC in acute A A B C intersect a t point E. Can you show in any way that m L B E C = m / B + m /C ?

1 6 . * Angle B is an obtuse angle in A ABC. T he altitudes to sides AB and AC intersect a t point E when they are extended. C an you show in any way that m /B E C = 180 — (m / B -j- m / C)?

1 7 . * Base BC of A ABC is extended to point D. T he bisectors of / Band /A C D intersect a t point E. Can you show in any way that

m / E = /A ?

t G iven: A B JL D F

AC JL DG Concl.: Z D c— /-A

3 , Given: DC X. AB

B E L A D Concl.; Z B

Given: / \ == / A Concl.: / B £* /C E D

Given: D E \\A B

D F || CB / \ — Z2

Concl.: / A = / C

306

5 , Given: A B S AC

ED L A B <-> <-»

EF 1 AC Concl.: ZC E F g* Z D E B

7 , Given: ZA C B is a right angle.

CD X AB Concl.: Z A C D ^ Z B

Using the diagram below with the suggested lines PQ and PR, prove the. theorem that the sum of the measures of the angles of a triangle is equal to 180.

| Given: AD and CE a re 6 .| altitudes.1 Concl.: Z B C E K / B A D

THE ANGLES OF A POLYGON

Using the d iagram below 8 . with the suggested line CE, prove the theorem th a t the sum of the measures of the angles of a triangle is equal to 180.

Given: A B = AC 10,Z \ S Z l M is the m idpoint

of M

Concl. : D M ~ E M

li

11, Given: ABCD is a O .

AF L C D <-> <->

AE L BC

~AF S A E Concl.: ABCD is a rhombus.

A ,

THE ANGLES OF A POLYGON

13. Given: BD L DC <-> <-+ CA 1 ABzi s n

Concl.: A B & Z D

15. Given: AC & AB_ B E & C H

Concl.

G iven-. ABCD is a O . <-> *-+A E X BC <-► <->CF X AD

Concl.: A E — C F

12.

307

EF 1 BG

G H L BG : D is the midpoint

of SB.

Given: BQ is the m edian

to AC.<-> <-*CJR X BR

AP L BR Concl,: APCR is a O .

A .

14.

Given: A D AC is isos, w ith 16.* DA S DC.D B ^ D C

C oncl.: Z B A C is a right angle. (H int: Prove m Z B A C = m Z B + rn Z C and see Problem 12, group A.)

3 0 8

17.* Given: A A B C is isosceles w ith A B — AU.(Diagram a t right.)<-* <->FG j| H J~ 4B F bisects Z A B H .

CG bisects ZAG J.Concl.: B F ^ i W

. 01. T h e a ltitude to the base of an isosceles triangle bisects the vertex angle.2 . T he perpendicular line segments from any point on the bisector of an

angle to the sides of d\ e angle are congruent.3. T h e altitudes to the legs of an isosceles triangle are congruent.4. Corresponding altitudes of congruent triangles are congruent.5. I f perpendicular line segments are draw n from th e vertices of the upper

base to th e lower base of an isosceles trapezoid, then these segments a re congruent.

6. I f perpendicular line segments are drawn from a pair of opposite vertices of a parallelogram to the diagonal joining the remaining vertices, then these segments are congruent.

7. I f perpendicular line segments are drawn from the m idpoint of the base of an isosceles triangle to the legs of the triangle, then these perpendiculars are congruent.

8. I f line segments a re drawn from any point on the base of an isosceles triangle to m ake congruent angles with the base, they will make congruent angles w ith the legs of the triangle.

9. If from the point of intersection of the altitudes to the legs of an isosceles triangle a line is draw n to the vertex of th e vertex angle, then this line bisects th e vertex angle,

10. I f a poin t is no t on the bisector of an angle, then the perpendicular line segments drawn from this point to the sides of the angle will not be congruent. (H int: Use the indirect proof.)

11. From the m idpoint of a side of a triangle perpendicular line segments are draw n to the o ther two sides. If the perpendicular segments are no t congruent, then the sides to which they are drawn are not congruent. (H in t: Use the indirect proof.)

12. In an isosceles right triangle one of the congruent sides can not be the hypotenuse.

THE ANGLES OF A POLYGONTHE ANGLES OF A POLYGON 309

| The Angles of a PolygonIt is just a short ju m p from finding the sum of the measures

of the angles of a triangle to finding the sum of the measures of the angles of any polygon. Perhaps our statem ent should not have been so broad, for we do not intend to determine the sum of the measures of the angles of the

polygon in Figure 10-7.

Figure 10-7.

Recall th a t early in our work it was agreed no t to concern ourselves with angles th a t were greater than a straight angle. In this polygon, Z E is greater than a straight angle; hence, polygons w here this occurs will be excluded from _eunvorE ?4^Nreality, we have lim ited ourselves to an in- vestigatioiy'of convex polygons onfyj.D efinition^ ; A epnytSTpoiygon is a polygon in which each of the angles

is less than a straight angle. —Henceforth, whenever the term polygon is used, it will be- understood

to imply only the convex polygon and no other.

T H E O R E M 54: T h e sum o f th e m easures of th e ang les o f a po lygon of n sides is 180(n — 2).

Given: Polygon A B C D E . . .contain ing n sides

C oncl.: m Z A 4- m Z B + m Z C + ....• = 180(n - 2)

A n a l y s i s : W e resort to dividing th e polygon into triangles so as to enable us to use the theorem on the sum of the measures o f the angles of a triangle. This is done by selecting a point P in the interior o f the polygon and draw ing lines from this point to each of the vertices. T here will then exist a one- tc-onc correspondence between each of the triangles, and each of the sides

of the polygon.

310THE ANGLES OF A POLYGON

Figure 10-8,

PRO O F s t a t e m e n t s '

1. L et P be a point in the interior of

A B C D E . . , and let PA be the line through P and A.

2. T he same is tru e of PB, PC, etc.3. m Z l + m Z w + m Z x = 180

m Z 2 + m Zy + m Z z = 180etc.

4. m Z l + m Z 2 + . . . + (ot Z x + m Zy) -f- (m Z z + m Zv) + . . . = 180n

5. But Z t + Z_y = Z AZ z + Z v ■= Z.B

etc.6 . .'. (m Z x + m Zy) + (m Z z + m Zv) +

. . . = m ZA + m Z.B -f- m Z.C + . . .

7. Also, m Z l + Z 2 + m Z 3 - f • • •= 360, or 2 straight angles

8 . 360 + (m Z /l + m Z B + m Z C . .)= 180n

.'. m Z A 4- m Z B + m Z C + . . .= 180n - 360or m Z A + m Z B + m Z C + . . .= 180(n - 2)

9.

REASONS

1 . There exists one and only one line through two points.

2. Same as 13. Why?

4. Addition postulate

5. Def. of the sum of two angles


7. Why?

8. Substitution postulate (See step 4.)

9. Subtraction postulate

This theorem paves th e way for the proof of a theorem th a t seems to b e unreasonable a t first glance, No m atter how m any sides a polygon m ay have, be it 3 o r 3,Q00( the sum of the measures of the exterior angles formed by extending these sides in the same order is always 360!

THE ANGLES OF A POLYGON 311

THEOREM 55: T he sum of the measures o f the exterior angles o f a polygon formed by extending the sides in the same order is equal to 360.

Given: Polygon ABCDE . ■ . containing n sides

Concl.: Sum of the measure of the exterior angles shown = 360.

A n a l y s is : Notice that there exists a one-to-one correspondence between each vertex of the polygon and each pair of angles marked in the diagram . Those m arked in red are the exterior angles; those in black, the interior. T hus, if we subtract the sum of the interior from the sum of all pairs, we will be left w ith simply the sum of the exterior angles.


1. Def. of the sum . of two angles


1. m Z l + m Z2 = 180, or m ZC D P m Z 3 + m Z 4 = 180, or m Z D E R etc.

2. (m Z l + m Z2) + (m Z3 + m Z4) +(m Z5 + m Z 6) + . . . = 180n or (m Z l + m Z 3 + m Z5 + . . .) +(tn Z 2 -j- m Z4 -{- tn Z 6 -f- . . = 180n

3. But m Z2 + 'm Z4 + m Z 6 + . . . =180(n - 2)

4. (m Z l + m Z3 + m Z5 + . . .) +180(n - 2) = 180n

5. Hence m Z l + m Z 3 + m.Z5 + . . . =180n - 180(n - 2) or m Z l + m Z 3 + m Z5 + . . . =180n - 180/t + 360or m Z l -b m Z 3 + m Z 5 + . . . = 360

For simplicity, polygons are given names in accordance with the num ber of sides, or angles, they possess. Thus, a three-sided polygon is called a 3-gon; a five-sided polygon is a 5-gon; a twenty-seven-sided polygon is a 27-gon; etc. Special polygons whose sides are relatively few in num ber a re m ore often referred to by the names listed below.

3. Why?

4. Substitution postulate

5. Subtraction postulate

312 THE ANGLES OF A POLYGONN um berof Sides Nam e

3 triangle or 3-gon4 quadrilateral or 4-gor.5 prntagon or 5-gon6 hexogir, or 6-gon8 octagon or 8-gon

10 dtcagcn or 10-gon12 duodecagon or 12-gon

Illustration 1:

Find the sum of the measures cf the angles of a 15-gon.

M e t h o d : £ = (n - 2)180= (15 - 2)180 = 13.180 = 2,340

Illustration 2:

Find the measure of each angle of an equiangular 20-gon.

M e t h o d : Since the measures of the angles of the polygon are equal, then the measures of the exterior angles will also be equal. In view of the fact th a t the sum of the measures of the exterior angles is equal to 360, then dividing 360 by 20 will give the size of each.

Each exterior angle: 360 -f- 20 = 18 Each interior angle: 180 — 18 = 162

EXERCISES

1. Find the sum of the measures of the angles of each of the following polygons:(a) 7-gon (d ) pentagon(b ) 15-gon (e) octagon(c) 100-gon ({) decagon

2. W hat is the m easure of each exterior angle of each of the following equiangular polygons?(a) 30-gon (c) hexagon(b ) 72-gon (d ) duodecagon

3. W hat is the measure of each interior angle of each of the following equiangular polygons?(a) 18-gon (c) quadrilateral |(b ) 40-gon (d ) octagon

A BRIEF JOURNEY INTO NON-EUCLIDEAN GEOMETRY 313

4. If the sum of the measures of the angles of a polygon is given by each of the following num bers, how many sides does the polygon have?(a) 900 (c) 9,360(b ) 1,260 (d ) 15 straight angles

5. If the measure of each exterior angle of an equiangular polygon is given by each of the following numbers, how m any sides does the polygon

have?(a) 30 (c) 72(b) 45 (d) k

6. If the measure of each angle of an equiangular polygon is given by each of the following num bers, how m any sides does the polygon have?(a) 165 (c) 172(b) 140 (d) k

7. T he measure of each angle of an equiangular polygon is five times as large as the m easure of an exterior angle of the polygon. W hat is the nam e of the polygon?

8. (a) W hat is the largest m easure an exterior artgle of an equiangularpolygon m ay have?

(b) W hat is the smallest measure that an angle of an equiangular, polygon may have?

9. How m any sides does a polygon have if the sum of the measures of its angles is five times as large as the sum of the measures of its exterior angles?

10. Tw o angles of a quadrilateral are supplementary. Show th a t the other two angles are also supplementary.

■ A Brief Journey into Non-Euclidean GeometryfE arlier we discussed the possibility of the existence of pos

tulates other than the parallel postulate:

“ T hrough a given point there exists only one line th a t is parallel to a given line.”

As a consequence of accepting this postulate we were able to prove that the sum of the measures of the angles of any triangle is equal to 180.

At this time we w ould like to explore briefly what m ight have happened had we not accepted the parallel postulate but instead had assumed either of the following statements'.

(1) Through a given point there exist tvvo lines parallel to a given line, one falling to the right of the given point, the other to the left. (Lobachevsky; see pages 245-247.)

t O p t io n a l to p ic .


(2) Through a given point there exists no lines th a t can be drawn, pa rallel to a given line. T h a t is, two lines will always have a t least one point in common. (Riem ann.)

Geometries that accept either of the above assumptions are called non-Euclidean geometries, for ihey have discarded the Euclidean postulate on parallelism. All propositions on congruency, other than the A.A.S. theorem, and all those on perpendicularity apply equally well in the three geometries, as the proof? of these theorems are not dependent on parallel lines. W hy was the A.A.S. theorem excluded? All theorems, however, whose proofs are based on the concepts of parallelism will differ in the three geometries.

O u r goal in this short section will be to try to determ ine w hat the sum of the measures of the angles of a triangle will be in the two non-Euclidean geometries. T o simplify our work, we are going to discard the above postulates and replace them with equivalent statements. T h e new postulates are based on the quadrilateral in Figure 10-12, whose properties are given a t the right of the diagram.

CB 1 AB

DA 1 A B DA ~ CB

Figure 10-12.

This quadrilateral, having two right angles and two congruent sides, is called an isosceles birectangular quadrilateral. T he properties of this figure were first investigated by a Jesuit priest nam ed Saccheri, who lived around the middle of the eighteenth century. He had hoped to show through his analysis of this- quadrilateral th a t the Euclidean postulate on parallel lines could be proved. Unfortunately, he m ade several errors in his work. H ad he not, he would probably have been the first person to publish m aterial in the field of non-Euclidean geometry.

T he new postulates to which we have referred concern themselves w ith the summit angles, C and D, in the Saccheri quadrilateral.

Lobachevsky: T he sumn.it angles, C and D, are acute angles.Euclid: T h e summit angles, C and D, are right angles.R iem ann: T he sum m it angles, C and D, are obtuse angles.

W ith these three postulates as our tools we will now prove that

In Lobachevsky’s geometry: T he sum of the measures of the angles of a triangle is less than 180.

In Euclidean geometry; T he sum of the measures of the angles of a triangle is equal to 180.

A „ B JOUKMEy HOH-EUCUDEAH GEO^ET.V 3 '»

In R iem annian geometry: T h e sum of the measures of the angles of a tr i

angle is greater than 180.O ur proof will be separated into two parts. In P a n A we will show th a t

there exists a Saccheri quadrilateral in the figure that we have draw n. In Part B we will show that the sum of the measures of the summit angles in this quadrilateral is equal to the sum of the measures of the angles of the

triangle.L et A CDE be any triangle.

PROOFPART A (The reasons will be left for you to supply.)

10.U .12.

13.14.15.

L et G and H be th e midpoints of D E and C E respectively,

4—IL et GH be the line through points G and H.

. L et E F be the line through E that is perpendicular to GH.

. Extend FG to A so that ~Fd = GA.

>. Extend FH to B so th a t P H — H B.H5. L et D A be the line through points D and A.

7. L et CB be the line through points C and B.Now by proving DA — CB and Z A and /LB to be right angles, we can

show th a t ABCD is a Saccheri quadrilateral.

8 . D G S lG E

9. AG Z*G F Z l Z2

A D AG K A EFG H ence, Z D AG ZEFG

V u l E F l G HThus, ZEFG is a right angle. ... Z D AG is a right angle.

.16. Also, DA ^ F EIn a similar manner, A C B H c& n be shown to be congruent to

A E F H .17. Hence, Z C B H is a right angle.

18. A i i d C B ^ F E ,

19. :.~ C B 5*TM20. Hence, ABCD is a Saccheri

quadrilateral.


P R O O F ! PART B (The reasons will be left for you to supply.) I

We will now show th a t the sum of the measures of the summit angles,ZA D C and /.BC D , is equal to the sum of the measures of the angles ofA G D £.

1. Since A DAG = A BFG, then Zh = Z42. Since A C B H ~ A EFH, then Z 6 S Z53. But, m Z l + m Z 8 4- m ZCED = the sum of the measures of the angles

of t\C D E ■4. m Z J + m Z 8 4- m Z 4 + m Z5 = the sum of the measures of the

angles of A CDE5. Hence, m Z l + m Z 8 + m Z i + m Z 6 — the sum of the measures of

the angles of A CDE6. But, Z 2 + Z l = ZA D C1. And, Z 6 + Z S = ZB C D8, m Z A D C + m ZB C D = the sum of the measures of the angles of

A CDE

Thus, we have shown that the sum of the measures of the angles of any triangle is equal to the sum of the measures of the sum m it angles of a Saccheri isosceles birectangular quadrilateral. However,

(1) In Lobachevsky's geometry we assumed that each of the summit angles is acute and, hence, the sum of their measures is less than 180.(2) In Euclid’s geometry we assumed th a t each of the summit angles is a right angle and, hence, the sum of their measures is 180.(3) In R iem ann’s geometry we. assumed th a t ear.h of the summit angles is obtuse and, hence, the sum of their measures is greater than 180.

Therefore, it now follows that in each of the three geometries the sum of the measures of the angles of a triangle will be respectively either,

(1) less than 180(2) equal to 180(3 ) ' greater than 180

A great m any of the theorems in the non-Euclidean geometries involve relationships between num bers that are unequal. Hence, before considering any problem s, we will have to establish two postulates concerning the relative size of quantities. .One of these postulates will indicate the existence and uniqueness of the order of size of two numbers, while the other will perm it us to transfer the relation of size of three num bers among themselves.

A BRIEF JOURNEY INTO NON-EUCLIDEAN GEOMETRY 317

P o s t u l a t e 30: Given any two num bers a and b, one and only one of these three relationships must be true: a < b, a — b, a > b. (Existence an d

Uniqueness of O rder)Postulate 31: Given any three numbers a, b, and c, w here a > b and b > c,

then a > c. (Transitivity of O rder)

Illustration:A

Given: m AC > tn AB Cone) ', tn Z A B C > tn ZC

Figure 10-14.

P P n O F I STATEMENTSREASONS

1. Since tn AC > m AB,

let ~AD = ~AB.

2. L et BD be the line through points B and D.

3. m Z A B D = tn Z \4 . m ZA B C > m ZA B D

5. m Z A B C > tn ZX6. But, m Z \ > m Z C

7 . m Z A B C > m ZC

1. A line can be extended as far as

desired.


3. Why?4. T he whole is g reater th an any of its

parts.5. Substitution postulate6 . T he m easure o f a n exterior angle of a

triangle is greater than the measures of either of the rem ote in terio r angles.

7 . Postulate on transitivity o f order

C an you justify th a t point D in the above proof m ust fall between A and C such th a t the order of points is A, D, C and n o t A , C, D?

EXERCISES

1. Prove th a t the sum m it angles of a Saccheri quadrilateral a re congru

ent. (H in t: D raw E F X A B a t the

m idpoint £ of 5fF.)2. Prove th a t the perpendicular bisector of the-/lower base o f a Saccheri

quadrilateral is also the perpendicular bisector of the upper base.


3.* Given: DA X AB

CB X A 3

m DA > tn CB Concl.: m Z C > m Z D

(H int: Find point E such that A E — BC. Since A BCE w ill be a Saccheri quadrilateral, we can use information proved in Problem 1. See the method in the illustration on page 317.)

4.* Given: DA J . AB, CB X AB f ------- ---------->Cm Z C > m Z D

C oncl.: m DA > m CB

(H int: Use the indirect proof by setting up the possibilities in term s of the assumption on existence and uniqueness of the order o f two

num bers. Eliminate two of the possibilities by applying Problem s 1 and 3.)

5. G iven: DA J . AB, CB _L ABZ C S i Z D

Concl.: ~ D A ~ V B

(H int: Use the indirect proof by setting u p the possibilities in term s of the assumption on existence and uniqueness of the order of two quantities. Eliminate two of the possibilities by applying Problem 3.)

6. Given: ABCD is a Saccheri quadrilateral in Lobachevsky’s

geometry. EF is the 1 bi

sector of AB:

C oncl.: m C B > m EF

(H in t: By Problem 2, ZE F C was shown to be a right angle. Hence, how do ZE F C and Z C compare? Now apply Problem 4.)

7. G iven: ABCD is a Saccheri quadrilateral in Riemann’s ge-

ometry. EF is the 1 bisec

to r of AB.

C oncl.: m E F > m CB.

TEST AND REVIEW 319

8. Given: ABCD is a Saccheri quad- ,^ n __________ r__________ rrilateral in Euclidean ge- J \

ometry. EF is the X bisec- / \

tor of AB. I________________________ ^

Concl.: ~EF ~.~CB A £Problems 6, 7, and 8 present a ra ther interesting point of departu re

for the three geometries:1. If in Lobachevsky’s geometry two lines have a common perpendicular,

that perpendicular segment will be the shortest oath between the two lines, for on either side of this common perpendicular the two lines will

diverge. (See Problem 6.)2. I f in R iem ann’s geometry two lines have a common perpendicular,

th a t perpendicular will be the greatest path between the two lines, for on either side of this common perpendicular the two lines approach each

other. (See Problem 7.)3. I f in Euclidean geometry two lines have a com m on perpendicular, they

will have m any, many, common perpendiculars, all of which will be

congruent. (See Problem 8 .)

■ Test and Review i-------1

1. U sing the information shown in each diagram , find the measures o f

th e angles m arked x and y.

320 THE ANGLES OF A POLYGON—4 .. ___

2. In A ABC, CR bisects Z A C B and A S is the altitude to BC. A S and CR intersect a t T. If m ZC A S — 40, then w hat is the measure of Z A T R ?

3. I f the measure of one of the acute angles of a rhom bus is 70, then w hat is the measure of an angle formed by a diagonal and one of the sides? (Give two answers.)

4. If the m easure of one of the acute angles of a right triangle is five times th a t of the other acute angle, w hat is the m easure of each?

5. W hat is the m easure of the acute angle formed by two of the m edians of an equilateral triangle?

6. I f the m easure of the vertex angle of an isosceles triangle is 66, w hat is th e m easure of the obtuse angle formed by the m edian to the base and the altitude to a leg?

7. I f the sum of the measures of two of the angles of a triangle is equal to the m easure of the th ird angle, then w hat can be said of the triangle?

8. I f two angles of a triangle are congruent and one-half the sum of their measures is equal to the measure of the th ird angle, then what can be said of the triangle?

9. If the measure of an exterior angle a t the vertex of the vertex angle of an isosceles triangle is four times as large as the measure of its ad jacent interior angle, how large is each angle of the triangle?

10. Find the sum of the measures of the angles of each of the following polygons:(a ) 20-gon (b ) hexagon

11. W hat is the measure of each interior angle of the following equiangular polygons?(a) 24-gon (b ) pentagon

12. T he sum of the measures of three angles of a quadrilateral is 290. W hat is the measure of the fourth angle?

13. (a) If the m easure of an exterior angle of an equiangular polygon is 12,how m any sides does the polygon have?

(b ) If the m easure of an exterior angle of an equiangular polygon is 18, w hat is the sum of the measures of the angles of the polygon?

14. (a) How m any degrees are there in each exterior angle of an equiangular6-gon?

(b ) How m any degrees are there in each exterior angle of an equiangular 12-gon?

(c) I f the num ber of sides of an equiangular polygon is doubled, how will this affect the measure of each exterior angle?

(d ) .If the num ber of sides of an equiangular polygon is quadrupled, how will this affect the measure of each exterior angle?

test a n d r e v ie w341

15. Justify why the measure of an angle of an equiangular polygon can

not be 130?16. * Can you show in any way that one of the angles form ed by the al

titudes to two sides of a triangle is congruent to the angle form ed b y

these sides?17.* In acute A ABC side BC is extended to poin t D. T he a ltitude B F is

extended to intersect the biscctor of ZA C D a t poin t E. C an you show

in any way that m Z E = \tn ZCT!

I

B

Prove each of the following.

t G iven: A ABC is isosceles, with I B & 'A C <-> *-*ED 1 BC <-v *■+

EF ± AC

Concl.-. / I —

3 . Given: A B C is isosceles

with A B SZAC. Z C & Z l

Concl.: A.ADE is isosceles.

Given:

Concl.

A A B C w ith any line 2 iA F intersecting BC

4-4B E L A F <-*

CF ± A F : Z E B D S IF C D

Given: & A B C is isosceles

w ith AB ==: AC.

BA is extended to D.

Concl.

D E J .B C A A D F is isosceles.

4 .


5 . Given: A A B C w ith AB S AC B E is the a ltitude to AC.

CD is the altiv--.de to AB.

FG 1 BC

Concl.: FG bisects ZBFC.

Given: E F and A B intersect at center of GO.*-r <->

AD 1 EF

BC ± EF

C oncl.: AD . = BC

cProve each of the following statements:

1. Perpendicular line segments are drawn from any point on a given side of a triangle to the o ther sides of the triangle. If these segments m ake congruent angles with the given side, then the triangle is isosceles.

2. If perpendicular line segments are drawn from the m idpoints of the legsto the base of an isosceles triangle, then the segments are congruent.

3. If perpendicular line segments are drawn from two vertices of a triangleto the median (extended) from the third vertex, then these line segmentsare congruent.

4. If two angles of a triangle are complementary, then the triangle is a right triangle.

5. If the sides of one angle are perpendicular respectively to the sides of a second angle, then the angles are either congruent or supplementary.

6. I f an altitude of a triangle does not bisect the angle from which it is drawn, then the sides forming this angle are not congruent.

Similar Triangles

W E W O ULD L IK E T O T U R N O U R A T T E N T IO N ' again to the polygon. Earlier we had devoted tim e to considering those conditions under which two polygons, particularly triangles, m ight have th e sam e “ size” and “ shape"; that is, be congruent. Now, we w ant to investigate those properties alone that give polygons the sam e "shape." Specifically,

" ---------u w tKp -nairs of polygons p ictured in Figure 11-1.

Figure 11-1.

Basic to the development of this un it is a theorem on parallel lines th a t

m ust be proved,

323

324 SIMILAR TRIANGLEST H E O R E M 56; I t th re e o r m ore p a ra lle l lines in te r c e p tf co n g ru en t

segm ents o n o n e transversal, th e y w ill in te rc e p t co n g ru en t segm ents on every transversal.

G iven: a II b II c II dA B S ~BC S CD

Concl.: ~ E F £ i F 5 ^ G H

A n a ly s is : By p ro v in g tria n g le s congruen t it is possible to prove th a t th e segm ents a r e co n g ru en t.

PR O O F (The reasons will be left for you to supply.)

1. L et EP be the line through E

th a t is parallel to AB.

2. Same for FQ and GR

3. /. £ ? II FQ II GR4. a It b II c II d5. ABPE, BCQF, a n d CDRG a re HI.

CD6. E P ^ A B , 7 :Q ^ ~ B C ,G R ^7. B u t~AB & ~ E U ^ 'U S8. /. E F s F iJ a * C ? 7 f ( j )9. Z P E F £*; ZQFG ^ Z R G H (a)

10. ZE F P S ZFGQ S ZG H R (a)11. ••• A EPF S A FQG S A G RH12. F G ^ G H

Establishing this theorem opens the way for the proof of two interesting theorems concerning a triangle.

T H E O R E M 57: I f a l in e bisects one side o f a trian g le and is p a ra lle l to a second side, th en it bisects the th ird side o f th e trian g le .

G iven: D E II BC<—> _____

D E bisects A S. <—►

Concl.: D E bisects AC.

t “ I n te r c e p t” w ill m e a n “ c u t o ff.”

I f: i ! ^

- I

I. f

IIriii

SIMILAR TRIANGLES 325A n a l y s is : By allowing line I to be the line through A th a t is parallel to

4-* -BC, we will have three parallel lines. Since the transversal A B is cu t into two congruent segments, we can immediately apply T heorem 56.

PRO O FT he proof is left for you to do.

TH EO R EM 58: I f a l in e bisects two sides of a tr ia n g le , th e n it is p a r a lle l to th e th ird side.

Given: D E bisects AB.<-> —

D E bisects AC.

Concl.: D E II BC

A nalysis: Since we have T heorem 57 a t our disposal, it w ould seem that an advisable approach would be the indirect proof.

PROOF

By the law of the excluded m iddle one of these statements '

must be true and no o ther possibility exists:

D E II B C or D E % BC

L et us accept the possibility th a t D E X BC. T hen by the parallel postulate

there exists only one line th rough D th a t is parallel to BC. L et this line be

DP. Hence, since D P bisects A B , by Theorem 57 it m ust also bisect ~ACl,

T his implies that P is the m idpoint of AC and, therefore, m AP — \m AC;

326 SIMILAR TRIANGLES

\ —* _____j T he Given D ata, however, informs us that D E bisects AC; therefore, tn A E —

’ AC. Thus, m A E = m I f . But this is contradictory to the postulate th a tthe whole is greater than any of its parts. Therefore, accepting the possibil-

£ ity th a t D E Jf BC led to the logical inconsistency of the tru th of bothjF m A E = m AF and m I E r- m I P . By the law of contradiction both cannot

| be true a t the same tinier Since m A B ^ m I F m ust be tru e as the result

i of a postulate, m AE = m AP m ust be false and, therefore, the statem entv < - » « - + < -» < ->

D E % BC is also false. Hence, D E II BC is true, for it is the only remaining | possibility.

EXERCISES

A

1, By using the diagram below in

which D E is extended so that

D E £= EF, prove TKeorem 58. (H in t: Show that DBCFis a parallelogram.)

By using the diagram below in 2 .

which E F iz the line through E

parallel to AB, prove T heo

rem 57.

SIMILAR TRIANGLES327

3 . Given: P is the midpoint of I B .Q is the midpoint of IC .

Concl,: m PQ = \m~ED (H int: Let R be the m idpoint

of SC and prove PBRQ to be a parallelogram.) S tate as a proposition w hat you have proved in this problem.

5 . Given: P is the m idpoint o f IS .Q is the m idpoint of AS.<-*• ___

C oncl.: PQ bisects AD.

Given: a II b II c 4.x II y II 2

Concl. : P $ S i QR

Given: B M is the m edian 6. to A S.P is the m idpoint

of IB .R is the m idpoint

of 5C.

PQ and AS II B M

Concl.; I Q ,& ( [ M & t t S & S C '

328SIMILAR TRIANGLES

. G iven; B M is th e median

to

UP is the m edian to A B

R is the m idpoint of 2RJ.

PQ and R S II B M Concl. i T and I f are trisection

points o f PC.A

_Q

-MS

9 . Given: D, C, an d E a re the

m idpoints of AB, AF,

and AU respectively. Concl.: Points D, G fand E arc |

collinear, I

Given: ABCD is a O . ,

E is the midpoint

of AD.F is the m idpointof W3.*i— <—>

Concl.: B E and D F trisect

AU. (H int: Prove B FD E to be a parallelogram ; then let

/ II B E and m II FD.)

E

8 :

G iven: a II 6 U c

A B S

Concl.: m S Z E F (H int: Let4—►A F be the line joining points A and F.) W hat is the statem ent of the space theorem that you have proved through this problem?

10.

¥

E1. A line th a t bisects one of the nonparallel sides of a trapezoid and is

parallel to th e bases bisects the remaining side.

RATIOS AND PROPORTION 329

2. A line that bisects the nonparallel sides of a trapezoid is parallel to the bases. (Hint: Use the indirect proof; follow the m ethod applied in Theorem 58.)

3. A line that bisects the nonparallel sides of a trapezoid will bisect either diagonal.

4. If line segments are drawn between the midpoints of consecutive sides of a quadrilateral, the quadrilateral formed will be a parallelogram .

5. If a line is drawn from the midpoint of the hypotenuse of a right triangle perpendicular to either leg, it will bisect the leg.

6. If line segments are drawn from the midpoints of the legs of an isosceles triangle to the m idpoint of the base, then these line segments will form a rhombus with the legs of the triangle.

7. If lines are drawn joining the midpoints of the three sides of a triangle, then four congruent triangles will be formed.

■ Ratios and ProportionBefore going further, we would like to review briefly a unit

that you studied in algebra. Necessary to an understanding of the work we plan to develop is the knowledge of a few simple algebraic principles.

Ratio■ ■ ■ ■ ■ ■ ■ ■ I A ratio is the quotient of the measures of two quantities if the quantities are measured in the same unit.

Thus, if we wanted to compare the length of a blackboard 24 feet long, to its width of 3 feet, we would merely divide 24 by 3 and express the answer by saying that the length is 8 times as long as the width. H ad 3 been divided by 24, the result would have been designated by saying that the width is i the size of the length. Both of these answers can be w ritten in what is known as the ratio form :

8 18:1 or - and 1:8 or ~

1 o

In the definition of a ratio we were careful to point out th a t the quail' titles have to be measured in the same unit before it is possible to compare them. Thus, suppose a blackboard is measured as 24 feet long and 36 inches . wide. Before these m easurements can be compared, they will have to beexpressed in terms of a common unit. T h a t is, either the 24 feet will haveto be changed to inches, or the 36 inches changed to feet. O n the other hand, the weight of a person, 180 pounds, can never be com pared to his height,6 feet, for it is not possible to express the weight and the height in terms of a common unit of measure.

Proportion

■ ■ ■ ■ A proportion is an equation in which the left side and the right side of the equation are single ratios.


Thus, the equation * = $ is a proportion, for it contains a single ratio on each side of the equal sign. This proportion can also be w ritten in the form 3:4 = 6 : 8. In either form it can be read in one of two ways:

' (1) 3 over 4 equals 6 over S.(2) 3 is to 4 as 6 is to 8 .

T he genernl proportion is usually written in terms of the elements a, b, c, and d, and is expressed as

a c . L ,- = or a:b = c'.d b a

In the proportion f = $, w hat 3re the values of o, b, c, and d? Each of the v terms of a proportion is called a proportional. By exam ining the proportion

a: b — c'.d it is easy to realize why the a is called the fa st proportional; the k, the second proportional', etc.

Special proportions arise in which the second and th ird term s are identical. Examples of these are

2 _ _6_ 1 = 3 a b6 18’ 3 9’ b ~c

y Proportions such as these are called mean proportions. In a m ean proportionthe c is considered as the third proportional to the terms a and b, while the 6 is referred to as the mean proportional to a and c.

In the general proportion a'.b = c'.d the first and fourth terms of the proportion—that is, a and d— are called the extremes of the proportion, while the second and third proportionals, b and c, arc known as the means.' > A very im portant theorem concerning the relationship between these quantities is

T H E O R E M 59: T h e p ro d u c t o f th e means of a p ro p o rtio n is eq u al to th e p ro d u c t of th e extremes.

Given: = - Concl.: ad - beb a


i . 2 = - b d2. bd = bd

3. bd • x = b d - C- b a

or ad = be

1. Given

2. Reflexive property of equality

3. If equals are m ultiplied by equals, the products will be equal.

This theorem gives us a handy way of finding any one of the terms in a proportion when given the rem aining three terms.

Illustration:Find the fourth proportional to 5, 7, and 4.

RATIOS AND PROPORTION331

5S o l u t io n :

... by T h e o r e m 59or

5* = 28* = 5§ (T he fourth proportional)

T he converse of Theorem 59 is also true; th a t is,TH E O R E M 60: I f th e p ro d u c t o f two n u m b ers is e q u a l to th e p ro d u c t

of two o th er n u m b ers , e ith e r p a ir m ay b e m ade th e m eans of a p ro p o rtio n , w h ile th e o th e r p a ir is m ade th e

extremes of the proportion., * t

Given: xy — w i Concl.: -

1 . xy = u*2 . wy = “V

, 2 1 - H Swy wy

1. Given2. Reflexive property of equality

3. If equals a re divided by equals, the quotients will be equal.

orxw

illustration:I f 2x - 3y, then what is the ra tio of x toy?

S o lu tio n : In the relation 2x = 3y the product of the two num bers 2 and x is equal to the product of the num bers 3 and y. Hence, T heorem 60 can be applied. Since x is to be one of the extremes, the other m ust be 2, Similarly,

since y is the second proportional, 3 will have to be the th ird .

Hence, ~ y ~ \THEOREM 61: If four numbers (a , b, c , d) are in proportion

, . . . . . . /a + b c + d\th ey a te in. p ro p o rtio n b y a d d itio n I — - — = —^— r

a + b e + dGiven: r = Concl.: —;— = — :—

b d i d

A n a l y s i s : Bv adding 1 to each side of th e equation ~ = th e conclusiono a

will follow.

PRO O FT he proof will be left for you to do.

332SIMILAR TRIANGLES

1. Find the ratio of the following:(a ) A qu arte r to a half dollar(b ) T w o feet to three inches(c) T en ounces to two pounds(d ) T he length of the rcctangle a t the

rig h t to its width.(e ) T h e m easure of a right angle to

the m easure of a straight angle.

2. F ind the value of x in each of the following proportions:

, . x 5 . . 3 122 = ?2 ( b ) ; = j

/ > 5 * + 1 , 3 2* — 1<c> 7 - — — (d ) 5 - 12

3. F ind the fourth proportional if the first three proportionals are(a ) 2, 3, 8 (b ) 5, 7, 9(c) a, b, c (d ) x, 2x, ix

4. F ind th e th ird proportional of a mean proportion if the first and second proportionals are

(a ) 2, 4 (b ) 4, 12 (c) 3, 5 (d ) a, b

5. F ind the m ean proportional of a mean proportion if the first and th ird proportionals are

(a ) 1 ,4 <b) 3, 12 (c) \ , 18 (d ) 2a, 8a

6 . W rite eight different proportions expressing a relation between 3, 4 , 2, and 6 if 3 ’4 = 2 ’6.

7. I f 5a = 74, then what is the ratio of a to 6?

8. Com plete each of the following proportions under the conditions g iven :

(a ) I f 2x = 4y, then - = ?

(b ) I f 3 -4 = 4*, then } = ?

(c) I f 5 • 7 = 2y, then | = ?

(d ) I f x5 = ab, then - = ?*

EXERCISES ________ 6 ft. 8 i;

(e) If * = then

CO I f f = i t h e n ^ = ?5 i

then X- ± I = ?

THEOREMS BASIC TO THE PROOFS OF SIMILARITY 333

(h ) Ifa + b 17

10 ’then 7 = ?o

■ Theorems Basic to the Proofs of SimilarityThe definition of a ratio insisted that the quantities being

compared be measured in the same unit. O ur concern is not with the nature of a unit of measure in general but, rather, with the unit th a t pertains specifically to line segments. In order to clarify this concept, we will have to re-examine the m uch earlier discussion we had on the m easure of a line segment.

As you recall, the measure of a line segment is simply the co-ordinate of the right endpoint of the line segment when the left endpoint coincides with the zero point on the num ber line. Thus, in Figure 11-6 the measure

0 1 2 3 4 5 6 7 0 , 1 2 3■ 0 ■ ■ 1 1 1 1-------- 1— , b . , i i---------A « A I

Figure 11-6. F igure 11-7.

of A B would be 6, for the co-ordinate of 3 is 6 on the num ber line. T he name given to the segment a whose endpoints are “ 0”. and “ 1” is the unit of measure on this num ber line. Thus, the unit of measure, a, is contained an exact, or integral, number of times, 6, in AB. H ad we doubled our unit of measure, as we did in Figure 11-7, then the measure of A B would now be 3 rather than 6. Again, the unit of measure, b, is contained an integral num ber of times in AB. T o avoid confusion as to whether the m easure of A B is 6 or 3, or any other num ber depending on the selection of the un it of measure, we say that m AB is 6 in a units or m A B is 3 in A units. In reality, when we consider the definition of the sum of line segments, it is often far better to express these relations as

A B = 6 a or A B = 3i

Pursuing this further, we see a common unit of m easure to two line segments will be one that is contained in both segments an exact, or integral, num ber of times. Specifically, if the inch was contained in UD 9 J times and in ~EP times, then a common un it of measure for these two line segments is the ^ inch. I t wili be contained 38 times in UD and 29 times in EF.

C D £


v

v

Unfortunately, all line segments do no t have a common un it of m easure. Line segments that do are called by the frightening nam e of commsnsu- rable segments. Those that do not have a common unit of m easure arc called incommensurable segments. T here are several pairs of incom m ensurable segments with which you have had contact in the past. The circumference of a circle and the diam eter of the same circle have no common unit of measure. The unit of measure that fits into the diam eter exactly 1 time will fit into the.circumference approximately 3.1416 times; it can not fit into the circum ference exactly. Another pair of line segments that are incom m ensurable are the leg and hypotenuse of an isosceles right triangle.

The proof of the next theorem necessitates finding a com m on u n it of measure to two line segments. As just explained, this does not always exist. Hence, our proof is true only under the condition that the line segments are commensurable. We will assume, how ever i-. th a t this theorem is also true when the line segments are not commensurable. This is proved in m ore advanced fields of mathematics.

THEOREM 62: If a line is parallel to one side of a triangle, then the ratios o f the measures o f corresponding segm ents o f the other two sides w ill b e equal.

A

G iven:

C oncl.: ■

AD and D B are commensurable.

D E II BC

m AE m liC

| PRO O F j STATEMENTS REASONS

1. A B and B B are commensurable. 1. Given2. Let a be the common unit of 2 . Def. of commensurable line

measure for A B and B B . segments3. Let a fit into A B k times and 3. Def.. of a unit of measure

into B B p times.

4. m A B = k 4. Def. of the m easure of a linesegment

5. m D B — p 5. Same as 4, m AD k6. — — «= -

m D B P6 . Division postulate

INow we will try to . ; ,W th a t the ratio of the measures of XE and EC

is also k over p.


7. At each point of division on AB<-+ \ t ' 7

draw a line parallel to AC. !

8. But DE II BC9. Hence, each of these lines is

parallel to each other and to

DE.10. These parallels intersect con-

congruent segments called 6 on

AC.

1. Parallel postulate

8 . Given9. If two lines are parallel to the

same line, then they are parallel to each other.

10. If three or m ore parallel lines intercept congruent segments on one transversal, they in tercept congruent segments on every transversal.

Note that there exists a one-to-one correspondence between each of

the “a” units and each of the “ A” units.11. Def. of the sum of line segments

12. Sam e as 4

13. Sam e as 11

14. Sam e as 4

11. Hence, ~AE = kb

12. m 1 E = k

13. And ~EC = pb14. m EC — fi

m~AE _ k m SC J

m AD m AE

15.

16.

15. Same as 6

16. Transitive property of equality

For the sake of simplifying our symbols, we will use the following

no tation :m ~AD = a m H E = c m A B — em D B = b m EC — u m ~AjC = /

Now the proportionm AD ^ m AE

m D B tn EC

can be rew ritten in the m uch simpler form of

336 SIMILAR TRIANGLESa _ c b ~ d

T h e above relation is bu t one of the m any different proportions that<—* 4—5will hold if D E II BC (see Figure 11-10). From the theorem on addition, Theorem 61, this proportion will lead to the truth of

a -j- b __ r -j- a' ! _ m A B mb d b d m £)£[ m EC

From th e theorem th a t the product of the means is equai to the product of the extremes, the m iddle proportion above will become

e X d ^ b X f (1)and, on the basis of Theorem 60, this product can be rewritten as eight different proportions. Similarly, the original proportion th a t was obtained from T heorem 62 can be w ritten as

a X d = b X c (2)T his product, too, w i^ lead to eight different proportions. Finally, by trans

form ing one of the proportions found in (2), - — by the use of the add ition theorem , it follows that

b - f a _ d -f- c e _ [ m A B _ m ACa c a c m AD m A E

From this proportion we obtain a third product:

< X c = a X / (3)th a t will also lead to eighl different proportions. Theorem 62 will be quoted when applying any one of these 24 different proportions.

At first glance it would seem almost impossible to try to recall these24 proportions. You will find, however, th a t you will almost autom atically sense proportions th a t are no t true. Thus, you will never cross the “ parallel” w hen setting u p a ra tio ; th a t is, the ratio a'.d will equal no o ther ratio, nor will b:c. Similarly, the proportion starting w ith i \c cannot be completed.

W e are m ore fortunate w ith the converse of Theorem 62; i t is no tnearly so lengthy nor so difficult. O n the o ther hand, its im portance is no t nearly so g reat either.

TH EOREM 63; I f a l in e intersects two sides o f a triangle j o that the ratios o f the measures o f corresponding segm ents are equal, then the line is parallel to the third side,’

THEOREMS BASIC TO THE PROOFS OF SIMILARITY 3374—> ►

A n a l y s i s : We will use the indirect proof by assuming th a t D E % BC and show that this leads to the logical inconsistency that AE = IF and A E ££ AP.

PROOFBy the law of the excluded m iddle one of the following

statements is true and no other possibility exists:» 4*4 <-4

D E W B C or D E X BC4-4 <-4 H

Let us accept the possibility that D E X B C ‘, then, through D let DP bea x

the line that is parallel to BC. By Theorem 62. - = -> where x is the meas

ure of IF . Hence, it follows th a t

e

Similarly, from the Given D ataa _ c* ~ f

. thereforec = a- * l

e

Thus, we can say th a t x — c by the transitive property of equality. B at we have a postulate stating th a t the whole is greater th an any of its parts.

This implies that x > c. Therefore, accepting the possibility th a t .Dis Jf BC led to the logical inconsistency of the tru th of both x — c and x ^ c. By the law of contradiction both cannot be true a t the same time. Since x ^ c must be true as the result of a postulate, then x — c m ust be false and, there-

fore, the statem ent D E X BC is also false. Hence, D E || B C is true, for it is

the only remaining possibility.

In view of the previous few proofs it.should be qu ite apparent th a t the symbolism involving the measures of line segments can become so in- : volved as to be distracting. However, since the measures of line segments are im portant to the development of the work on similar polygons, i t was imperative that our symbols be simplified. I t was for ju s t this occasion th a t we had been waiting to introduce the symbol A B w ithout anything above it. Each time AB was w ritten earlier in our work, i t was either

4~4 ^(1) a line, A B w ritten with the double arrow above i t

■4or (2) a ray, A B w ritten with the single arrow above it

o r (3) a segment, A B written w ith a segment above it

i I I Henceforth, should A B appear with no m ark above it, it will signify that ' 1 it represents the measure of the line segment AB. Thus,

A B — m AB

Numerical Illustration:


Given: D E II A BCE = 4, CD = 6, BD = 2

T o find: AC

An a ly sis : I t is suggested th a t each of the six segments be labeled as was was done in the triangle above. Always labeling the measure of the segm ent th a t you are asked to find with the symbol x will avoid one of th e possibilities of error. You are now free to choose any one of the 24 different proportions that follow when a line is parallel to one side of a triangle. T w o are shown here.

First Solution

jc _ 84 “ 6

6 x — 22 x = 5$

EXERCISES

Second Solution

4 = 6x - 4 2

8 = 6 * — 24 5J = x

Figure 11-12.

A

1. Given: D E II BC

For the diagram at the right complete each of the following proportions:


G iven : S T I PQFor the diagram a t the right complete each of the following proportions:

(a ) RP: RS = ? (d ) P S'.Q T = ?

(b) Q T :T R = ? (e) Q R :P R = ?

3. G iven: VW || Y ZFor the diagram at the right complete each of the following proportions. If any of them can not be completed, indicate this by drawing a line through the equality sign.

(a)

<d)

YVW ZX Vxw

<b)X WVY

= ?

(h)V YW Z

4. G iven: D E II BCIn terms of the diagram a t the right find the measures requested.

(a ) AD = 2, D B = 4, A E = S, EC = ?(b ) D B = 6, EC = 8, A E = 5, A D = ?(c) A E = 5, AD '= 4, EC = 7, A B = ?(d ) A B = 12, AD = 3, AE = 4, EC = ?(e ) A B = 10, AC = 12, EC = 8, D B - ?(f) D B = 8, A B = 18, A E = 12, AC = ?

5. G iven: R S ] \X ZIn terms of the diagram a t the right find the measures requested.

(a) X R = 4, R Y = 6, KS = 8 , S Z = ?(b) YX = 12, YZ = 16, SZ = 4, F/? = ?(c) - O = 5, W = 10, S Z = 6; X Y = ?(d ) y f l = YS, R X = 1, S Z ~ 1(e ) FA" = YZ, YS = 9, YR = ?


6. U nder which of the following condi-

tions will D E be parallel to AB in the diagram a t the right?

(a) AC = 10, CD = 4, EC = 2, BC = 5(b ) AC = 12, AD = 8, EC = 3, BC = 9(c) £ C = 6 , BC = 14, /ID = 12, DC = 8(d ) <4D = 6, BC = 14, BC = 18, DC = 21(e ) = 20, DC = 10, AC = 25, BC = 36

1. G iven: /ID II E F II BC

C o n c l, A * D F

3 . Given: A B \\ CD

C oncl.:

E B II FD PA PC

PEPF

B

G iven:

Concl.:

AD _ D S £ C

A B AQ

G iven: D E II AC

Concl.

D C II AP B E _ BC EC CP

^H in t: Prove both ra-

i . B D \ tios equal to


5 . Given: AD bisects ZBAC .

Concl.

BE \\ AD CA _ CD AB D B

7 . Given: ABCD is a trapezoid

with AD II BC. D R'.RC = D Q :Q B A P:PB = D Q :Q B

Concl.: Points P, Q, and R are collinear.

9 . Given: DFW AC

D E II AB

C oncl.: E F II BC

Given: A R bisects exterior /.Q A B .

Concl.

BP II RC R B '

RA

_ 4 £~ AB

QR I A B

Concl.: RS II BC

Given: plane a ll plane 4 I plane c

_ , AB ADConcl.: BC D E

6.

10.


l£l1. ff a line is parallel to the bases of a trapezoid, it divides the two diagonals

into two eqaal ratios.

2. If a line is drawn intersecting the legs of an isosceles triangle so as to form a second isosceles triangle, then it is parallel to the base.

3. I f a line intersects two sides of a triangle bu t is not parallel to the th ird . side, then it does not divide the first two sides into two equal ratios.

4. I f a line intersects two sides of a triangle but does not divide these twc sides into two equal ratios, then it is not parallel to the th ird side.

5. In space geometry t.Lree parallel planes will divide two segments in to two equal ratios.

| Similar Triangles

W e have drifted so far from the goal th a t we had set for ourselves a t the outset of this chapter that it would seem th a t we may have lost sight of where we are going. O riginally our objective was to establish conditions under which polygons would have the “sam e shape,” a lthough not necessarily be congruent. O ur detour, though seemingly lengthy, wasindeed imperative for the proofs th a t are to follow_____

The m athematical term for the wor^s “ same shape” is the w 4id similar] But to show polygons similar by trying to prove that they have the “ same shape” would be as difficult as trying to prove polygons congruent by somehow establishing the fact that they “ fit exactly.” T he terms “ same shape” or “ fit exactly” are far too vague to lend themselves to a precise m eaning. Hence, similar polygons will be defined in a m anner m uch the same as used when defining congruent polygons, f

D e f in i t i o n 69: Similar polygons are two polygons in which there exists a one-to-one correspondence between the vertices such that

(1) All the corresponding angles are congruent.(2) All the ratios of the measures of the corresponding sides are equal.

How do we interpret this in the case of our most im portan t polygon, the triangle? There are six correspondences that can be arranged between

K

2c ^ > \ 2o

F i g u r e 1 1 -1 3 .

t I t m ig h t b e a d v isa b le to rev ie w p a g e s 114 th ro u g h 120.

SIMILAR TRIANGLES 343

the vertices of two triangles. If any one of these leads to the corresponding angles being congruent and the ratios of the measures of the corresponding sides being equal, then these triangles will be similar. To illustrate, examine the two triangles above. The six correspondences are

1 2 3A B C ^ P Q R ABC <-> PRQ ABC*->QPR

4 5 6ABC<->QRP A B C «-> RPQ A B C ^> R Q P

Any other correspondence that can be set up for the sets of elements{A, B, C} and {P, Q, R} by rearranging the elements in {A, B, C} will simply be equivalent to one of the sin correspondences above. From the m arkings in the triangles, since /CA is not congruent to ZP, it follows immediately that neither the first nor second correspondence can be a similarity correspondence. In the same way, we can discard the th ird and fourth correspondence. In both the fifth and sixth, Z A is congruent to its corresponding angle, ZR ', however, in the sixth, Z B is no t congruent to its corresponding angle, ZQ . Hence, the sixth correspondence is ou t also.

Although the corresponding angles are congruent in the fifth correspondence, we must investigate still further to see if the ratios of the meas-' ures of the corresponding sides are also equal.

ABC «-> RPQ

From our understanding of a correspondence between the vertices of two polygons, the three pa,irs of corresponding sides will be

A B and HP ~BC and TQ AC and RQ

T he m arkings in the diagram indicate th a t their ratios will be

AB = s, £ £ * = £ , AC _ 5RP 2 PQ 2 RQ 2

HenceA B _ BC _ AC RP PQ RQ

Now, finally, since there exists a correspondence in which both the corresponding angles are congruent and the ratios of the measures of the cor- . responding sides are equal, we can say th a t the triangles are similar. H ad the fifth correspondence, too, failed to hold under these two conditions, it would follow th a t these triangles were not similar.

T o have to prove triangles congruent by resorting to the definition would be a long and tiresome process. Hence, a num ber of' m ethods were developed (S.A.S., A.S.A., etc.) to drastically shorten this process. So, too, is the case for similar, triangles. W e plan, now, to prove several theorems th a t will enable us to show triangles to be sim ilar under less trying conditions than those needed for polygons in general. In fact, our theorems will show th a t if bu t one of the two requirements for similarity can be shown to hold, triangles will be similar. This is not true for any other polygons.

T H E O R E M 64: T w o tr ia n g le s a re sim ilar i f th e re exists a co rrespondence be tw een the vertices in w h ich th e co rresp o n d in g an g les a re congruent.


Given: Z A ~ ZD Z B ~ Z E Z C S Z F

Concl.: A A B C ~ A D E F \

A n a ly sis : Since the corresponding angles a re already congruent by th e G iven D a ta , i t is merely necessary to prove th a t

A B = A C = BC D E D F E F

I f this is true, then by the reverse of the definition of sim ilar polygons we can conclude th a t the tri vagles are similar.


1. L et P be a point on A B, extended if necessary, such th a t I P ^ "HE.

2. Let ZA P Q S Z B

3. PQ II BC4. Z E S Z B5. ZA P Q £* Z E6. Z A S Z D7. /. A D E F ^ A A P Q8. Since I F ~~ 25E

. 9. A nd A ^ ^ S F

in n + d C10' ~AP ~ AQ

REASONS □

11." D E D F

1. A line can be extended as far as desired.

2. A t a given point on a line there exists an angle that is congruent to any given angle.

3. Why?4. Why?5. Why?6. Why?7. Why?8. See step 1.9. Why?

10. I f a line is parallel to one side o f a triangle, then the ratios of the m easures of the corresponding segments of the o ther sides will be equal.

11, Substitution postulate

t T h e s y m b o l fo r th e w o rd s im ila r is


In the same way, had point R been __ __

found on BC so th a t B R = E F and Z B R S was m ade congruent to ZC, it

u , A B B Ccould be shown that — = — •

D a EjP

12. Hence,A B _ AC BC

' D E D F EF 13. A A B C ~ A DEF

12. T ransitive property of equality

13. Reverse oC the def. of similar polygons

As though this m ethod for proving triangles to be similar was not simple enough to apply, there is yet a shorter m ethod. By an 'ea rlie r theorem we know that if two angles of one triangle are congruent respectively to two angles of a second triangle, then the th ird angles will be congruent. Hence, ra ther than prove three pairs of corresponding angles to be congruent to show th a t triangles are similar, we need merely prove two pairs congruent. Having proved these to be congruent, the congruence of the third pair will immediately follow by the theorem just quoted.

T H EO R EM 65: T w o tr ian g les a re sim ilar i f th e re exists a correspondence be tw een th e vertices in w hich-tw o- p a irs of co rre sp o n d in g angles a re congruen t! (A.A. Theorem on Sim-, ilarity)

PRO OF

T h e proof will be left for you to do.

There is a likeness between the statem ents on congruence and those on similarity. To prove triangles congruent, we need not only two pairs of corresponding angles to be congruent bu t also a pair of corresponding sides. If merely the two pairs of angles are congruent, the triangles will be similar, not congruent. Thus, the A.A. theorem on sim ilarity is com parable to the A.A.S. theorem or A.S.A. postulate on congruency. In the sam e way there will be two further theorems on similarity to com pare w ith the S.S.S. and 6'./4.5. m ethod for proving triangles congruent.

T H E O R E M 66: T w o trian g les a re sim ilar i f th e re exists a correspondence b e tw een th e vertices in w h ich th e ra tio s of th e m easures of two pairs of co rresp o n d in g sides a^e equal an d th e ang les in c lu d ed b e tw een each p a ir o f sides a re co n g ru en t. (S.A.S . Theorem on Similarity)

346SIMILAR. TRIANGLES

G iven: ZA

Concl.:

d i = AC D E D F

A A B C ~ A D E F

A n a ly sis: Since we already know th a t Z .4 = ZD , by showing th a t Z B iscongruent to Z E the two triangles will be sim ilar upon the basis of the A.A. theorem on similarity.

P R O O F 1 STATEMENTS

1. Let P be a point on A B , extended if

necessary, such that AP — UE.

2. L et Q be a point on AC, extended if

necessary, such th a t /RJ = 25F.3. -ZA £* Z D (a)4. A A P Q £* A D EF5. Z E ~ ^1

A B AC.DE. D F v

A S AC AP AQ

8. PQ || BC9. Z B a - Z \

10. Z B ~ .Z E (a)11. A A B C ~ A D E F

6.

7.

REASONS

1. Why?

2 . Why?

3. Given4. S.A.S.5. Why?

6. Given

7. Substitution postulate (See steps 1 an d 2.)

8 . Why?9. Why?

10. Why? (See steps'5 an d 9.)11. A.A. theorem on sim ilarity

T H E O R E M 67: Tw o tr ian g les a re s im ila r i f th e re exists a co rresp o n d ence b e tw een th e vertices in w hich th e ra tio s o f th e m easures o f co rre sp o n d in g sides a re eq u al. (S.S.S. T h e orem on Similarity)

A

G iven: AB AC BCE E D F ~ E F

Concl.: & A B C '~ A D E F

Figure 11-17,

An a ly sis: We will try to prove the two triangles similar by showing th a t two pairs of angles are congruent. This can be "'one by showing th a t both Z B and Z E are congruent to Z l , and also that both Z C and Z F are. con

gruent to Z2.


PRO O F STATEMENTS

1. Let P be a point on AB, extended if necessary, such that

2. L et Q be a point on AC, extended if necessary, such that

A B = AC D E DF

. A B _ A C 4> AP ~ AQ

3.

5. PQ II BC

6. Z B S Z Z l and Z C ^ Z 27. A A B C ~ AAPQ

AB _ BC8’ AP PQ9. AB-PQ = BC-AP

BC-AP 10. PQ = - j j f

™ BC-D E U ' P<2 ~ AB

r, AB12. But = pp

13. Hencc, EF

BCEF

BC-DEAB

14. PQ = EF15. Hence, AAP Q = A D E F16. Z E S Z l and Z F & Z217. ZjS Z E and ZC £■ Z F18. A A B C ~ A D E F

REASONS

1. Why?

2. Why?

3. Why?


5. I f a line intersects two sides o f a triangle such th a t the ratios of the measures of corresponding segments are equal, then the line is parallel to the third side.

6. Why?7. A.A. theorem on similarity

8. Def. of similar polygons

9. Why?

10. Division postulate


12. Given

13. Same as 9 and 10

14. Why?15. J.J.S .16. Why?17. Why? (See steps 16 and 6.)18. A.A . theorem on similarity

I


Illustration:

G iven: IE and- BD a r t altitudes in A ABC.

Concl.: A A E C ~ A B D C

A n a l y s i s : Y o u w ill f in d t h a t in m o s t c a se s t h e b e s t a n d u s u a l ly th e o n ly m e th o d to a p p l y to p r o v e t r i a n g le s s im i la r is th e A.A. t h e o r e m o n s im i l a r i t y .

T h i s is p a r t i c u l a r l y t r u e w h e n no in f o r m a t io n is g iv e n c o n c e r n in g e q u a l

r a t i o s a s is t h e c a s e i n th i s p r o b le m .


1. I E and 5 5 are altitudes in A ABC.

2. A E X BC3. Z A E C is a right angle.

4. B D J_ AC5. A B D C is a right angle.6 . Z A E C = A BD C (a)7. Z C 9 * Z C (a )8. A A E C ~ A B D C

1. Given

2. Def. of an altitude3. Def. of perpendicular lines

4. Same as 25. Same as 36. Why?7. Why?8. A.A. theorem on similarity

EXERCISES

W1. T he similarity correspondence between A R S T and A X Y Z is of the

form JUST *-*YXZ. W hat three ratios will be equal?2. T he similarity correspondence between A A B C and A D E F is of the

form A B C * ^F D E . If A B = 6, BC = 5, AC = 9, and F E = 3, then

w hat are the measures of ~FD and DE?3. W hat conditions would have to exist under which it would be possible

to' have two distinct similarity correspondences between the vertices of two triangles?

Figure 11-18.

SIMILAR TRIANGLES4. In A ABC below find the lengths of the rem aining sides.

349

F i n d t h e l e n g t h o f s id e BC in th e d i a g r a m b e lo w .

6. T h e stick 6 feet long casts a shadow 4 feet long. T h e flag pole casts a shadow 24 feet long a t the same time. How high is th e flag pole? C an

you explain w hy th e triangles are similar?

shadow

7. In the d iagram below; the m an is 6 feet tall, while his shadow is 9 feet long. If th e lam p post is 30 feet high, how far is the m an from the lam p

post? C an you explain why the triangles a re similar?

350 SIMILAR TRIANGLES8. In the diagram below find the distance across the pond.

9. (a ) By using the theorem that if four num bers are in proportion, they are in proportion by addition (Theorem 61), can you show th a t the ratio of the perimeters of two similar triangles is equal to the ra tio of the measures of any two corresponding sides. (The perim eter of a triangle is the sum of the measures of its sides.)

(b ) The perimeters of two sim ilar triangles are 24 and 18 respectively. If the measure of a side of the first is 8, find the measure of the side corresponding to this one in the second triangle.

[F1, Given: D E II AB

Concl.: A A B C ^ A E D C

3 . Given: A B i . BC "

ED L A C Concl.: A A E D ~ A A B C

Given: A B L B E

CD X BE Concl.: A A B E —■ A C D E

Given; UD and B E are altitudes of A ABC.

Concl.: A D P B ~ AEFC.A

4 .

T f

SIMILAR TRIANGLES

5 . Given: ABCD is a parallelo- i—)■

gram in which AD has been extended to E.

Concl.: A A B E ~ A C F B

7 . Given: A A B C is isosceles w ith

I B S * A ?.

D E L A B

D F L A C Concl.: A D E B ~ A D F C

9 . Given: A B = A C *-*AD X BC <-*■EF X AC

Concl.: A A B D - A E C F

Given: A B II EF

AC II DF Concl.: A A B C ~ A P E D

351

6.

Given: D E I! AB

DF II AC Concl.: A D E F ~ A A B C

G iven:B E _ BC_AC BD Z l S / 2

C oncl.: A A BD ~ A EBC

8.

10.


11. Given: A A B E ^ A A C D Concl.: A A D E ~ A A B C

(H int: Use S.A.S. theorem on similarity.)

Given: A D C C = A E F B Z l S Z2

C o n c l: A A F G ~ A A B C

12.

01. Two right triangles are similar if an acute angle of one is congruent to

an acute angle of the other,

2. Two triangles th a t are similar to the same triangle are similar to each other.

3. Equilateral triangles are similar triangles.4. I f from each of two points on the bisector of an angle perpendiculars are

drawn, one to each side of the angle, two similar triangles will be formed.•5. If two triangles have their sides respectively parallel, then the triangles

are similar.

6. An altitude to the hypotenuse of a right triangle will divide the triangle into two triangles such that either of these triangles will be similar to the original triangle.

7. I f from two points, one on each leg of an isosceles triangle, perpendiculars are dropped to the base, the two triangles formed will be similar.

8. I f two parallel lines intersect two intersecting lines and form two triangles, these two triangles will be similar.

9. Tw o isosceles triangles are similar if a base angle of one is congruent to a base angle of the other.

10. If two similar triangles have a pair of corresponding sides congruent, then the triangles are congruent.

11. Two isosceles right triangles are similar.12. T he line segmfents joining the midpoints of the sides of a triangle form a

' triangle that is similar to the original triangle.

PROVING RATIOS EQUAL AND PRODUCTS EQUAL 353

B Proving Ratios Equal and Products EqualAfter we had proved two triangles to be congruent, it was

possible for us to conclude th a t the corresponding sides and corresponding angles were congruent. In the same way, once triangles have been shown to be similar, the definition of similar polygons leads us to the fact that the ratios of the measures of the corresponding sides will be equal. As an example, consider the triangles in Figure 11-19. W ere it necessary to prove th a t A B BC= 7^ ;, we would show, in some m anner, th a t A A B C ~ A R S T , T h erO 0 1

definition of similar polygons would then enable us to conclude that the two ratios were equal.

T here are times when the proportion m ay be w ritten as

H = (See Figure 11-19.) (1).

th a t is, where the terms in each ratio come from a single triangle ra ther than being corresponding sides in the two triangles. W e know, however, th a t the above proportion leads to the product

A B - S T = B C -RS

This product, in turn, can be rew ritten as any one of eight different proportions. O ne of these is

M = m (2)R S S T w

which is identical to the proportion discussed earlier and involves the ratios of the corresponding sides of the triangles. In proportion (1) the corresponding sides were the first and th ird proportionals; also the second and fourth. From our analysis above we see th a t this can v e rewritten in form (2),where the corresponding sides appear as the first and second proportionals;also the third and fourth. Hence, we will consider th a t either form (1) or form (2) follows from the fact th a t triangles are similar.

354

Illustration:

SIMILAR TRIANGLES

Given: ABCD is a parallelogram . 0 <-> «-*

AE 1 BC«-+ *->

AF 1 CD

f ig u re 1J-20.


1. ABCD is a parallelogram.2. Z B = / .D

3. A E L BC4. / A E B is a right angle.

5. A F 1 CD

9.

6. ZA F D is a right angle.7. Z A E B =~ ZAFD8. A A B E ~ A A D F

A B _ AD A E AF(Def. of similar polygons)

Situations also arise in which we m ay be asked to prove th a t two product relations are equal. Thus, in the illustration above consider the conclusion o f:

A B -A F = A E -A D

Since these four numbers can be expressed as the proportion

A B AD A E ■ AF

the proof of this problem becomes identically the same as the one illustrated w ith one additional s tep :

10. A B -A F = A D -A E 10. T he product of the m eans of a proportion is equal to the product of the extremes.

As a word of suggestion, should a conclusion call for showing that the product of the measures of two segments is equal to the product of the measures of two other segments, rewrite these four numbers in the form of a proportion. By examining the terms of the proportion, the triangles th a t must be shown to be similar can m ore readily be determined th an from the product form.

Remem ber that, as before, in most instances the first and second proportionals will be the measures of line segments in the first triangle, while the third and fourth proportionals will be the measures of line segments in the second triangle. If this is not so, then the first and third proportionals ■will be the measures of line segments in the first triangle, while the remaining proportionals will be the measures of segments in the second triangle.

PROVING RATIOS EQUAL AND PRODUCTS EQUAL

EXERCISES

355

1 . Given: AB X BD

D E I BD AB _ BC

C oncl.' DE ~ DC

3 . G iven: Z C is a right angle.

D E 1 AB „ . AD AEConcl.: AB AC

5 . Given: ABCD is a parallelogram.A B _ DE

Concl.: A F FE

Given: AD _L CD

Concl.

CB 1 AB AD _ BC D E BE

Given: ABCD is a trape

zoid w ith AD II BC. A E AD

C oncl.: EC BC

G iven: ~SD and V E are altitudes in A ABC.D F _ CF

C oncl.: l B p

4.

6.

356

Given: A E and BD are altitudes in A ABC.

Concl.: ~ = ~ /‘CD CB

9 . G iven:

Concl.

A A B C ~ A R S T S B is an altitude to 2V .

S W is an altitude to R T .BD _ AC S W ~ R T

11. G iven: EG II BC

C oncl.:

SIMILAR TRIANGLES

8.Given: A A B C ^ ^ R S T •—►

AD bisects ZBAC .

R W bisects Z SR 1

Concl.:A B R S

Given: E F || BC

Concl. EG _ GF B D ~ DC (H int: Prove both ratios equal to the same ratio.)

G iven: A B CD is a parallelogram.

E F and GH inter

sect diagonal a t P.

10.

12.

PROVING RATIOS EQUAL AN D PRODUCTS EQUAL 357

13. G iven: D F || A B Z \ S Z l

Concl.: BC'.AC = B D :AF

15. Given: A B J . BD

DC L AC Concl.: B E -E D = A E -E C

17. Given: Z A C B is a right angle.

CB is an altitude

to AB.Concl.: (,4C)1 = A B -A D

(H in t: R ew rite as c AC-AC.)

Given: A A C D is equilateral. 14. C and D a re trisection

points of BE.Conci.: (A B Y = B C -B E

A

G iven: A BCD is a parallel

ogram.

T e l b c

A F L C DC oncl:

(a) A E -A D = A F -A B(b ) A E -B C = AF-C D

Given: ZA C B is a right angle.PQRS is a rec-

. tangle.Concl.: (PQ)1 = A Q -B R

16.

18.

358

*19. Given: A B ^ A C

A B and B F are aldtudes.

Conci.: A B '.B E — A D :D C

21. Given: A B ^ A C Cb ^ U B

Concl.: A B -D B = (C B )2

o * < - » < - +Z J . Given: A B ± B F

<-4 <-+CD 1 B F *-* <-*

EF 1 B F Concl.: B D -E F - D F -A B

Given: All points are 20.coplaner.<-> <->

BC || EF <--» <-»

j| D EConcl.: A ABC ~ A D E F

SIMILAR TRIANGLES

Given: A A B C is isosceles 22.* with i4.fi S 2(7.

CD L A B Concl.: (BC)1 = 2 A B -B D

(H int: L et A E be the perpendicular

from A to BC.)

G iven: A B X BC 24*<-+

DC X BC <-> «-♦

B D 1 A C Concl.: (BC)1 = A B -D C

PROVING RATIOS EQUAL AND PRODUCTS EQUAL 359

2 5 . Given: BC II FE

AC II FDC oncl.:

(a) A A B C ~ A D E F

(b ) AB II ED

G iven: M is the m idpoint 26 . of AB.<-> «->

CQ II AB M , R, Q, P are collinear.M R _ PM RQ PQ.

C oncl.:

2 7 . Given: AD II BCA A B E ~ A D C E

C oncl.: ABCD is an isosceles trapezoid.

G iven: plane a II plane b Concl.: A A B C ~ A P Q R

2 8 .

B1. T h e diagonals of a trapezoid divide each o ther into two equal ratios.2. T riangle ACB is a right triangle with angle C the right angle. If from

any point P on leg BC a perpendicular is drawn to the hypotenuse A B m eeting it at D, then B P-BC = B D -B A .

3. A m edian to a side of a triangle bisects all line segments parallel to th a t side and term inated by the other two sides.

4. T he ratio of the measures of corresponding medians of two sim ilar


triangles is equal to the ratio of the measures of any two corresponding sides.

5. I f three lines pass through a common point and intersect two parallel lines, then the three ratios of the measures of the segments cut off by the parallel lines are equal. :If, in a right triarg ie an altitude is drawn to the hypotenuse, then the m easure of the altitude is the mean proportional between the measures of the segments of the hypotenuse.

7. If two triangles are similar, the ratio of the measures of corresponding altitudes is equal to the ratio of the measures of corresponding m edians drawn from the same vertex.

8. If an altitude is drawn to the hypotenuse of a right triangle, then the product of the m east^is of the altitude and the hypotenuse is equal to the product of the measures of the legs of the right triangle.

9.* If a line segment term inating in the nonparallel sides of a trapezoid is parallel to the bases and passes through the point of intersection of the diagonals, then it is bisected by the diagonals.

\ j ■ The Right Triangle

If an altitude is drawn to the hypotenuse of a right triangle, several theorems result th a t have numerical application. In addition, these theorems will enable us to prove the paost widely known property in the field of geometry, the Theorem .of Pythagoras.

T H E O R E M 68: In a r ig h t trian g le w ith an a ltitu d e to th e hyp o ten u se th e sq u are of the m easure of e ith e r leg is equal to th e p ro d u c t of th e m easures of th e e n tire hypotenuse a n d th e segm en t of th e hypotenuse ad jacen t to th a t leg.

Given: ZA C B is a right angle.S B is the altitude to AB.

Concl.: (a) (BC) 5 = B A -B D (b ) (AC)2 = A B -A D

Figure 11-21.

A n a ly s i s : P a rt (a) of th e conclusion can be proved ra ther easily by showing A B C D ~ A B A C . Part (b) of the conclusion is very m uch the same except th a t A C A D m ust be shown to be similar to A BA C . You may have noticed th a t Part (b) is the same as Problem 15, page 357,

THE RIGHT TRIANGLE 361

PRO OF j (The reasons will be left for you to supply.)

6. Z B S Z B7. A BCD ~ A B A C

BC BD8' BA " BC9. (BC)1 = BA ■ BD (Part (a))

1 . ZA C B is a right angle.

2. CD is the altitude to AB.

3. CD _L AB4. ZC D B is a right angle.5. Z A C B ^ ZC D B |

T he proof of P art (b) is left for you to do.

T H EO R EM 69: In a r ig h t tria n g le w ith an a lt i tu d e to th e h y po tenuse th e sq u a re of the m easure of th e a lt i tu d e is e q u a l to th e p ro d u c t o f th e m easures of th e segm ents o f th e h ypo te

nuse.

Given: ZA C B is a right angle.UD is the a ltitude to AB.

Concl: (CD)2 - A D -D B

Figure 11-22.

A n a l y s i s : After rewriting the conclusion in terms of the proportion

CDAD

DB '' CD

we find that it is necessary to prove A ACD ~ A CBD. T h e two angles a t D can readily be shown to be congruent. I t is also evident th a t we will no t show Z B to be congruent to Z A , for if this were so, A ACB would be isosceles, which it is not. Hence, this implies th a t we will have to prove Z A to be congruent to Z \ .


CD is an altitude to AB.«-» *->

2. CD 1 ABZCD A and ZC D B areright angles.ZC D A S ZC D B ZA C B is a right angle.Z \ is complementary to Z t .

1.

3.

4.5.6 .

7. Z A is com plem entary to Z t .

(Theorem 53)8. Z A ^ Z \9. A A C D ~ A C B D

CD _ D B10' AD ~ CD 11. (CD)2 = A D -D B

As a point of interest, in the proofs of the last two theorems not only were the two small triangles proved similar to each o ther bu t they w ere a lso both proved similar to the large triangle.

362 SIMILAR TRIANGLESQ uite frequently the sides of a triangle are labeled with small letters,

while the angles opposite them are labeled with the same capital letters. W hen the right triangle is so lettered, it becomes

F i g u r e 1 1 -2 3 .

This iettering enables us to see that Theorem s 68 and 69 have given us information concerning the squares of a, h, and b. This is

(1) O'

In try ing to find the solution to num erical problems, it is best to label all six segments before deciding which of the three equations listed above is appropriate .

Illustration: ^

Given: A ABC is a right triangle with TOD the altitude to the hypotenuse AB.AC = 9, AD = 6

T o Find; D B = ?

fig u re 11-24.

M e t h o d : As suggested, label all six segments a t the outset. For this prob-

Figure 11-25.

lem the three equations listed above become

( 1) a2 = x (x + 6) (2) h2 = 6x (3) 9! = 6 (x + 6)

Since We are looking for the num ber labeled “x,” we will use the equation in which x is the only placeholder that appears. In this case it is equation (3):

THE RIGHT TRIANGLE 363

81 = 6x + 3645 = 6x 71 « x

T here may be situations in which it is necessary to find first a segment o ther than the one for which we are looking. T his will depend on which of the three equations contains a single placeholder. Once this placeholder has been found, substitute its value in the equat.' _.n containing the m easure you have been asked to find.

EXERCISES

c

1. G iven: A A B C is a right triangle with CD the altitude to the hypotenuse.

(a ) AD = 4 ,D B = 16, CD = ? (b).£>5 = 4, A B = 9, BC = ?(c) A B = 27, D B = 24, AC = ? (d ) AC = 6, AD = 3, A B — ?(e) CD = 10, AD = 5, DB = ? (f) BC = 8 ,A B = 16, AD = ?(g) A B = 12, AD = 4, CD = ? (h ) AD = 6, D B = 8, BC = ?(i) BC = 4 ,A B = 8, CD = ? ( j) A B = 10, AC = 6, BC = ?

R2. Given: A R S T is a right triangle with

T W the altitude to the hypotenuse.(The last three problems in this group will involve the solution of quadratic equations.)

T(a) R T = 12, R S = 16, R W = ? (b ) SW = 6, R W = 8, S T = ?(c) T W = 5, R W = 6, RS = ? (d ) S T = 4, R W = 6, W S = ?(e) T W = 6,R S = 13, R W = ? (f) R T = 9, W S =* 24, R S = ?(Two answers)

3. Given: A A C B is a right triangle withCD the altitude to the hypotenuse.BC = 6, BD = 4

T o Find: AB, AD, CD, AC

A


4. Given: A Q P S is a right triangle with

F R the altitude to the hypotenuse.QR = 2, P S = 8

T o F ind : QS, PR, PQ, PS

5. In an isosceles right triangle the measure of the altitude to the hypotenuse is 4. W hat is the measure of each leg?

i-tI

3 . Given; A B i . BC

EG iven: AC JL CB

<-+ <-»D E L CB *-* «->

CD J . A B Concl.: (CD)* - f (D E)1 =

A D -D B + C E-EB

DC 1 B C

B E ± AC Concl.: B E -B D = CE-CA

10

Given: is a rectangle. 2 .R W JL QS

R W intersects Q P a t T- Concl.: (PS)1 = R W - R T

G iven: ZA C B is a right angle. 4 .

CD ± A B

C o n c l . : ® - 5 5 (CA)3 AD

THE THEOREM OF PYTHAGORAS 365

5 . Given: Circle 0 with 0 on AB.

AC L C B

AB L D C Concl.: A E -E B = CE-ED

A

7 . Given: In space geometry >

PA .J . plane m «-> *-»

AD J . PB <-> <->

A E JL PC Concl.: P D -P B = PE-PC

Given: AD X BC

D E L AC

D F L A B Concl.: A E -A C = A F -A B

■ The Theorem of PythagorasThe relationship between the legs and the hypotenuse of a

right triangle was known for certain special right triangles long, before it was proved for every right triangle. During the tim e of the(ajicien_t_£gypti^s^ this relationship was held in secret trust by their religious leaders. Each tim e a building was to be erected, these men, called “ rope stretchers,” had to be employed to lay out the corners of the structure so as to form a right angle. .

D uring the sixth century B.C. a secret'm athem atical society was organized by the Greek m athem atician Pythagoras, jind for obvious reasons it was called the Pythagorean Society;'Story lias it th a t any m athem atical discoveries made by a member of the group could not be to ld to the world a t large without permission having been granted by Pythagoras. Supposedly, a Pythagorean developed a proof for the theorem on the right triangle and announced it to friends ..before consulting Pythagoras. For breaking faith with the organization, the m an was “ done away w ith" in true m odern gangland style!

M any, m any proofs have since been given for the T heorem of Py

thagoras. In fact, recently someone has collected and published close to 300 of these proofs, one of which was developed by President Garfield. Those presented by the Greeks were probably based on areas o f squares and are quite a bit m ore difficult than the one given below.

T H E O R E M 70: T h e squ are o f th e m easure of th e hyp o ten u se of a r ig h ttr ia n g le is eq u al to th e sum of the squares o f th e measu res o f th e legs.


Given: A ACB is a right triangle with ZAC B the right angle.

Concl.: c! = a1 + b2

A nalysts: Theorem 68 enables us to draw conclusions about both a1 and b2.Hence, it is merely a m atte r o f adding these two quantities and hoping that th e ir sum will turn ou t to be c3.

PR O O F STATEMENTS

1 ■ Let CD be the perpendicular from C to AB.

2. A ACB is a right angle.3. b2 = cx4. a* — c (c — x)5. w + c (e _ x)

or a' + b* = c x + c * - c x or a* + b* = c*

REASONS

1. T here exists one and only one

perpendicular from a given point to a given line.

2. Why?3. Why?4. Why?


e x e r c is e s

AA1. Given: A ABC is a right triangle i

with Z C the right angle. |


(a) AC = 5 ,B C = 12, AB = ? (b ) A B = 20, BC = 16, AC = ?(c) AC = 10, AB = 26, BC = ? (d ) AC = 5, BC = 6, A B = ?

✓ 2. Determ ine the length of the diagonal of a rectangle whose sides are6 inches and 8 inches respectively.

( / 3. (a) Find the length of the diagonal of a square one of whose sides is 4 inches.

✓ (b ) Find the length of the diagonal of a square one of whose sides is aunit? long.

4. Find the altitude to the base of an isosceles triangle if one of its legs is 26 feet long while the base is 48 feet.

<■' 5. (a) Determine the length of one of the sides of a square whose diagonal is 6 inches.

(b ) Determine the length of one of the sides of a square'whose diagonal is 2a units long.

•' 6. (a) Determ ine the length of a side of a rhom bus whose diagonals are18 inches and 24 inches respectively.

i / (b ) Determ ine the length of one diagonal of a rhom bus if the otherdiagonal is 30 inches while a side is 25 inches.

/ 7. (a) How long is an altitude of an equilateral triangle if one of its sides is 10 inches?

^ ' (b ) How long is an altitude of an equilateral triangle if one of its sidesis 2a units long?

1/ 8. (a) O ne of the legs of an isosceles right triangle is 6 inches. How long is the hypotenuse?

>■' (b ) O ne of the legs of an isosceles right triangle is a units long. How long is the hypotenuse?

9. (a) T he hypotenuse of an isosceles right triangle is 8 inches. W hat is the length of each leg?

(b ) T he hypotenuse of an isosceles rie^ht triangle is a units. W hat is the length of each leg?

10. (a) Find the side of an equilateral triangle whose altitude is 6 inches,(b ) Find the side of an equilateral triangle whose altitude is 3a units

long./ 11. O ne of the nonparallel sides of a trapezoid is perpendicular to the bases.

T he other of these sides is 10 inches, and the upper and lower bases are 9 inches and 17 inches respectively. How long is the side whose length was not given?

y 12. T he upper and lower bases of an isosceles trapezoid are 16 feet and46 feet respectively. If one of the equal sides is 17 feet, how long is the altitude? (The altitude is the common perpendicular to the bases.) .

368 SIMILAR TRIANGLES/ •' .

13, A ladder 41 feet long lpans against a building and reaches a point on the building that is 40 feet above the ground. How far from the bottom of the building is the foot of the ladder?

■yj 14. A rope 35 feet long is attached to the top of a high vertical pole. W hen stretched tight, the rope reaches a point on the ground that is 13 feet from the foot of the pole. How long ii

15. T he box a t the right has dimensions as shown. How far is it from .4 to C?(H int: Find A B first.)

12 fe e t

16, T he dimensions of the box a t the right are all equal. Using the dimension shown,(a) Find the distance AB.(b ) Find the distance AC.

17. If, in the d iagram a t the right, P is

the m idpoint of BC, find AD.

18. In the diagram a t ’the right find the

m easure of CD.

19. D uring a wind stor.-.i a pole 20 feet long snapped in a way such th a t the topmost point of the upper section leaned over and reached a point on the ground that was 10 feet from the foot of the pole, while the lowest point of this section still remained attached to the pole, How long was this upper section?


s

1 , Given: A B J. BC

DC 1 BC Concl.: {A O 1 “ {ABY =

3 , Given: ABC is a rightangle.

Concl.: (AC)* + (DE)* (A E )' + (DC)2

Given: D B _L AC Concl.: (A D )2 + (B C )2 ■

(A B )2 + (D C )1

Given: AC _L CB

D E 1 CB *-+ *-*

D F 1 AC <-> «-*■

CD ± A B Concl.: (CD)1 = C E -E B +

CF-FA

4 .

5 . Given: BD X AD ___C is the m idpoint of AD.

m/LA=* 45 Concl.: (AB)- = 8(/4C)5


6. Using the diagram at the right, prove that c2 = a2 + b2 + 2ax.

8 . Prove the converse of the theorem of Pythagoras: If c2 = a2 + b2, then Z C is a right angle. M ethod w Let Q be a right angle, then let QR = a, and QP = b\ prove the triangles congruent by S-S.iS1.

V 9 . (a) Referring to Problem 8, show that the numbers 2n, n2 — 1, and n2 + 1 can represent the measures of the sides of a right triangle. Three num bers such th a t the sum of the squares of any two is equal to the square of the third are called Pythagorean Numbers,

(b) Show that the num bers 2b + 2, b2 + 2b, and b2 + 2b + 2 are Pythagorean Numbers.

10. (a) Prove that in a right triangle whose acute angles are 30° and 60°the ratio of the measure of the hypotenuse to the measure of the side opposite the 30° angle is 2:1.

(b ) Prove that in a right triangle whose acute angles are 30° and 60° the ratio of the measure of the side opposite the 60° angle to the

measure of the side opposite the 30° angle is v^3 : 1. (H int: For the proof of both (a ) and (b), start with an equilateral triangle.)

11. Using the theorem of Pythagoras, prove th a t if in space geometry two oblique lines and a perpendicular are drawn to a plane from a point such that the foot of the perpendicular is equidistant from the feet of the oblique lines, then the oblique lines are congruent.

12. State and prove the converse of Problem 11 by applying the theorem of Pythagoras.

TEST AND REVIEW

■ Test and Review

371

A

1. Find the value of x in each of the following proportions:

4 - I ' 6 - — 1—W 9 " * ( ' * - 2 * + 3

2. Find the fourth proportional to 3, 7, and 5.3. Find the third proportional of a m ean proportion if the first and second

proportionals are 2a and 34.4. Find the m ean proportional of a m ean proportion if the first and th ird

proportionals are 2a and 18ab2.5. W rite eight different proportions expressing the relation between 2, 3,

a, and b if 2a = 3b.6. If a = 2b, then what is the ratio of a to b?7. If the measures of the angles of a triangle a re in the ratio of 1 :2 :3 ,

then w hat is the measure of each angle? a

_ r„AB 4 , , A B + BC8‘ ~BC = 5* then f in d — g g —

9. Use the diagram a t the right to find the measure of each of the following

segments given that D E II BC.(a) 1 U (b ) D E

10. Use the diagram at the right to find the measure of each of the following

segments given that SV II R W.

(a ) TW (b ) 7TWw

11. T he similarity correspondence between the vertices of A A BC and A D E F is A BC <-» EFD. Nam e the pairs of angles in. the two triangles th a t are congrucnt,

12. I f the measures of two corresponding sides cf' two similar triangles are 8 and 12 respectively and the m easure of the m edian to the first side is 6, w hat is the measure of the m edian to the second side?

13. T he measures of two of the altitudes of a triangle are 3 and 5 respectively. If the measure of the side to which the second is draw n is 8, w hat is the m easure of the side to which the first is drawn?


14. Using the information given in the diagram at the right, find the measure of each of the following segments:(a) W S

W

15. If the altitude to the hypotenuse of a right triangle divides the hypotenuse into segments whose measures are 2 and 8 respectively, then w hat is the perim eter of the triangle?

16. (a ) T he upper and lower bases of an isosceles trapezoid are 16 and 24respectively. If one of the legs makes an angle of 45° w ith the lower base, w hat is the perimeter of the trapezoid?

(b ) If the angle had been 60°, what would the perim eter have been?(c) If the angle had been 30°, what would the perim eter have been?

B

1, Given:

Concl.


Given:A A C B with Z C the right angle,<-* *-*

D E ± CB BA = BD BC ~ BE

3 . G iven : D is the m idpoint of ~AU.E is the midpoint of HU.

C oncl.: A PAB ~ A P E D

C oncl.: -77-,

AFF B ‘ AD A H '

A

AGGCA EA J

Given: B E bisects ZA B C . Z \ S Z l

. A E AD nc ~EC = D E

TEST AND REVIEW 373

5 . Given: Z l ^ ZA C BConcl.: (BC)2 = BD X BA

7 . Given: ABCD is a trapezoid with the nonparallel sides extended to in tersect a t P.

~ , PF A BConcl’: PE = DC

Given: AD arid B E are alti- 6.tudes in A ABC.

Concl.: C A X C E = CB X CD

G iven: R ectangle AC BF

CE JL AB Concl.: (B F ) J = A E X AF

Prove each of the following statem ents:

1. If two triangles are congruent, then they are similar.2. If the midpoints of the sides of an isosceles trapezoid are jo ined in order,

the quadrilateral formed will be a rhombus.3. If line segments are draw n joining the m idpoints of the opposite sides

of a quadrilateral, they wili bisect each other.4. T he ratio of the measures of corresponding angle bisectors of two similar

triangles is equal to the ratio of the measures of any two corresponding sides.

5. If a line joins the m idpoints of two opposite sides of a parallelogram , i t will bisect both diagonals.

374 SIMILAR TRIANGLES6. If the vertex angle of one isosceles triangle is congruent to the vertex

angle of a second isosceles triangle, then the ratio of the measures of their bases is equal to the ratio of tiie measures of their legs.

7. Tv/o medians of a triangle intersect at a point such that the ratio of the measures of-the segments of one is equal tc the ratio of the measures of the segments of the other.

8. I f two parallel planes intersect tv/o intersecting lines, then the ratio of the measures of the segments of one will be equal to the ratio of the measures of the segments of the other.

■ Try This For FunPrinciples in m athem atics frequently bear the nam e of the

person who first noted them. Thus, we are already familiar with E uclid ’s Fifth Postulate, T he Theorem of Pythagoras, Aristotle’s Laws of Logic, Pasch’s Axiom, and others. Am ong the theorems in geometry th a t bear their originator’s name, perhaps the most intriguing is Ceva’s Theorem :

If from the vertices of a triangle concurrent lines are drawn intersecting a t a point in the interior of the triangle, then of the six segments formed by the intersections of these lines with the sides of the triangle the product of the measures of three alternating segments is equal to the product of the measures of the remaining three segments.

Given: A M , BQ, and C N are concurren t a t P.

Concl.: A N X B M X CQ = N B X M C X QA

Can you prove this theorem? Suggestion: Through B and C draw lines

parallel to M A .

Coordinate Geometry An Introduction

T H E PR IN C IPL E S O F S IM IL A R IT Y O F T R IA N G L E S and the theorem of Pythagoras have opened before us a broad new field— the field of a nalytic, o r coordinate, geometry. U ntil now we have concerned ourselves w ith a rather stilted and cum bersom e tool developed some 2,000 years ago by Greek mathematicians. U nfortunately, they had neither the algebra nor the arithmetic that we currently possess and , hence, were forced to form ulate their mathematics in term s of lines, or angles, or solids thatthey actually perceived.

T he development of algebra led a seventeenth century m athem atician and philosopher, Ren6 Descartes, to take the first giant step in geometry since the time of the Greeks. By showing how algebraic m ethods can be applied to geometric situations, he m ade possible a great deal of the growth of m odern mathematics. This tool is not only very powerful bu t also re latively simple to understand. W e shall try to develop some of its elem entary concepts and show how they can be used to prove a few of the propositionswe had proved earlier.

This is not m eant to belittle the value of synthetic geometry, the geometry of the ancient Greeks, for as we shall see, there a re a num ber o f situations in which a “ synthetic” proof will be far easier than an “ analytic” one. However, other than possibly for a brief course in w hat is called “college

376 COORDINATE GEOMETRV: A N INTRODUCTION

geom etry,” synthetic geometry “ goes nowhere,” nor does it have extensive practical application. Analytic geometry, on the other hand, forms the basis of m any areas of mathematics and has wide application.

3 Plotting Points

In the study of algebra and also as p a rt of the early work in geometry we become.-'familiar with a line th a t was called either the num ber line, o r the num ber scalc, or, possibly, the num ber axis. Basically, we assumed th a t a one-to-one correspondence existed between th e points of this line and the real numbers. The real num bers were the whole num bers, fractions, mixed numbers, decimals, .'rrational num bers (v^2, y / l , r ) , and, incidentally, all of these m ight have been either positive or negative. T hey d id no t include, however, num bers such as V —25 th a t w ere called im aginary and expressed as Si. And so, the assumption th a t was m ade was th a t for each real num ber there exists a unique point of the num ber line and, conversely, for each point there exists a. unique real num ber.

T h e num ber line was frequently drawn horizontally w ith an arrow head

- 4 - 3 - 2 - 1 0 +1 + 2 + 3 + 4

F ig u r e 12-1.

a t the right to indicate the direction in which the num bers increased. T here is nothing sacred about its being drawn horizontally, for equally well the num ber line m ay have been drawn as shown in Figure 12-2 (again, the

arrow head shows the positive direction). T hrough custom, however, th e line is drawn horizontally and the direction in which the num bers increase is to the right.

T h e num ber from which we start counting is called zero, while the point on the qum ber axis representing this num ber is referred to as the origin. By asking the question “ How is the position of each of these num bers determ ined on the num ber line?” , we bring to light a rather im portant feature. T h e position of each num ber is determ ined by its distance from the

PLOTTING POINTS 377

origin. Thus, the position of the point representing the num ber + 3 is determined by the fact that this point will be 3 units to the right of the origin. Hence, we can say that the distance from the point + 3 to the origin is 3 units. Notice that the name for the point is given by the num ber th a t leads to the point. This num ber is called the coordinate of the point. In this case, the coordinate of the point is + 3 . Thus, the coordinate of a point on the num ber line represents not only the name for th a t point bu t also the distance from that point to the origin. Similarly, knowing th e coordinate of a point, we a re really being told a direction through which the poin t can be locatedon the number line.

T o illustrate, the point whose coordinate is —4 (think of this as both the name for the point as well as its distance from the origin) is found 4 units

to the left of the origin.

- 4 - 3 -1 0

F ig u r e 12-3.

T he geometry we are learning, however, is not lim ited to figures th a t exist on a line, but rather, to figures existing in k plane. T his confronts us squarely with the need to represent points of a plane in some m anner similar to that used to represent points of a line. T o do this, we set up two num ber

se co n d q u a d ra n t+ 1

-3

th ird q u a d ra n t

-1

-1

first q u a d r a n t

+1

F ig u r e 12-4.

+2 +3

fo u rth q u a d r a n t

lines that are perpendicular to each other. Each of these divides the plane into- two half-planes, while the two together will divide the plane into four quarter-planes called quadrants. T he quadrants are num bered as shown in the diagram.

Each point of the num ber line was represented by a real num ber in dicating the distance from th a t po in t to the origin. In keeping with.this idea, a point of the plane will be represented by a pair of real numbers representing the distances from th a t point to each of the num ber lines. But w hat represents the distance from a point to a line?

In the discussion of distance earlier in the tex tf it was pointed out that

t S ee p a g e -179.

i 378 COORDINATE GEOMETRY: AN INTRODUCTION\

the distance between two ohjects was always measured along the shortest £ path , or geodesic, between the two objects. W hen these two objects arei; points, w e assume th a t the shortest path will be the line segm ent joining| these two points. At this tim e w e will further assume

I, P o s t u l a t e 32: T h e s h o r t e s t p a t h b e t w e e n a p o i n t a n d a l i n e is t h e p e r p e n -l dicular line segment from the point to the line,t <j In view of this, the distance from a point to a line is the m easure of the

perpendicular segment betw een the point and the line. In each of the dia-r grams of Figure 12-5 the m easure of the red segment is the distance from the

point P to the line /.

1fp ^

.r'

X .

j Tr

1 P

Figure 12-5.

R e tu rn now to the points in a plane and their representation by a pair o f real num bers. As in the case o f the num ber line, the pair of real num bers a r e called the coordinates of the point. T hey represent no t only the nam e for th e point but also the distances from that point to each of the num ber lines, called axes.

T o illustrate, consider the point whose name is (+ 3 , + 5 ) . In reference t o a num ber line, the + 3 indicated that the point was located 3 units to the r ig h t of the origin on the num ber line. On a plane the direction + 3 will in d ica te that the point is 3 units to the right of the vertical axis. U nfortun a te ly , th a t direction is n o t sufficient to locate the point in the plane, for t o be 3 units to the right of the vertical axis is to be anywhere along a line p a ra lle l to th e vertical axis and 3 units to its right. Each of the points on the

Figure 12-7.

PLOTTING POINTS a/yred line in Figure 12-6 above fulfills the condition th a t i t is 3 units to the right of the vertical axis. Thus, the need for the second coordinate. T he + 5 indicates that the point m ust also be 5 units above the horizontal axis.- T h e point of intersection of the two red lines represents th e point ( + 3 , + 5 ) . Since two lines can have but one poin t in comm on, the pa ir of num bers ( + 3, + 5) represents one and only one point.

A pair of num bers such as ( + 3 , + 5 ) is called an ordered pair, for the order in which the num bers a re w ritten is extremely im portant. T he first coordinate, by agreement, will always represent the direction w ith reference to the vertical axis; in this case, it states th a t the point is 3 units to the right of the vertical axis. T h e second coordinate will always represent th e direction w ith reference to the horizontal axis; in this case, it states that the point is 5 units above the horizontal axis. W ere the orde:; of these two num bers reversed—th at is, ( + 5, + 3)—to w hat point would these directions lead us?

As in the case of the direction of the single num ber axis, there is nothing but custom that insists that these lines be d raw n horizontally and vertically. Frequently the horizontal axis is called the x-axis, while the vertical one is called the y-axis. T h e combination of the two are known as the coordinate axes. T h e representation of points in a plane by the m ethod described is called the Cartesian Coordinate System.

P(+3,+5)

Figure 12-8.

T he distance from the point to the vertical axis is called the abscissa of the point. In the diagram above the abscissa is the measure of line segment QP, which is given by the first coordinate, + 3 . T he distance from the point to the horizontal axis is called the ordinate o f the point. In this d iagram the ordinate is the m easure of line segm ent RP, w hich is given by the second coordinate, + 5 . Thus, since one of the functions of the coordinates of a point is to indicate the distances from the point to each of the axes, the coordinates are often called the abscissa and ordinate o f a point.

I t is extremely important th a t we realize th a t the pair of directions—the point is 3 units to the right of the vertical axis and 5 units above the horizontal axis—can be given in any one of three ways: '

380 COORDINATE GEOMETRY! A N INTRODUCTION( 1) the ordered pa ir of num bers: ( + 3, + 5)

(2) the pa ir of equations: ^

(3) by its absciss?, and ordinate:T he abscissa is + 3 and the ordinate is + 5 .

Note th a t w hen the directions are given as a pa ir of equations, the x element represents the first coordinate in the ordered pair, or the distance from the vertical axis; th e^ elem ent represents the second coordinate in the ordered pair, or the distance from the horizontal axis.

Should one and only one direction be given—such as, x = —2— this would uniquely establish not a point bu t an entire set of points all located on the line th a t is two units to the left of the vertical axis and parallel to it. R em em ber that the direction x = —2 is read as, “ T h e point is two units to the left of the vertical axis.” (See Figure 12-9.) Similarly, the single direction

uK

Figure 12-9. F igure 12*10.

“ T he.ord ina te of the point is —4” will lead to a set of points all on a line parallel to th e x-axis and 4 units below it. W ere this direction restated, it would simply be that the point is 4 units below the horizontal axis. (See F igure 12-10.)

Before leaving this topic, it would be well to point ou t th a t w e have been assuming that

P ostula te 33: For a.specific pair of axes there exists a o ne-to -one correspondence between the set of ordered pairs of real num bers an d the set of points of a plane.

EXERClSESt1. D raw a pa ir of coordinate axes and locate each of the following points;

(a ) (+ 5 , + 7 ) (b ) ( + 6,9 ) (c) (7, + 4)(d ) (2 ,1 0 ) (e) ( - 3 , 6) (f) ( - 7 , 2)(g ) (4, - 6) (h ) (8, - 3 ) (i) ( - 4, - 9)( j ) ( —12, —3) (k ) (0 ,5 ) 0 ) (7 ,0 )(m ) ( 0 , - 1 4 ) (n ) ( - 1 5 ,0 ) (o ) (0 ,0 )}

f Use coordinate (or squared) paper for these exercises.X This point is called the origin.

PLOTTING POINTS 381

2. Draw the triangle having the following vertices:(a) (0, 0), (6, 0), (0, 8) (b) (2, 14), (2, 2), (7, 2)(c) ( - 4 , 7), (8, 7), (8, - 9 ) (d ) (7, 4), (3, 6), (8 , 6)(e) (2, 5), ( - 4 , 7), ( - 2 , - 3 ) (f) ( - 6 , - 4 ) , (0, 0), (2, - 7 )

3. T he triangle in exam ple 2a is a right triangle. How long is the hypote

nuse?4. T he triangles in examples 2b and 2c are right triangles.

(a) Find the length of the legs of each of these triangles.(b ) Find the length of the hypotenuse o f each of these triangles.

5. Determine the set of points located by each of the following directions:(a ) x = 6 (b ) y = 8 (c) x = — 4 (d ) y = —2 (e ) y = 0

6. Determine the set of points located by each of the following directions:(a ) T h e abscissa of the poin t is always 10.(b ) T he ordinate of the point is always —7.(c) T he ordinate of the point is always 0.(d ) The ordinate o f the point is always equal to the abscissa of the point.

7. Represent each of the following points by a pa ir of equations:(a ) (5, 9) (b ) (4, —7) ( c ) ( - 3 , 0 )(d ) The abscissa is —5, while the ordinate is —7.

8 . Represent each of the following statements by an equation:(a) T he abscissa of a point is always 4.(b ) T he ordinate o f a point is always —12.(c) T he abscissa of a point is always 0 .(d ) T he abscissa of the point is always equal to the ordinate of the point.(e) T he ordinate of the point is always five more than the abscissa of

the point.(f) T h e abscissa of th e point is always greater than 5.(g ) T he ordinate of the point is always greater than the abscissa of the

point.

9. In each of the following problem s three of the vertices of a parallelogram are given. Find the fourth vertex so that one pa ir of sides is parallel to cither the *-axis or the^-axis. T here are two possibilities in each problem , Find the coordinates of both points.(a) (0, 0), (2, 3), (8, 3) (b ) (0, 0), (0, - 1 0 ) , (6, 2) ,(c) (4, 5), (2, 3), (8, 3) (d ) (2, 10), (2, 1), (7, 3)(=) ( “ 5, 4), ( - 7 , - 2 ) , (6, 4) (f) (0.. 0), (a, 0), (c, d)

10. (a ) The length of a side of a square is b. T h e coordinates of one of thevertices is (0 , 0), while the vertex opposite this is in the first quadrant. Find the coordinates of the three vertices not given.

382 COORDINATE GEOMETRY.- A N INTRODUCTION(b ) I f the vertex opposite the origin is in the second quadran t, find the

coordinates of the three vertices.

(c) I f the vertex opposite th e origin is in the fourth quadran t, find the coordinates of the three vertices.

11. Two of the vertices of an equilateral triangle are (0, 0) and (10, 0). I f the third vertex is in the first quadran t, And its coordinates.

12. T h e vertices of th e base of an isosceles triangle are (0, 0) and (2a, 0).T hrough th e means of an equation give a description for the abscissa of the th ird vertex.

13. T h e lower base of an isosceles trapezoid is 12 units, while th e upper base is 10 units in length. T h e distance between the bases is 4 units. If the lower base lies on the x-axis with its m idpoint at the origin, w hat a re the coordinates of the four vertices?

14. (a ) A set of points lie on a line th a t is parallel to the x-axis. W hich of thecoordinates is the sam e for all points?

(b ) A set of points lie on a line that is perpendicular to the vertical axis. W hich of th e coordinates is the same for all points?

I Distance Between Two Points and Dividing a Line Segment into A n y Given Ratio

I t would be ra th e r confusing if in a problem involving m a n y points these points w ere nam ed (a, b), (c, d), (« ,/) , and the like. L abeling th e points in this way, w e would soon become bewildered as to ■whether the coordinate m represented the abscissa of a point or the ordinate. I n addition, there is no way of knowing whether this is the abscissa or ordin a te of the twelfth point in question, the fifteenth point in question, o r p e rh a p s even the first. T o overcome this difficulty, points a re very often la b e le d w ith th e symbols x and y only; for the abscissa, of course, we always u s e th e letter while the letter for the ordinate is y. In addition, to disting u ish one point from another subscripts are used. Thus, the first point in our discussion will be written..as (x,, yi), the second as (* , yi), and the like. I t t h e n becomes imm ediately evident th a t the symbol xu appearing in any p ro b le m will represent th e abscissa of th e fifteenth point.

W hich of th e coordinates of two points will be the same if these two p o in ts appear on a line th a t is parallel to the^-axis? If the first of these points i s w ritten as (xi, yi), how will you express the second of the points? If, in the

Ir ‘ < * , - y , ) ( x j , / , )

X

Figure 12-11.

DISTANCE BETWEEN TWO POINTS 383

same way, two points are on a line th a t is parallel to th e x-axis, the ordinates of these points will be the same. T his is so since the ordinate of a point is the distance from th a t point to the x-axis. But the two points are equidistant from the x-axis; hence, their ordinates m ust be equal. Therefore, should we label the first (xi, yi), the ordinate of the second will also have to be yi. The abscissa of the second point, however, can not be xt, for this would signify th a t the two points were the same distance from the 7-axis, which they are not. In view of this we label the abscissa of the second point Xi.

(2 7 )

Q (2,1)

( v y 2) ( x , / ^

1 1Figure 12-12. Figure 12-13. Figure 12-14. Figure 12-15.

In Figure 12-12 the points P and Q are on a line parallel to the ^-axis; hence, their abscissas are the same, T o find the distance from Q to P, we m ust first realize th a t the o rdinate 1 represents th e distance RQ, while the ordinate 7 is th e distance RP. Thus, by subtracting 1 from 7 we Obtain the distance QP. I t appears then th a t

If two points are on a line parallel to the ^-axis, the distance between them can be found by subtracting the ordinate of the first po in t from that of the second point.

Let us suppose, however, th a t we had selected P as our first point and Q as the second, then it would appear th a t th e distance from P to Q would be

1 - 7 o r - 6

Thus, we seem to be faced with an inconsistency. W hereas in one case the distance between the points Was 6, in th e o ther i t was —6 ! T o overcome this difficulty, it has been agreed that,

P ostula te 34 : T h e distance betw een two points on a line parallel to the ji-axis is th e absolute value of th e difference of their ordinates; th a t is,

b i . - y i l -Thus, i t w ould make no difference w hether P o r Q was selected as th e

first point, for|7 — 1| = |6| = 0 and |1 — 7| = | —6|. = 6

In the event you m ay have forgotten, the absolute value of a num ber, ' briefly, is th a t num ber devoid of its sign. As in the Illustration above, both

the absolute value of 6 (written as |6 |) and th e absolute value of —6 is 6 ,

384 COORDINATE GEOMETRY: A N INTRODUCTION

Is Postulate 34 true, though, for the two points in Figure 12-13 where P is above the x-axis, while Q is below? In this case the distance jHP is 7, the distance QR is 1, and, hence, the distance QP is 8. Now it appears that one ordinate is being added to the other rather than being su b trac ted ! This is not sc, however, for the ordinate of the first point is - 1, and should we express the difl'erence between the ordinate of the second point and that of the first point, it would be

| 7 - ( - l ) |

From our knowledge of algebra, we realize that we must change the sign of the subtrahend and hence this difference becomes

\1 + 11, or 8

T hus, Postulate 34 applies whether P and Q are on the same side of the x-axis or on opposite sides.

In Figure 12-14 RQ = y u RP = yh and, hence,

QP = \p - y>\

In Figure 12-15 QP will still be |y, - ji,|, for as in Figure 12-13 y , is a negative value. Therefore, when subtracting y\ from y^ its sign will be changed to a positive one and then it will be added to y t . This is exactly w hat occured in Figure 12-13. H ad both P and Q been placed below the x-axis, we would find th a t Postulate 34 still held.

T h e discussion concerning Postulate 34 led to the conclusion that this principle will hold equally well no m atter where the points P and Q hap pened to fall. In general, the selection of the quadrants in which to: place the points in any discussion does not in any way affect the proof of any theorem in analytic geometry. In view of this, when proving any theorem, we shall always place the points in the first quadrant. By so doing, we reduce the possibility of m aking errors in the signs of the numbers.

If two points lie on a line that is parallel to the x-axis, their ordinates will be the same. T hrough an analysis very m uch the same as th a t ju st completed, we realize, (see Figure 12-16) that RQ is the abscissa of point Q-, th a t is, 3. R P is the abscissa of point P (10). Hence, the distance QP is

,10 — 3, or 7. In Figure 12-17 this distance is xj — xi. How would you find the distance between tw o jic in ts on a line th a t is parallel to the x-axis?

(3,2) (10,2 ) ----->P

fig u re 12-16.

P o st u l a t e 35: The distance between two points on a Jine parallel to the


x-axis is the absolute value of the difference of their abscissas; that is,

|X2 — Xl|.

Postulates 34 and 35 enable us to find the distance between two points if the points are on a line that is parallel to either axis. But how is the distance between two points found if the points are on a line that is not parallel to either axis? This, we shall investigate now.

TH EO R EM 71: T h e d istance be tw een two po in ts Q a n d P w ith coordinates (Xi, t/i) a n d (* 2, j/2) is g iv en by th e fo rm ula

d = V (x2 - xd1 + (y2 - j/0! f {*2^2 )

(y rt>R(*2,y,)

G iv e n P o in ts Q and P w ith coordinates(*i, and (x8,f t )__________

Concl.: d = V (x 2 — xi)2 + (y% — y iY

F ig u r e 12-18.

A n a ly s is : By draw ing lines through P and Q th a t are parallel to th e y andx axes respectively, it can easily be shown that Z R is a right angle; the proof of this is left for you to do. T he abscissa of point R will have to be the same as that of point P since both points are equidistant from the -axis. Similarly, the ordinate of point R is the same as that of point Q, as both points are equidistant from the x-axis. Hence, the coordinates of R are (X2,>i). Since it is possible to find RP and QR by Postulates 34 and 35, we can determine the measure of QP by the theorem of Pythagoras.


1. Let PR be the line through P that is 1 . Parallel postulate

parallel to the >-axis.► .

2. Let QR be the line through Q that is 2 . Same as 1parallel to the x-axis.

3. Coordinates of P are (x2, yi) . 3. Given, 4. Coordinates of Q are (xi, yi) 4. Given5. Coordinates of R are (x2, yi) 5. See analysis.6. Z R is a right angle. 6. Proof left for you to do.

1II£ 7. Postulate 348. QR = |x8 - xi| . 8. Postulate 359. (QR)1 -= (QR)* + (RP)* ■ 9. Theorem of Pythagoras

10. d} = |x2 — xil! + \yi — y t f or 10. Substitution postulate

d - V ( x 2 - xi)2 + (>2 - n )2 .

Notice that we did not include the negative sign of the radical when wc took the square root of both sides of the equation. T he distance between two points will always be considered to be a positive num ber. Similarly, the

386 COORDINATE GEOMETRY; AN INTRODUCTION

absolute bars were dropped from \xz — *i|2 and |y% — ji, |2 for the square of any num ber is always positive.

Illustration 1;Find ihe distance between the points (2, 3) and ( — 4, 11).

M e t h o d : By the distance form ula

d = V (2 + 4)! + (3 ^ T T j~8 d = V '(6)2 + ( — 8)* d = V 3 6 + 64 </ = VlOO d = 1 0

T o save time, we im m ediately changed-the sign of xh or —4, and avoided writing the distance in the cumbersome form of

d = V [ 2~ - ( - 4 ) ] » + (3 - l l )1

Illustration 2:Show that the points (2, 3), (17 ,10), and (8 , —5) are the vertices of

an isosceles triangle.

M e t h o d : By the distance formula

dl = (17 - 2)! + (10 - 3) 2 dt = (17 - 8)2 4- (10 + 5) 2 = (15)5 + (7) 2 = (7)2 + (15)’= 274 = 274

S in ce th e m easures of tw o of th e sides of th e trian g le a re equal, th en by th e rev erse o f th e definition o f a n isosceles triang le w e ca n conclude th a t th e t r i an g le is isosceles.

O u r next objective will be to determ ine the coordinates of the point th a t divides a line segment with known endpoints into segments whose m easures have any desired ratio.

THEOREM 72; T he coordinates of the point that w ill d iv ide the Jin: segment whose endpoints are (xj, yi), (*!, Vs) into the ratio o f m '.n are g iven by the formulas

m xt + n x-i m + n

a — b _ my, + n Vi m + n

y n /r ( o . b y x

m /r> /

S (*2,b)

T (x j-X i)(*1 'Y\ )

G iven: Points P and Q with coordinates (*a, y>) and f o y i )

0 3 = 2R P n

Concl.: a =

Figure 12-19.

m x i + nxi m + n

m yt + n yt m + n


A n a l y s is : As in the previous proof, lines are draw n through P parallel to the y-axis and through R and Q parallel to the x-axis. T he abscissas of S and T a re the sam e as that o f /3. T he ordinate of S is the same as the ordinate of R, while the ordinate of T is the same as th a t of Q. T he theorem concerning a line being parallel to one side of a triangle and hence dividing the other two sides into two equal ratios will enable us to set up an equation through which we can find the coordinate b.

P R O O F STATEMENTS REASONS |

i 2 * = ™ R P n

1. Given

4r->2. R S II Q T 2. See analysis.

3. If a line is parallel to one side of aR P SP triangle, then the ratios of the meas

ures of corresponding segments of theother two sides will be equal.

. T S m ~SP ~ "n


5. T S '= b 5. Postulate 34

SP = yt - *>t

6.m 6. Sam e as 4

yt — b nT he product of the means of a p ro7. .'. nb — nyi = myt — mb 7.

mb + nb = myi + nyi portion is equal to the product of the

b(m + n) - myi + nyi extremes, and the laws of algebra.

my2 + ny im + n

In the same way, had a line been draw n through R parallel to the y-axis, it.could have been shown that

mxj -j- nxi m + n

Although these formulas m ay appear quite foreboding, they are really very simple to apply. This will be shown in the following illustration.

Illustration:D eterm ine the coordinates of the poin t th a t will divide the line segment

joining the points ( —3, 7) and (4 ,1 ) into the ratio of 2 :3 . T he point ( —3, 7) will be one of the endpoints of the sm aller segment.

M s t h c d : T he very first thing co do is to draw a rough sketch of the figure. In addition to w riting down the coordinates of the two points, write in the

f T h e absolu te bars w ere not used for bo th of these num bers are positive.

388 COORDINATE GEOMETRY A N INTRODUCTION


num bers 2 and 3 on those segments that will bear the ratio of 2 : 3. Then, applying the formula for finding the abscissa, w e obtain

= 2(4) + 3 (—3) = 8 - 9 = ^ 1a ' '2 + 3 5 5

N otice how in Figure 12-20 the arrows point to the num bers th a t are usedin each product. In Figure 12-21 we see th a t the value of the ordinate, b, is found in the very sam e way.

2(1) + 3(7) 2 + 216 ~ 2 + 3 “ 5 ~ 4*

H ence, the coordinates of the poin t are (-%!> 4| )

Probably the most im portant application of Theorem 72 is to enable us to determ ine the coordinates o f the m idpoint of a line segment, when the coordinates of the endpoints of the line segment are known.

THEOREM 73: T he coordinates o f the m idpoin t o f a lin e segmentwhose endpoints are f a , y>) and (Xs, jfa) are g iven b y the formulas

0 = & d KL±JL«2 - 2

A n a l y s i s : Since the point (a , b) is the m idpoint of the line segment, the values of m and a in the formulas given in T heorem 72 are identically the

same. Hence, the ratio of — can be written as —> or 7 ' W hen we replace m n m l r

a n d n by 1 and 1 in the formulas of Theorem 72, they become

Ixt + l xi *2 + Xi‘ ~ 1 + 1 2

h = + fyi _ y> + yi1 + 1 2

Illustration:

I f ^ ( — 4, 3), B( — l , 1), and C( — 9, —3) are the vertices of a triangle, determ ine the length of the median from A to BC.


F ig u r e 12-22.

M e t h o d : A s always, a rough sketch is made o f the figure; notice that there is no need to draw the coordinate axes. Finding the coordinates of M can be done m entally by adding —7 to —9, then dividing this sum by 2. Similarly, adding 1 to —3 gives us a sum of —2; dividing this by 2, we obtain the —1 that is shown as the ordinate of the m idpoint. T he length of the median, AM , is simply the distance from A to M . This is found by using the distance formula.

d = V ( — 4 + 8)2 + ( 3 + 1)! = V (4)! + (4)2

' d = V l 6 + 16 = V 32 d = 4 ^ 2

Before giving you an opportunity to apply the formulas just developed, we would like to illustrate the power of our new tool. Although we have but scratched the surface of coordinate geometry, it is possible for us to prove a t this early stage that the diagonals of a parallelogram bisect each other.

M e t h o d :

Since we are free to place the parallelogram in any position we desire, it is best to so place it that one vertex will be at the origin, while one side will lie along the x-axis. Vertex B is labeled as (xi,‘0). T urn ing next to point D, since it has no relation to point A, we express its coordinates as (xj,^2). W ith point C, however, we m ust be more careful. Since its distance from the x-axis is the same as that of point D, 'its ordinate m ust also be yi. Being

a parallelogram, the length of U S must be the same as that of A 3 . But A B - xy. In order th a t DC be equal to xi also, the abscissa of C will have to

be xi + xi. By applying Postulate 35, page 385, the length of DC can be verified as being xl.

390 COORDINATE GEOMETRy; A N INTRODUCTIONBy the m idpoint formulas

coordinates of the m idpoint of TTB are

coordinates of the m idpoint of AC are

Hence, D B and A V biseci each other.

Your first reaction to this “ proof” is very likely to be a shrug of your shoulders followed by a long period of bewildered silence. L et’s backstep to see ju st w hat has been accomplished. W e have shown that the m idpoint of TTB is exactly the same point as the m idpoint of AC. Hence, this poin t is comm on to the two line segments. This, in tu rn , implies th a t these segments ' m ust intersect a.t each of their m idpoints. Therefore, they bisect each other!

EXERCISES

X. Determ ine the distance between each of the following pairs of points. Leave answers in radical form unless you have a square roo t table available.

(a ) (6, 8), (6, 12) (b ) ( - S , - 2), ( - 1, - 2)(c) ( - 2 , - 1 5 ) , (4, - 7 ) (d ) ( - 1 4 , 2), ( - 2 , - 3 )(e ) (0 ,7 ), ( - 6 , 4 ) (f) ( - 5a, a), (3a, a)

2. (a ) Find the distance from the point ( — 12, 5) to the origin.(b ) Find the lengths of the sides of a triangle whose vertices are

(2 ,4 ), (6 ,1 ) , ( U ,U ) .

(c) Find the lengths of the sides of a triangle whose vertices are ( - 3 , 5), (2, 0), ( - 1 , - 2 ) .

3. Show that the points ( —5, 2), (7, 4), and (2, —3) are the vertices of an isosceles triangie.

4. Show th a t the poirits ( —3, 7), (3, —1), and (4, 6) are the vertices of a right triangle. (See problem 8, page 370.)

5. Show that the diagonals o f a rectangle whose vertices are (0, 0), (0, 6),(8, 6), and (8, 0) are congruent.

6. (a ) F ind the fourth vertex of the rectangle if three of its vertices are(0, 0), (a, 0) , and (0, b).

(b ) Prove th a t the diagonals of this rectangle are congruent.

7. D eterm ine the coordinates o f the m idpoint of each of the following line segments.

(a ) (0, 0), (8, 6 ) (b ) (5, 4), (11,14).<c) ( -2 ,7 ) , (6,3) (d) ( - 9 , -4 ) , ( - ! , 6)


(e) (5, 8), (14,17) • (f) ( - 3 , 9 ) , ( - 2 , - 1 2 )(g) ( - 7 , - 1 0 ) , ( - 7 , - 1 ) <h) ( - a , - 6), (0, 0)0 ) (a, 6), ( 2 a , -2 6 )

8 . T h e vertices of a triangle are ( - 2 , 6), (0, 10), and (8, - 3 ) . Find the m idpoint of each of the sides.

9. T he vertices of a triangle are 4(6 , 4), B (2 ,1), and C(8,1 ).(a) Find the lengths of the sides of the triangle.

(b ) F ind the coordinates of the m idpoint of BC.

(c) Find the length of the m edian from A to TiC,

10. T he line segment joining the points (6 ,1 4 ) and (2, —2) is divided into four congruent parts. F ind the coordinates of the points of division,

11. T h e vertices of the parallelogram ABCD are j4(—1 ,4 ), B (2 ,8 ), C(7, 3), and D{4, —1). Find the point a t which the diagonals bisect each other.

12. O ne endpoint of a line segment is (7, —2), while the coordinates of the m idpoint are (1, —5). Find the coordinates of the o ther endpoint.

13. T h e line segment with endpoints *4(3, —2) and B {— 5, 6) is extendedits own length through point B to point C. W hat are the coordinatesof point C?

14. A(3, 5) is the midpoint of the diagonals of a parallelogram . If B ( — 5, 2)and C(7, —4) are two of the vertices of this parallelogram, find thecoordinates of the other two vertices.

15. Find the coordinates of the point that will divide each of the following line segments into the ratio indicated. T he ratio will be of the m easure of the left segment to the measure of the rigjit segment.(a ) (0 ,0 ), (5 ,0 ); 2 :3 (b ) (6,1 0 ;, (12, - 2 ) ; 1:2(c) ( - 2 , 5), (6, - 4 ) ; 3:1 (d ) ( - 1 0 , - 8), (2, - 3 ) ; 5 :2 .(e ) ( - 6 , 0 ) , ( - 5 ,1 0 ) ; 3:5 (f) ( - 5 , 4), (3, 4 ); 2:1

16. T h e vertices of a triangle are ^ (5 , 8), 5 (1 , —2), and C(7, 6).

(a) F ind the coordinates of the m idpoints of A B and AC.

(b ) Find the length of the line segment joining the m idpoints of A b and AC.

(c) F ind the length of BC.(d ) W hat conclusion can you draw in terms of your answers to b and c?

17. Using A(2xi, 2yi), B(D, 0), and C(2*i, 0) as the vertices of a triangle, prove th a t the measure of the line segm ent joining the m idpoints o f

th e sides AB and i4C'is equal to one-half the measure of side £C.

392 COORDINATE GEOMETRY: AN INTRODUCTION18. T he vertices of an isosceles trapezoid are /i(0, 0), B ( \2 , 0), C(10, 6),

and 1){2, 6).(a) Show that the measure of the iinc segment joining the midpoints

of the nonparaiic! sides is equal ;o one-half the sum of the measures of the bases. (This line segment is called the median of a trapezoid.)

(b ) Show th a t the diagonals of this trapezoid are congruent.

(c) Show that the line segment joining A to the m idpoint of BC is

congrucnt to the line segment joining B to the m idpoint of AD.(d ) Show th a t the line segments joining the midpoints of the sides i n .

order form a parallelogram. (H int: Use Theorem 40.)

19. A (0, 0), 5 (6 , 4), and C(8, 2) are the vertices of a triangle,(a ) W hat are the coordinates of the point th a t divides the m edian from

A to BC into the ratio of 2:1? (The ratio is taken in the direction from A to 5 £ \)

(b ) Determ ine the coordinates of the point th a t divides the m edian

from B to ~AC into the ratio of 2:1.(c) D eterm ine the coordinates of the point th a t divides the median

from C to I B into the ratio of 2 :1.(d ) In view of the answers that you have found for a, b, and c, what

conclusion seems to be true concerning the m edians of a triangle?

20.* (a) Following the methfldoisedjnJProblem 19, prove that the medians of a triangle are(\concurrent i^/hat is, that they meet a t a point. (H in t: Use (xt, yt), (xi,yih and (x3, y 3) as the vertices of the triangle.)

(b ) After exam ining th e coordinates of the po in t of concurrency th a t you have found in answer to (a), form ulate a statem ent expressing how to determ ine the coordinates of the poin t of concurrency of the medians of any triangle if the coordinates of the vertices are known.

21. (a) j4 (—10, —3) and B (5, 7) are two points. Find the coordinates ofa th ird point, P, on the segment AB such th a t A P :P B - 2 :3 .

(b ) Find the coordinates of a third point, Q, on the line AB bu t not on the segment A B such that AQ'.QB = 2 :3 . T h e point P found in(a) is said to divide the line segment internally into the ratio of 2 : 3, while the point Q found-now divides the line segment externally in to the ratio of 2:3 .

22. T h e vertices of an isosceles triangle are given by the points 4(0, 0),■ B (4 i, 0), C(2a, 26).

(a ) W hich of the three points A, B, and C are the endpoints of the base?:(b ) Prove that the medians to the legs of this isosceles triangle are

congruent.

PARALLELISM A N D PERPENDICULARITY 393

(c) Prove that the line segments From the m idpoin t o f the base to the midpoints of the legs are congruent.

(d ) Can you give any reason why the coordinates of the vertices in this problem were chosen as they were?

23. Two of the vertices of an equilateral triangle are (0, 0) and (2a, 0). W hat will the coordinates of the third vertex have to be if it falls in

the first quadrant?

| Parallelism and PerpendicularityTwo of the im portant topics discussed in synthetic geometry

concerned themselves w ith parallelism and perpendicularity of two lines. We would like to take another look at these topics, bu t from an analytic standpoint. This can be done, though, only after an understanding of the term slope has been established.

W e have often heard people speak of a hill as having a “ steep” slope or a “ gradual” slope; yet precisely w hat did they mean? Are the slopes of the hills in Figures 12-24 and 12-25 “steeper” than th a t o f F igure 12-26?

Figure 12-24. F igure 12-25. F igure 12-26.

Is it possible to distinguish between the slopes of the hills in Figures 12-24- and 12-25; that is, in Figure 12-24 we seem to be going uphill, while in Figure 12-25 our direction is downhill? To answer these questions, the m athem atician has defined the slope of a hill in terms of a ra tio :

slope of a hill “riserun

'L e t usl examine our th ree hills to see exactly w hat this m eans.• T h e ^ “ rise” of ajh ill is the change in our vertical position as we move-from the bottom to' the top of the hill or from the top to the bottom. T he “ru n ” is the..di£tance we moved horizontally in traveling from the bottoin to the top--

^ t ) f the hill or from the top to the bottom. By measuring the “ rise’ C ntf'"run” in Figure 12-27 we discover that the “ rise,” or vertical change, is only twe

runFigure 12-27. F ig u r e 12-28. Figure 12-29.

n$e


units, while the “ ru n ,” or horizontal change, is four units. Hence, we say th a t the slope of the hill in Figure 12-27 is

slopei =

In the same way, by m easuring ihc "rise” and " ru n ” ir. Figure 12-29 we note th a t they are 2 units and 6 units respectively, and, hence, the slope of ihat hill is

, rise 2 1slopes = ---- = 7 = rru n 6 3

In term s of these numbers, we can conclude th a t the slope of the first hill is g reater than th a t of the third, for the fraction % is greater than J.

T he hill in Figure 12-28 presents a bit of a problem, for here the “ rise” is actually negative. In going from the top of the hill to the bottom, the vertical position of the person has decreased two units, while his horizontal position has increased four units. Thus, the slope becomes

slope2 =-2 _ -1

4 2

H ence, it appears th a t the “ rise” m ay be either positive or negative depending on w hether the vertical change has been positive or negative. This, in tu rn , will determ ine the sign of th e slope.

Now, how does all this tie in with our coordinate system? R ather than

considering hills and their slopes, we now investigate line segments and their slopes. Hence, we must ask ourselves two questions:

(1) How great is the “ rise” between P\ and P2; that is, w hat is the extent of the vertical change?(2) How great is the “ run” between P i and P2; that is, what is the extent of the horizontal change?

T o answer both of these questions it is necessary to find the coordinates of the point where the line through P2 parallel to the j>-axis intersects the line th rough Pi th a t is parallel to the x-axis.

PARALLELISM AND PERPENDICULARITY 395

Exam ining this diagram , we realize th a t the coordinates of Q will have to be (xt,yi). From this it follows th a t the “ ru n ” m ust be (x2 — *(}, whiie the “ rise” is (y2 — y i). This brings us to the point where we can safely form ulate the definition of the slope of a line segment.

D e f in it io n 70: T he slope of a line segment whose endpoints are (xi,y{) an d (x2l yi) is defined by the form ula .

m ~* “ *»

Notice th a t this definition says no m ore th an d6es our parlier description in terms of the “ rise” and “ ru n ” of a hill. T h e l.etter “ m” Was used to represent the w ord slope; this is comm on practice. Also, in raorp advanced courses in m athem atics'you will find the symbol Ay (rjead delt^> ) replacing >2 — yi. Similarly, Ax is/used to replace x2 — xt . T h m r^ ll- 'd iree o f the expressions below represent the slope of a line segirient.

m = X U Z J l = ^Xi — Xi Ax

illu stra tion :If A (3, 5), B ( —2 , 7), and C(8, —3) are the vertices of a triangle, then

find the slope of the m edian from A to BC,

M e t h o d : By applying the m idpoint formula, we find th a t the coordinates

of the m idpoint of BC arexm = ( - 2 + 8 )/2 = 3

= (7 - 3 )/2 = 2

Therefore the slope of the m edian from A to BC is

« = (r» - y i)/{xs - x,)= (3 - 5 )/(2 - = 2

Now we are prepared to prove several theorems on parallelism and

perpendicularity.

THEOREM 74: The slope of the line segm ent jo in in g any two points on a lin e is equal to the slope o f the lin e segm ent jo in in g any other two points on the line.


V Given: PR || AC ||

Concl.:

•>c-axis

QR .11 BC || ji-axis

BC _ QR AC P R

F i g u r e 1 2 - 3 2 .

BCA n a l y s is : T he slope of the line segment A B is given b y the ratio while

0 Rth a t of PQ is th e ratio t - r 1 By proving the triangles similar we can show

A

the slopes to be equal.

PR O O F

T he proof is left for you to do.

Theorem 74 enables us to prove th a t if we start w ith a line, the slope of th e line segment joining any two points will be equal to the slope of the line segment joining any other two points. Equally as im portant to us is a question closely related to this. I t is, “ U nder w hat conditions will points be collinear?” T his we plan to answer by the next theorem.

THEOREM 75: Three points are collinear i f the slope of the line segm ent jo in ing two of the points is equal to the slope o f the lin e segm ent jo in ing either o f these points to the th ird point.

G iven: CD II B E fj

BD II A E || *-axis

B E A E '

CDBD

Concl,: A, B- and C are collinear.

Figure 12-33,

A nalysis: T h e proof will be o f the indirect variety, By assuming tha t A , B , and C are not collinear, w e w ill be led to the inconsistency th a t Z F B D S Z C B D and ZF B D gg ZCBD.

PARALLELISM AND' PERPENDICULARITY 397

PRO O F


A, B, and C are collinear.or

A, B, and C are not collinear.

Let us accept the possibility th a t A, B, and C are not collinear. Extend A B

to some point F; in doing so A B will not pass m rough C, since we have

accepted the possibility th a t A, B, and C are not collinear. Since BD is

parallel to AE, Z F B D = Z B A E . However, A B D C ~ A A E B by the S.A.S. theorem on sim ilarity; hence ZC B D = ZB A E . Therefore, it follows th a t Z F B D = ZC B D . However, from the postulate th a t the whole is greater than any of its parts, m Z F B D > m ZC BD . Thus, accepting the possibility that A, B, and C were not collinear has led to the logical inconsistency of the tru th of both Z F B D S ZC B D and Z F B D 3£ ZC BD . By the law of contradiction both cannot be true a t the same time. Since Z F B D 9= Z C B D m ust be true as the result of a postulate, ZF B D ~ ZC B D m ust be false, and, therefore, the statem ent “A, B, and C are not collinear” is also false. Hence, the statem ent "A , B, and C art collinear” is true, for it is the only rem aining possibility.

Illustration:Are the points A{2, 1), B(5, 0), and C(26, —7) collinear?

M e t h o d : By applying Theorem 75 to show that the slope of A B is equal to the slope of BC, we will be able to conclude that A, B, and C are collinear.

r X 5 0 - 1 - 1slope of A B = mi = g _ ~2 =

- 7 - 0 - 7 - 1slope of m = i r = T = -2i - = ~

.\ mi = m2; hence, A, B, and C are collinear.

D e f in it io n 71: T h e slope of a line is the slope of the line segment joining any two points on th a t line.

In view of Theorem 74 do you see why it is possible to say th a t the slope of a line is the slope of the line segment joining any two points ra ther th an two specific points on the line?

THEOREM 76: If two lines are parallel, their slopes are equal.


A n a l y s i s : As in the proof of T heorem 74, th e slopes can be shown to b e equal by proving A A B C ~ A D E F .

PROOF (The reasons will be left for you to supply.)________

1. D E II A B II x-axis2. Z 3 S Z l and ZA S Z23. h II h4. Z \ S Z l5. Z 3 S Z 4 (a)6. Z $ S Z l

1. F E II CB8. Z 6 S Z l9. Z 5 S Z 6 (a)

10. .-. A A B C ~ A D E F n C B _ F E

' A B D E

T H E O R E M 77: If two lines h av e equal slopes, th e lines a re p a ra lle l.

G iven: D E II A B || *-axis

F E II CB II y-axis

CB _ F E A B D E

Concl.: /j || /j

A n a l y s i s : will be shown parallel to lt by proving ZA ~ Z5 . T o do this,th e two angles will be shown congruent to A 6 and 7 respectively. Hence., the proof revolves around the need to show th a t Z 6 and Z l a re congruent This can be done by proving A A BC ~ A DEF.



1. FEW CB2. Z l Z2>

3. D E II A B II x-axis4. Z A B C = Z l and Z D E F — Z l5. Z A B C SZ Z D E F {a)

CB FEA B D E

7. A A B C ~ A D E F8. Z 6 - J. Z l9. ZA S Z6, Z 5 S Z l

10. .-. Z 4 S Z 5U . 4

Illustration:A (—2, —5), 5(4 , —7), C(10, 9), and D ( — 8 , 13) are the vertices of a

quadrilateral. Prove that the line join ing th e m idpoints of AB and 5 2 is

parallel to the line joining the m idpoints of Cl) and B A .

M e t h o d : A sketch is m ade of the figure, and the coordinates of the m idpoints are w ritten on the diagram at sight,

slope of SR = mi = = -

i ■ t 1 + 6 7slope of PQ = m5 = = g

/. mi = mi*, hence, by Theorem 77 the lines a re parallel.

I t should be pointed out th a t it does no t m atter which of the points is considered the first point-and which is the second. In either event the slope of the line segment will be the same. Thus, in the illustration above we. found

the slope of SR by considering S as the first point and R as the second. H ad • this been reversed, we would have obtained,

400 COORDINATE GEOMETRY, A N INTRODUCTION

Thus, the slope of SR is the same as th a t of RS.W hen the slope of one line is J, while another is -J, the slope of the

second is said to be the reciprocal of the slope of the first, for the fraction -J is the reciprocal of -f. Similarly, if the slope of one line is § and the other — | , then the second slope is the negative reciprocal of the slope of the first. T hus, to find the negative reciprocal of a fraction, it is simply a m atter of inverting the fraction and changing its sign. This information is needed for the next theorem. W hat is the negative reciprocal of f? O f O f — J?

THEOREM 78: If two lines are perpendicular, the slope of one is the negative reciprocal of the slope of the other.

Given: F E II B C \\ y-axis

d e i i a b i

x-axis h J- li

njj

A nalysis and P r o o f : By show ing th a t th e co rre sp o n d in g angles a re co n g ru e n t as m arked in th e d iag ram , i t can be p roved A FED ~ A ABC, an d , hence , i t w ould follow th a t

F E _ _ A B D E CB (1)

FEFortunately, 7 -= is the slope of k, and, therefore, is mi.

JJJh

FEThus,

D E= m 1 (2)

T h e slope of W, however, involves m ore difficulty, for this slope is negative since the “ rise” is a negative quantity. By placing a negative sign in front

CBof m2, it will become positive and, hence, have the same sign as -j-g

. CB " A B

-mi

A B 101 CB m3 (3)


Hence, by substituting both (2) and (3) in (1), we obtain1

888 “ Imi

THEOREM 7 9 ;C:in h e slope o f orieHne is the negative reciprocal o f the slope of a second lin e, the lin es are perpendicular.

G iven: F E II BC II y -axis

<-> <-> D E II A B I x-axis

mi * m2

Concl.: k -L k

A n a l y s is : In a m anner sim ilar to that used in the proof of T heorem 77 we can show A FED ~ A ABC. T h en to prove l\ J . 4 , Z l will be shown to be congruent to Z E , which is a right angle. W hy is Z E a right angle? W ith Z \ a right angle, the lines will be perpendicular.

PRO OF(The reasonTwill be left for you to suppiy.)

Since the triangles are similar, Z F _ Z A

But AB II x-axis, Hence,

But D E I! x-axis, Hence,But,

Hence,

Z2 £* ZA Z F == Z2 {a)

Z D S Z l (a) Z l S i Z E Z E is a right angle. Z l is a right angle.

l\ -L k

Illustration:Show that the diagonals of the rhom bus whose vertices are .4(0, 0),

5(10, 0), C(16, 8), and D (6, 3) are perpendicular to each other.


> i—►M e t h o d : U sing Figure 12-39, we find the slopes of AC and D B to see if one is the negative reciprocal of the other.

i f T r 8 - 0 1slope of AC - mi = J g T 'o = 2

, . £ + 0 - 8 - 8 - 2slope of D B = m2 = = — = —

mi — ------; hence, by Theorem 79, AC X D B .TKj

There are several ra ther im portant omissions th a t we have m ade when discussing the slope of a line in the coordinate plane. T hese concern the slopes of lines th a t are parallel to either the x- o r j>-axis.

Q U2/y,)

Figure 12-40.

In Figure 12-40 we see that the slope of the line segment jo in ing the points P and Q will be y

yi — yi 0 m = = ----------Xi — *i xs — Xi

This will be interpreted to mean that there is na “ rise” or vertical change in the line for any two points that may be selected on the line. T h a t is, no m atter how m any units the abscissa m ay increase between two points on- th e line, the ordinate will not increase a t all. In view of this, we will express a slope such as this in its simplest form of


01

for, as was said, whether x increases 1 unit or 1,000 units between two points on this line, the increase in y between these same two points will always remain 0 .

Similarly, in Figure 12-41

„ _ yg ~ _ yt ~ >>*1 - *1 0

As before, this will simply be interpreted to m ean th a t no m atter how m any units the ordinate may change between two points on this line, the change in the abscissa will always remain 0. As before, the simplest w ay of expressing a slope whose denominator is zero will be in the form of

10

We must impress upon you the fact that -J is not to be interpreted as a division problem, for division by zero is impossible. I t should be interpreted only in terms of a “ rise” and “ run ,” as was done in the preceding paragraph.

The negative reciprocal of i will be considered to be Similarly,

the negative reciprocal of f will be This in terpreta tion will simplify

a number of problems we will encounter.

EXERCISES

1. Determine the slope of the line that passes through each of the following pairs of points:(a) (0, 0), (5, 4) (b ) (0 ,0 ), ( - 3 , 6 )(c) (2, 5), (10,14) <d) ( - 3 , 4 ) , (6, - 8)(e) (0 ,4 ), (8, - 6) ( 0 (5, - 7 ) , (2, - 1 0 )(g) ( - 4 , - 3 ) , ( - 1 , - 8) (h ) (3, 7), (9, 7)

2. Find the slope of the line th a t is perpendicular to the line th a t passes through each of the following pairs of points ;(a) (2, 3), (5 ,14) (b ) ( - 4 , 7), (1, - 6) (c) ( - 8, - 5 ) , ( - 2 , - 1 0 )

3. Which of the following sets of points are collinear?(a) ( 0 , - 7 ) , ( 3 , - 1 ) , ( - 2 , - 1 1 )(b) (0 ,3 ), ( 1 5 , - 8 ) , ( - 1 2 ,1 0 )

4. W hat conclusion can be draw n concerning a line whose slope is 0?

404 COORDINATE GEOMETRY: A N INTRODUCTION5. *4(8, 6), 5 (0 , 0), and C( 12, 2) are the vertices of a triangle.

i —i(a ) W hat is th e slope of the Sine through C parallel to AB?.(b ) W hat is the dope of the aititude from B to AC?

(c) V /hat is the slope of the m edian from C to AB?

(d ) W hat is the slope of the perpendicular bisector of AB?

6. /4(1, 5), B (5, 9), and £ (1 1 ,1 ) a re the vertir.es of a triangle.(a ) Find the midpoints of AB, AC, and BC”.

(b ) Show th a t the line joining the m idpoints of A B and A C is parallel to BC.

(c) Show th a t the line joining th e m idpoints of AC and BC is parallel to AB.

7. R ( — 2, 5), £ (0, 11), and T ( —6, —7) are the vertices of a triangle. Show

th a t the line passing through the m idpoints of SR and S T is parallel to R T .

8 . Prove th a t the line passing through the midpoints of two sides o f a triangle is parallel to the third side. (H in t: Use the coordinates (2a, 2b), (2c, 2d), and (2e, 2j ) as the vertices of the triangle.)

9. Show th a t the line segments joining the points ( —4, —3), (8, 2), (11, 6),and ( — 1, 1) taken in order form a parallelogram. (H int: Use the reverse of the definition of a parallelogram ,)

10. Show that the vertices in each of the problems below are the vertices of a right triangle. Do not use the distance formula.(a ) (5, 2), (8, 9), (10, 4) (b ) ( - 3 ,1 ) , ( - 1 , 20), (5, 4)

11. A{ —12, 0), B (2, 0), C (l, 5), and D ( — 11, 5) are the vertices of an isosceles trapezoid.

(a ) Show th a t the line th a t passes through the midpoints of the nonparallel sides is parallel to the bases.

(b ) Show th a t the line th a t passes through the midpoints of the parallel sides is perpendicular to them.

12. (a ) If .<4(0, 0), £ ( 12, 0), and D {2, 5) are three vertices of an isosceles• ^ (

trapezoid in which A B is the lower base, find the fourth vertex C so th a t the vertices will read in the order /!, B, C, and D.

(b ) I f <4(0, 0), B(2a, 0), and D(2b, 2c) a re three vertices o f an isosceles . ^

trapezoid where A B is the lower base, write the coordinates of thefourth vertex C so th a t the vertices will read in the order A, B , C, and D. r"

(c) Using the vertices in b, prove th a t the m edian of a trapezoid is parallel to the bases and equal to one-half their sum . (See Problem 18, page 392.)

13. /3(0, 0), 5(8, 0), C(12, - 6 ) , and D( — 4, —4) are the vertices of a quadrilateral.(a) Show that if the midpoints of the sides are joined in order, the

quadrilateral will be a parallelogram .(b) Show that if line segments are draw n between pairs of opposite

midpoints, these line segments will bisect each other.

14. Prove that if the midpoints of the sides of any quadrilateral are joined in order, the quadrilateral formed will be a parallelogram . (H int: Use (2x i ,2yi), (2x2, 2yi), (2x3, 2y 3), (2x4, 2>4) as the vertices of the quadrilateral.)

15. /4(0, 0), B(8, 0), C(10, 4), and D (2, 4) are the vertices of a parallelogram^ If segments are draw n from B to M , the m idpoint of ZJC, and from D to N, the midpoint of AB, then B M D N will be a parallelogram .

16. ABCD is a parallelogram where M and N are the midpoints of 5 C aiid.

A B respectively. Prove that B M D N is a parallelogram . (H int: Use (0, 0), (2a, 0), and (2b, 2c) as the coordinates of A, B, and D \ then find the coordinates for C in the same m anner as you had for Probleta 12b.)

17. T he coordinates of a parallelogram are *4(3, 6), 5 (6 , 6), C(12, 15), an d Z)(9, 15). Prove that if segments are draw n from D to the trisectiori.

points of AC and from B to these sam e points, th a t the quadrilateral formed will be a parallelogram.

18. Prove that if segments are draw n from a pa ir of opposite vertices of a parallelogram to the trisection points of the diagonal jo in ing the rem aining vertices, the quadrilateral formed will be a parallelogram . (H in t: Use (0, 0), (3a, 0), and (34, 3c) for the vertices A, 5 , and D; then find the coordinates of vertex C.)

19 Test and Review1. Represent each of the following statem ents by an equation:

(a) T h e ordinate of a point is always 5.(b ) T he abscissa of a point is always —7.(c) T he ordinate of a poin t is always equal to twice the abscissa.(il) T he abscissa of a point is always less thao 10.(e) T h e sum of the square of the abscissa and the square of the ordinate

is always 25.

TEST AND-REVIEW 405

406 COORDINATE GEOMETRY,- A N INTRODUCTION

2. (a ) A set of points lie on a line perpendicular to the >-axis. W hich-ofthe coordinates is the same for all points of the set?

(b) How could you give a description for the points in this set by using symbols?

3 . D e t e r m i n e t h e d i s t a n c e b e t w e e n e a c h o f t h e f o l l o w i n g p a i r s o f p o i n t s .

L e a v e y o u r a n s w e r s i n r a d i c a l f o r m .

(a) (3 ,7 ), (9 ,2 ) (b ) ( - 5 , - 6), (8, - 3 )

4. Determ ine the coordinates of the m idpoints of the line segments whose endpoints are

(a) ( - 2 , - 9 ) , (8, - 3 ) (b ) (5a, b), (2 a ,U )

5. Find the coordinates of the point that will divide each of the following line segments into the ratio indicated. T he ratio will be the measure of the left segment to the measure of the right segment.(a ) (0, 3), (7 ,1 2 ); 2 :1 (b ) ( - 5 , 1 ) , ( - 2 , - 3 ) ; 1 :4

6. T h e three vertices of a triangle are .<4(0, 2), £ ( 4, 10), and C(12, —4).(a ) Find the length of AB.

(b ) Find the coordinates of the m idpoint of BC.

(c) Find the length of th e m edian from A to BU.

• (d ) Find the length of th e line segment-joining th e m idpoints of C/4 and US.

7. T he coordinates of two consecutive vertices of a parallelogram are (0, 2) and (10, 4). If the poin t of intersection of the diagonals is (6, 5), w hat are the other two vertices of the parallelogram?

8. Determine the slope of the line that passes through each of the following pairs of points.(a ) (0, 8), (5, 2) (b ) ( - 3 ,4 ) , ( - 2 , - 1 )

9. T he vertices of a quadrilateral taken in consecutive order are <4(0, 0), £ (1 2 ,2 ), C(10, 6), and D (4, 8).

(a ) Show th a t the line segment joining the midpoints of S A and BU is

congruent to the line segment joining the midpoints of D A and DU.(b ) Show th a t the above line segments are parallel,

10. i4(l, 4), £ ( —3, 8), and C(5, 6) are the vertices of a triangle.

(a) W hat is the slope of the line through A parallel to BC?(b ) W hat is the slops of the altitude from C to J l ?

(c) W hat is the slope of the m edian from B to AU?

(d ) Show th a t the line join ing the midpoints of A B and AU is parallel


11. (a ) How do you in terpret the statem ent th a t the slope of a line is 4?- 3

(b ) How do you in terpret the statem ent th a t the slope of a line is — ?

(c) How do you in terpret the statem ent th a t the slope of a line is

■ Try This For FunI t was noted earlier that the distance between two "objects”

was the m easure of the shortest path between them . T his p a th was investigated when the “ objects” were two points of a plane or a point and a line of a plane. But w hat if the surface is not a plane? Is there anything that we, as students of plane geometry, can do to find the distance between two objects on a surface such as this? M ore often th an not, the answer to this question is “ No.” T here are, however, certain special situations in which, it is possible for us to press the surface flat, then trea t the problem as though it existed on a plane. O ne such case is described below.

P

For reasons known only to himself, a spider built his web on the side of a cone. O ne ra ther hot day he emerged from his shelter and, in spite of his better judgem ent, decided to em bark on a walk around the cone. N ot wanting to overdo this, however, he felt th a t it would be best if he restricted his path to the shortest route that would enable him to see the back of the cone and re tu rn him to his home. T o do this, he p ictured the cone as being cut down the back, then pressed flat on the surface of a table. The path he decided to take was along the perpendicular frtV»i the nest to the cut ~PB

P

408 COORDINATE GEOMETRY: AN INTRODUCTION

and back along a second perpendicular to PB'. Remember, of course, th a t A and. A ' are the same point a t the back of the cone.

(1) Can you prove that this is the shortest route?(2) W hat pa th should the spider take if, when the cone was cut and pressed flat on the plane, th e Z.B 'PB turned out to be a straight angle?(3) Is there still a third situation that might arise when the cone is laid ou t flat in the plane?

13Coordinate Geometry The Graph

O U R F IR S T E N C O U N T E R W IT H A S E T WAS during the very first few weeks of the course. A set, as we learned, may be any collection of elements where the elements are found by some rule. T he m athem atician prefers to say th a t the elements are determined by the conditions presented in the problem. T hus, the rule conditions presented in the problem may be that the elements in the set must be the boys in the Conway family. T he set would then be w ritten as {Joe, Fred, William}. T h e elements listed in this set comply now with two requirem ents:

(1) All the names listed— that is, the elements in the set—are the boys in the Conway family.(2) All the boys in the Conway family have their names listed in this set, th a t is, are elements in the set.

Notice th a t requirem ent 2 is the converse of requirem ent 1.T h e elements in a set are frequently num bers. Specifically, if the con

ditions of the problem are that the elements be even numbers greater than 0 bu t less than 10, then the set would be

{2, 4, 6, 8}A nd again we can say that

(1) All the num bers in the set are even num bers greater than 0 and less th an 10.

409

410 COORDINATE GEOMETRY: THE GRAPH

(2) All the num bers th a t are even numbers greater than 0 and less than 10 are num bers in the set.

Frequently there will be sets of elements where all the elements, or members, cannot be listed. These are called infinite spts. Thus, had the rule in the previous prpblem been to determ ine the set whose members were even num bers greater than zero, it would be impossible to list all of them, To overcome this, we merely list a few and indicate th a t there are m ore to follow by using three dots:

{2, 4, 6, 8, . . .}

In order to make certain th a t there is no doubt as to how the< elements in the set were determined, we will express the description of the elements in the set as

{x]x is an even num ber greater than zero}T his is read as

“ T he set of all x such th a t x is an even num ber greater than zero.”

W hen expressed in this m anner, we can see th a t the symbol x is designed to represent any element in the set. I t may be the element 2, or 6, or 38, or 2,594; in fact, any even num ber greater than zero.

O ur present concern is with sets whose elements are ordered pairs of num bers. Consider the set

{(1, 5), (1, 6), (2, 5), (2, 6)}

T he conditions leading to this set of elements were that the first num ber in each pair had to be either 1 or 2, while the second num ber had to be either5 o r 6 . This same set could have been written in the form,

{(x,y)\x is either 1 or 2 a n d ^ is either 5 or 6}We would read this as

“ T he set of all ordered pairs of num bers (x, y) such th a t x is either 1 or 2 and y is either 5 or 6.”

T he im portance to us of sets whose elements a re pairs of num bers rests in the fact th a t every ordered pair of num bers represents a poin t in the

A

V

Figure 13-1.

COORDINATE GEOMETRY: THE GRAPH 411

coordinate plane and, conversely, every point in the coordinate plane represents a pair of ordered numbers. Hence, all sets consisting of ordered pairs of numbers can be represented by points in the coordinate plane. T he set of elements {(1, 5), (1, 6), (2, 5), (2, 6)} is pictured as A, B, C, and D in the plane.of Figure 13-1. T he points A, B, C, and D are called the graph of the set of elements {(1, 5), (1, 6), (2, 5), (2, 6)}.

. D e f in it io n 72: T he graph of a set consisting of ordered pairs of numbers is the set of points whose coordinates are members of the original set.

Now let us examine all of this in terms of the following se t:

{(*.>)!> “ 2x + 1}As before, this is read as

“ T he set of all ordered pairs of numbers (x, y) such th a t the second num ber in each pair is always 1 more than twice the first num ber.”

W hen you studied algebra, you learned to determ ine such pairs of numbers by substituting values for x, then finding the corresponding value for y. Thus, when x is 1, the value of y will be 3. Some of these.pairs of values were tabulated as

Now, we can express these pairs of values as the set of elements

{(0, 1), (1, 3), (2, 5), ( - 4 , - 7 ) , . . .}

where each of the elements in this set is restricted by the fact that it must satisfy the equation y — 2x + 1. T he set of pairs of values th a t satisfy an equation—that is, make it true—is called the solution set of th a t equation. The pair of values (5, 1) when used as replacements for x and y in the equation

y = 2x + 1

will make the left side 1 while the right side will be 11. Hence, (5 ,1 ) is not a mem ber of the solution set of the equation y = 2x + 1. O n the other hand, the elements listed in the set above are m embers of the solution set of this equation, for they will m ake this equation true.

T he graph of the elements thus far found in the solution set of the equation y = 2x + 1 is pictured in Figure 13-2. Although this is a set consisting of infinitely m any pairs of numbers, we have found only four of these pairs. This question then arises, “ W hat would the graph of the solution set have been had we been able to express all the pairs of values in the solution set?” Examination of the four points leads us to suspect that the graph will consist of collinear points. This is what we now intend to prove.


F ig u r e 13-2.

Actually, our problem is twofold:

(1) Will all the pairs of values in the solution set of y = 2x + 1 represent collinear points?(2) Will all the points on this line be elements in the solution set of y = 2x + 1?

W e will attack each of these questions individually. T o answer the first, we know by Theorem 75 that three points will be collinear if the slope of the line segment joining any two is equal to the slope of the line segment join ing either of these to the third. Hence, we will let A(xi,y\), B(xt, y2), and C(xz,y j) ' be any three pairs of values th a t are elements in the solution set. W e wouldnow like to show that they represent three collinear points. T o do this byapplying Theorem 75, we m ust show that

ya — >i __ >3 - y%X i — X i X j — X j

for these are the slopes of A B and SC,Since the three points are elements in the solution set of y — 2x + 1,

then(1) = 2x, + 1( 2) yi = 2xs + 1(3) yi = 2xj + 1

Therefore, slope of AB = m, = ~ (2*‘ +Xi — X i X2 — XI

— 2x2 + 1 — 2xi — 1 _ 2x; — 2xiXi — X I X2 - X I

_ 2 ( x 2 - X l ) _ 2

COORDINATE GEOMETRY: THE GRAPH 413

.. , . r ys - yi ( 2 x 3 + 1) - ( 2 x 2 + 1)Similarly, slope of BC — w? = ------- - --------------------------------’ • x j — x 2 x j — x 2

_ 2 x a + 1 — 2 x 8 — 1 _ 2xa — 2 x 2

x j — x 2 xs — Xi

= 2(x3 ~ ■*») = 2X i — x2

Thus, mi = m2, and hence, A, B, and C are collinear.T o answer question 2, we change its form to one that we can handle

more easily. T h at is, will all the points on the line th a t passes through twomembers of the solution set also be members of the solution set? f T o prove this, let A(xi,yi) and B (xi,y2) be two points in the solution set of the equation y = 2x + 1, We are now called upon to show that a point such as D (xt, yi) that is on the line determined by points A and B will satisfy the equation / = 2x + 1. This will be true if

>4 = 2 x , + 1 ( 1 )

By Theorem 74 we know th a t the slope of the line segm ent jo in ing any two points on a line is- equal to the slope of the line segment jo in ing any o ther two points. Hence,

slope of D B = slope of BA

y< - y> = y%. - >1* 4 — Xi X i — Xl

But points A and B are in the solution set; therefore,

yi = 2x i + l and yi = 2x2 + 1Hence

y* - yi _ (2x2 + 1) - (2xi + 1)x4 — x2 x2 — Xl

y< ~ yt _ 2x2 + 1 ~ 2xi — 1* 4 T~ X2 ■ X j — X i

>4 — y2 2x2 — 2xiX4 — X2 X2 — X l

X ~ _ oX4 “ X2

Thereforey* — yt = 2x4 — 2x2

>>4 — (2x2 + 1) = 2 X4 — 2x2yt — 2x2 — i = 2x4 - 2x2

y< = 2 x 4 + 1

And this is what we had planned to prove. (See (1) above.)

t Notice that we are using the words “points” and “pairs of values” interchangeably. This is often done since there is a one-to-one correspondence between them.


EXERCISES

1. Give the rule or description for finding the elements in each of the following sets:(a) { 1 ,3 ,5 , 7, 9} (b ) {S, 10, 15, 20}(c) {7, 8, 9, 10, 11} (d) {3, 6, 9, 12, . . .}(e) {(1, 1), (1, 2), (1, 3), (1, 4)} (f) {(3, 1), (3, 2), (4, 1), (4. 2)}

2. How would you read each of the following expressions?(a) {x | x is a whole number}(b ) {x I 2x + 1 = 6}(cr) {x j x > 3}(d ) {x | x < 5 and x is a natural number}(e) {* j 3x > x + 2}<0 {(*,>) I * + ^ = 5}(g ) { (* ,y ) \y = 3* - 2}

3. Find four ordered pairs of num bers in each of the following sets:(a ) {(*>jy) I * + y = 4} (b ) {(x ,y) | 2x - y = 5}(c) {(*. J1) I 2* + 3y = 6} (d ) {(*, ji) | x 1 + y* = 25}

4. Find four ordered pairs of num bers in the solution set of each of the following equations and graph these four po in ts:(a ) y = * + 3 (b ) * + 2y = 4(c) 3* + y = 7 (d ) x2 + y* = 100

5. For each of the equations below, find three points in its solution set. Using Theorem 75, determ ine whether these points are collinear.(a ) y = 3x - 2 (b ) x = 2j> + 3(c) 2x — 3y = 6 (d ) x* — y 1 = 16

■ The Straight LineT he analysis in the preceding section leads us to the fol

lowing definition:

D e f in it io n 73: T h e graph of an equation is the graph of the solution set of th a t equation.

Having accepted the property that a definition is reversible, we have a t our disposal, through Definition 73, two im portant principles:

(1) T he coordinates of every point of the graph of an equation is an element of the solution set of that equation.(2) Every element of the solution set of an equation is the coordinates of a point of the graph of the equation.

THE STRAIGHT LINE 415

One of the simplest and yet one of the most im portant graphs is th a t of the equation of the form

ax + by = c

In the previous section we proved that the graph of a particular form of this equation, y = Zx + 1, was a straight line. W hat are the values of a, b, and c in this equation? Now we plan to show that every equation having the form ax + by = c will be a straight line. T he pattern of proof will be identical with that used in showing the graph of y = 2r..+ 1 to be a straight line. Since this is so, our first objective will be to write the equation ax + by = c so that it resembles the equation^ = 2x + 1.

ax + by = iby = — ax + c

y = —7 x + 7 (Assume b ^ 0.) b o

As —7 and 7 are constants, we will simplify the appearance of this equationb b

by replacing them by m and k.y = mx + k

Again, we are confronted with proving two properties:

(1) All points in the solution set are collinear.(2) All points lying on the line joining any two points in the solution set are also in the solution set.

P R O O F -P A R T 1

Let A(xi, y 1), B(x2, yt), and C{x3, ^3) be points in the solution set of the equation y - mx + k. We will now show th a t these points are collinear.

(1) y 1 = mxi + k (2) yt = mx2 + k (3) y 3 = mx3 + k

slope of A B = = fr1*? + *) - 0"«i + *) =' Xi - Xl x2 - X: Xi — Xl

m{x2 - xi)— --------------= m

xi - Xi

slo e of BD = ^ ~ ~ (mxt ~t~ _ w*3— mx2* ! — * 2 ' X3 — Xi Xi — Xi

m(*3 ~ xi)= --------------- tn

X i~ X 2

Hence, slope of A S = slope of SCTherefore, the three points A, B, and C are collinear. (SeeTheorem 75.)


P R O O F— PART 2

Since this proof is identical to that presented on page 413 (except the symbols m and k replace the symbols 2 and 1), it will be left for you to do.

Now let us examine what would have happened had 4 been equal to 0 in the equation ax + by = c. W ere this so, then

ax + Oy — c ax = c

and thereforex = c/a

T h e equality x = cl a is simply a direction stating that the point will always be c /a units from the vertical axis.j For this to be so, the point will have to be on a line parallel to the_y-axis. Hence, even were 4 = 0, the equation ax + by = c would still represent a line.

Before leaving the proof of this theorem, it is imperative to point out th a t as a by-product of Part 1 of the proof we have shown that the slope of the line segment joining any two points in the solution set of the equation y = mx -j- k is m. By definition, though, the slope of a line is the slope of the line segment joining any two points on the line. Hence, the slope of the line y — mx + k is m. But how is m found? It is the coefficient of x when the equation is solved for y. To illustrate, to find the slope of the line

3y + 2x = 6solve the equation fory:

3y = - Z x + 6

> - . - • § * + 2

Hence, m = —2/3, and therefore theslope of the line 3y + 2x = 6 is —2/3. In sum m ary, we have proved

THEOREM 80: (1) T he graph of any equation of the form ax + by = C w ill be a line.(2) Every lin e can be represented by an equation of the form ax + by = c.

Now we come to the mechanics of finding the equation of a line. No m atte r w hat the conditions of the problem, the equation of a line will always be found by applying Theorem 74; that is, the slope of the line segment, join ing any two points of a line is equal to the slope of the line segment jo in ing any other two points of the line. This will be indicated henceforth by writing mi = mj.

t S e e p a g e 380.

THE STRAIGHT LINE 417

Illustration 1:Determ ine the equation of the line that passes through the points

A (2, 3) and B(4, - 1 ) .M e t h o d : Let C (x,y) r e p r e s e n t the coordinates of a third point on the line

jo in ing the points 4 and B. T hen, sincem i = mj

slope BC = slope AB

y + 1 = 3 + 1 x - 4 2 - 4

y 4- 1 = _4_ x - 4 - 2

L ± 1 = J Lx - 4 - 1

~ y - t = 2x - 8

7 = 2* + / or Zx + y = 7

Illustration 2:Determine the equation of the line that passes through the point

A ( — S, 1) and is perpendicular to the line 2x + 3y = 4.M e t h o d : T he slope of the line 2x + 3_y = 4 is found by solving the equation foT y and examining the coefficient of x:

iy = — 2x -(- 4 (1)

y = - ! * + I

slope of (1): —2/3slope of a line perpendicular to (1): 3 /2

Let B {x,y) be a second point on the Jine that passes through A ( — 5, 1) andis perpendicular to 2x + 3/ = 4. Then, since

mi = m2

slope of A B — slope of line perpendicular to 2x -{- iy = 4

X + 5 2

3* + 15 = 2_y — 2

3x — 2y «= - 1 7

Illustration 3:

Prove th a t the points h(1 , 5), B (—2. 6), and C(0. —3) a re no t collinear.

M e t h o d : Let D {x,y) bc any point on the line joining the points A and B.

T hen ,

418 COORDINATE GEOMETRY: THE GRAPH771] — 77)2

slope D B = slope A B

y - 6 5 - 6x -f- 2 1 4- 2

y - 6 = M x + 2 3

3y - 18 = - ,v - 2

x + Zy = 16

If point C(0, —3) is on the line passing through A and B, it will be an dem en t in the solution set of its equation. Hence,

0 + 3 ( - 3 ) J , 16.- 9 ^ 1 6

,\ C, A , and B are not collinear.

EXERCISES

1. In each of the following problems, determ ine the equation of the line th a t passes th rough the two points th a t are given.(a ) (0, 2), (3, 4) (b) (4, 6), (8, 2)(c) (1, 5), (0, 9) (d ) ( - 2 , 8), (3, 10)(e ) ( - 5 , - 2 ) , (0, - 5 ) (f) ( - 1 , 7), (3, - 3 )

2. Determ ine the equation o f the line th a t passes through each of the following points and has th e slope indicated.(a ) (2, 3); slope = 2 /5 (b ) (0, 5); slope = - 1 / 2(c) ( — 4, 6) ; slope = —3 /4 (d ) ( — 3, —4); slope = 4(e) (5, —3); slope = 0 (f) (0, 2 ); slope = — 1

3. Determ ine the equation of the line under each of the following conditions: *

(a ) Passes through the origin and the point (a, b).(b ) Intersects the x-axis a t x — 3 and the 7-axis a t y = —3.(c) Parallel to th e x-axis an d passes through (2, 8).(d ) Parallel to they-axis and passes through ( —5, —2),(e ) Passes through the origin and has a slope of 1.(f) Intersects the x-axis a t x = 4 and has a slope of — 5/ 3.

4. Find the equation of the line th a t passes through the po in t (4 ,1 ) and is parallel to the line th a t passes through the points ( —5, 7) and (2, 3).

5. F ind the equation of the line th a t passes through the point ( —2, —2) and is perpendicular to the line that passes through the points (1, 3) and ( —5,0).

INTERSECTION OF TWO SETS 419

6 . /1(0, 3), B(6 , - 1 ) , and C(4, 7) are the vertices of a triangle.(a) Find the equation of the line that passes through A and is parallel

to BC.(b ) Find the equation of the line that passes through C and is perpen-

dicular to AB.(c) Find the equation of the altitude from B to AC.

(d ) Find the equation of the median from A to BC.(e) Find the equation of the perpendicular bisector of AB.(f) Find the equation of the line that passes through the m idpoints of

A B and AC.7. Which of the following sets of points are collinear? Justify your answer,

(a) {(2, 3), (5, 8), ( - 4 , -7 )} (b ) {(2, 1), ( - 3 , 5), ( - 6, - 1 ) }

8. Determ ine the slope of each of the following lines:(a) y = 3x + 4 (b) y = —2x + 5 (c) x + y = 7(d ) 2y - 3x = 5 (e) 2x + y = 3 (f) 3x - 4y = 5(g) ax + by = c (h) 3y = 4 f (i) 5x = 2 f

9. Find the equation of each of the following lines:(a) Passes through (2 ,1) and is parallel to the line 5x + y ' — 6.(b ) Passes through ( — 3 ,1) and is perpendicular to the line 3x — 4y — 5.(c) Passes through ( - 2 , - 3 ) and is parallel to the line 2y = — 5.(d ) Passes through (4, —1) and is perpendicular, to the line x = —7-

10. Prove that the lines lQx — 4y = 3 and 5x = 2y a re parallel.

11. Are the lines 3x + 2y = 5 and 3x — 2y = 1 perpendicular to each other? Justify your answer.

12. Prove th a t the point (5, —4) lies on the line 4x — 3y = 32.

13. W hat is the equation of the perpendicular bisector of the line segm ent

joining the points (6, —2) and (4, 8)?

14. A (2, 4), 5 (6 ,' 10), and C( 10, 3) are the vertices o f an isosceles triangle

where C is th e vertex of the vertex angle.(a) Show th a t the perpendicular bisector o f the base passes through C.(b ) Show th a t the altitude to the base bisects the base,(c) Show th a t the median to the base is perpendicular to the base.

6 Intersection of Two SelsKnowing how to determ ine the equation of a line places us

in the position of being able to prove several im portan t geometric relations.

t S ec p a g e s 4 0 2 -4 0 3 .

420 COORDINATE GEOMETRY-THE- GRAPHW h at these relations are can best be illustrated through the use of the problem below.

Illustration:

Prove th a t the perpendicular bisectors of the sides of the triangleA ( — 3, 5), S(3 , 3), and C'(U, 19) are concurrent; th a t is, have a po in t in com m on.

M e t h o d : A rough sketch is drawn at the very outset. The coordinates of

the m idpoints of the sides are determ ined m entally and placed in th e d iagram . T o find the slopes of Li, Lt, and Lz, we must first determine the slope of A S , SU, and UA.

slope of A S — j ; slope of I i = 3/1

slope of BC = | = 2; slope of Lt = - 1 / 2

slope of U I = \ = 1 ; slope of L% ~ — 11 1 “r J

I , : = f ; 3 j t = > - 4 ; 3* - y = - 4

U-. = {■, 2 > - 2 2 = - x + 7 ; x + 2> = 29

A : Lx ~ ~ = - 1 ; y - 12 = - x + 4 ; x + y = 16

Using the methods of algebra, we find the common element to L\ and Li\

INTERSECTION OF TWO SETS 421

L\\ 3x — y = — 4Lt: x + 2y = . 29

L\\ 6x — 2y = — 8L i\ x + 2y = 29

I x = 21

.-. x = 3and hence

j. .= 13

You may have called this pair of values—(3, 13)— the sim ultaneous solution to the two equations "ix — y = —4 and * + 2y = 29. At the present time we prefer to refer to this pair of values as the intersection of the solution sets of these equations. This simply means that (3, 13) will be a common element to the solution sets of both 3x — y = —4 and x + 2y - 29: Geometrically, however, the graph of the solution set of each of these equations is a straight line. Hence, the pair of values (3, 13) represents the coordinates of the in tersection of the two graphs of the equations. Thus, the intersection of the solution sets of two equations has two interpretations:

(1) Algebraic: I t is a pair of values that is comm on to both solution sets.(2) Geometric: I t is a point of the graphs of both equations.

Thus far we have shown that L\ and 1? intersect at the poin t (3 ,13). We are now faced with the problem of. dem onstrating that Li also passes through this point. I f (3 ,13) is a point on Lt, then this pair of values m ust be an element in the solution set of the equation of L i. Hence, it is simply necessary to determ ine whether (3, 13) satisfies the equation x + y = 16.

x + y = 16 3 + 13 4 16

16 = 16

Since (3, 13) does satisfy the equation of I 3, we can now conclude that Li, Lt, and Li are concurrent.

There are times when the intersection of two sets is sought bu t can not be found. This will occur w hen the lines are parallel. If the lines are parallel, there will be no point th a t is common to the two graphs; and hence, there will be no element that is common to the two solution sets. It is possible to tell at a glance whether the solution sets of the equations of two lines have no intersection. Should the slopes of these two lines be equal, then, except lor one condition, the lines will be parallel. Therefore, there will be no common elements to the solution sets of these equations. Equal slopes, however, may imply th a t the lines coincide; and hence, further investigation m ay be necessary. If the lines coincide, how m any elements will there be in the intersection of the two sets?

Geometrically, three possibilities exist when trying to find the in tersection of the solution sets of. two equations whose graphs are straight lines:

422 COORDINATE GEOWETRy: THE GRAPH(1) T he lines intersect a t one point; these are called consistent-equations.(2 ) T he lines are parallel; these are called inconsistent equations.(3) The lines coincide; these are called dependent equations.

How many elements will there be in the intersection of the solution sets in each of these situations?

co n sis ten te q u a t i o n s

in co n sis ten te q u a tio n s

t v

Figure 13-4.

d e p e n d e n te q u a tio n s

EXERCISES

1. D eterm ine th e in tersec tion o f th e solution sets o f each of the follow ing pairs o f equatio n s:

(a) x + y <= 1 (b ) x hy ~ 17 (c) 2x — y — 22x — y *= 2 x -h 2y = 8 x 2y = — 4

2. Determ ine the p o in t of intersection of th e graphs of '■ach of the following pairs of equations:(a) 2x + y = — 6 (b) x - —2\y

2y - x = 8 2.v - 2y = 21

3. S ta te w h e th e r e a c h o f th e follow ing pa irs o f equations are consistent, inconsistent, o r d e p e n d e n t:

(a ) 6x — 9y = 5 (b ) 5* + 2y = 1 (c) x + y = 54x — 6y = 7 y 3x — 4 y — x — 3

. (d ) 5x = 9 - lOy (e ) 7* - 21 y = 14 (f) 3x - <by = 4by = 13 — i x 2x — 6y = 4 5* = 3j> + 4

4. Show that the lines x — y — 4, 2x — 5y = 17, and x - f 2y — —5. are concurrent.

5. Determ ine the equation of the line that passes through the point (4, — 1) and the intersection of the two lines 3x -r y = 4 and * + 2y = 13.

6. A ( 6 ,10), 5 (0 , 0), and C(6, 4) are the vertices of a triangle. Prove th a t the perpendicular bisectors of the sides of this triangle are concurrent.

7. A(0, 0), 5 ( 8, 6), and C(4, 12) are the vertices of a triangle. Prove that the medians of this triangle are concurrent.

ANALYTIC PROOFS OF PROBLEMS 423

8. /4(0, 0), 5 ( 8, 0), and C(6, 2) are the vertices of a triangle. Prove th a t the altitudes of this triangle are concurrent.

9. A (2, 5), 5(12, 3), and C(8, 9) are the vertices of a right triangle where Z C is the right angle. Prove that the m idpoint of the hypotenuse lies

on the perpendicular bisectors of the legs.

10. T he vertices of an isosceles triangle are .<4(2, 3), 5 (6 , 11),-.and C(8, 5) where C is the vertex of the vertex angle. Prove that the altitude to the base, the m edian to the base, and the perpendicular bisector of the base coincide. (Hint: Show th a t the equations of all three lines are

the same.)11.* Determine the distance from the point (5, 6) to the line x + 2y - 2.

(H int: Determine the equation of the perpendicular from the point

to the line.)12.* /4(5, —1), 5(1,1), and C(5, —11) are the vertices of a triangle. D eter

m ine the length of the altitude from A to SC.

■ Analytic Proofs of Problems from Synthetic GeometryAmong the problems that you were called upon to investi

gate in this chapter were a number that you had proved earlier by synthetic methods. T he purpose of this unit is to present to you a great m any problems for which you have given synthetic proofs. Now, however, you will be asked to give analytic proofs for these same problems. T hough not necessarily true

■ in all cases, you will find that in most of the situations presented here the analytic proof is somewhat simpler than the synthetic proof had been. J

Illustration:Prove that the line drawn from one of the vertices of a parallelogram

to the m idpoint of one of the opposite sides is a trisector of a diagonal of

the parallelogram.

D (6b,6c) C (6o + 6b,6e)

M e t h o d : Let A (0, 0 ), B{6a, 0), C(6a -f- 65, 6c), and D{6b, 6c) be the four vertices o f the parallelogram. O u r attack will be to find the coordinates of poin t P, one of the trisection points ol DB, and to show th a t this poin t satisfies

424 COORDINATE GEOMETRY: THE GRAPH<—► __

the equation of CM. T he coordinates of M , the midpoint of AB, were found m entally.

„ 2(6a) + 1(66) .12a + 6b . ,abscissa o f P t = —- ----- ------ • - ------ ;----- = 4a f 2b

I -+* 1 j

. 2 ( 0 ) + 1 (6 c ) 6c „o rd in a te ol l ’\ = ^ = -- = 2c

co ord inates of P: (4a 4- 2b, 2c)

r A t , y ~ 0 6c ~ 0equation of C M : ;— r—~,------- —x — ia oa + nb — ia

y_________ 2cx — 3a a + 2b

Substituting th e coordinates of P in this equation,

2c , 2c4a + 2b — 3a = a + 2b

2c _ 2ca -f- 2b a ~h 2b

Since P satisfies the equation of CM, CM will pass through a trisection point of IT5 and, hence, be one of the trisectors of S s ,

EXERCISES

1. Prove that the diagonals of a square are perpendicular to each other. Let *4(0, 0), B(a, 0), C(a, a), and Z>(0, a) be the vertices of the square.

2. Prove th a t the diagonals of a rhombus are perpendicular to each other. Let A(0, 0), B(5a, 0), C(8a, 4a), and D(3a, 4a) be the vertices of the rhom bus. (Can you prove that these points must be the vertices of a rhombus?)

3. Prove th a t the m idpoint of the hypotenuse of a right, triangle is equidistant from the three vertices. Let A{2a, 0), B(0, 2b), and C(0, 0) be the vertices of the triangle.

4. Prove th a t the line joining the midpoints of the diagonals of a trapezoid is parallel to the bases. Let /!(0, 0), B(2a, 0), C(2b, 2c), and D(2d, 2c) be th e vertices of the trapezoid.

5. Prove th a t the measure of the line segment joining the m idpoints of the diagonals of a trapezoid is equal to one-half th e difference of the measures of the bases. See Problem 4 for the vertices.

6. Prove th a t the diagonals of an isosceles trapezoid are congruent. L et A ( —2a, 0), B (2a,0), C(2c, 2d), and D( — 2c, 2d) be the vertices.

t See page 386.

THE GRAPHS OF INEQUALITIES 425

7. Prove that if line segments are drawn joining the m idpoints of con- sccutive sides of an isosceles trapezoid, these segments will be congruent. See Problem 6 for the vertices.

8. Prove that if a line passes through the midpoint of one side of a triangle and is parallel to the second side, then it will pass through the m idpoint

of the third side.9. Prove that if a line bisects one of the nonparaliei sides of a trapezoid

and is parallel to the bases, then it bisects the o ther of the nonparallel sides. See Problem 4 for the vertices and use the m ethod illustrated on pages 423-424.

10. Prove that if line segments are drawn between the m idpoints of opposite sides of a quadrilateral, they will bisect each other. L et A (0, 0), B(2xi, 0), C(2*3, 2y3), and £1(2x4, 2><) be the vertices of the quadrilateral.

11. Prove that the perpendicular bisector of the lower base of an isosceles trapezoid is also the perpendicular bisector of the upper base. See Problem 6 for the vertices.

12. Prove that if the nonparallel sides of an isosccles trapezoid are extended, they will be concurrent with the perpendicular bisector of the lower base. See Problem 6 for the vertices.

13. Prove that if the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram. (H int: Use as vertices the values forA, B, and M shown below; then prove th a t the opposite sides are parallel.)

14. Prove that the sum of the squares of the measures of the sides o f a parallelogram is equal to the sum of the squares of the measures of iu diagonals. Let A (0, 0), B(a, 0), C(a + c, d), and D(c, d) be the vertices of the parallelogram.

15.* Prove that the altitudes of a triangle are concurrent. L et A(0, a), B (0, 6), and C(c, 0) be the vertices of the triangle.

■ The Graphs of InequalitiesThus far wc have considered only those situations in which

the elements in the solution set yielded a set of points th a t fell on a line. T here are conditions, however, under which the graph of the solution set

will consist of the points in a half-plane or possibly the points enclosed within a region of the coord inate plane.

T o illustrate, consider th e open sentence

x > 3 ,

As before, we can look upon this as a direction stating th a t the point must be more than three units to the right of the vertical, or 7-axis. Since no requirements were placed on how far the points m ust be from the x-axis, the second e lem en t o r o rd inate, in each ordered pair can be any value we desire. Thus, a few of the pairs of elements in the solution set of x > 3 will be

{(4, 0), (4, - 1 0 ) , (5, 17), (5,246), (9, - 5 7 8 ) , . , .}

W hat m ust be true of the first coordinate in each ordered pair? W hat m ust be true of the second coordinate?

To answer the question “ W here do these points lie?” , we first graph the equation x = 3.


y

n1><

____ .P(incomplete graph)

Figure 13-6.

Any point on this graph, the red line, will be 3 units from the 7-axis. O n the other hand the distance from point P to the 7-axis is greater than 3 units; hence,' the coordinates of P will be an element in the solution set of the inequality

x > 3

Furtherm ore, the point P itself is a point of the graph of the solution set. Similarly, the distance from Q to the 7-axis is greater than 3 units, thus making Q a point of the graph. In addition, the distance from any point to the right of the line -x = 3 to the 7-axis will be greater than 3 units; therefore, all points in this half-plane will be points of the graph of x > 3. T his graph is shown as the shaded area of the coordinate plane in Figure 13-7,


To indicate the fact that the points on the line x = 3 are not elements in the solution set, the red line was drawn as a dotted line. I f the points on this line were to be included in the graph, it would have been draw n as a solid

line.In the same way, the solution set of the open sentence

' ' j > < 7will consist 0/ all ordered pairs of num bers in which the second coordinate is equal to or iess than 7. T he sentence7 < 7 implies that there are two types of replacements for y under which this sentence will be true. These replace

ments are(1) T h at 7 be 7.(2) T h a t 7 be less than 7.

T he graph of the solution set of 7 < 7 is the set of points in the half-plane below the lin e7 = 7 and this line itself. Each of the points on the lin e 7 = 7

Figure 13-8.

conforms with the first replacements for 7 , while the points in the half-plane below 7 = 7 fulfills the second requirem ent. Notice th a t in Figure 13-8 the line g raph 7 = 7 was drawn as a solid line, for it is to be included in the

graph. .W e have examined graphs consisting of half-planes th a t fell to either

the right or left of a vertical line and half-planes above or below a horizontal line. W hat would occur, however, if the graph of the equality were no t parallel to either axis? Before it is possible to answer this question, it will be necessary to form ulate several assumptions on inequalities similar to those developed earlier for equalities.

P o s t u l a t e 36: If a > b

then a + c > b + c

This symbolic relation is often expressed as

“ If equals are added to unequals, the sums will be unequal in the sam e order.”

As an illustration, we know that 9 is greater than 7. W ere we to add 3 to


both 9 and 7, the sum of 9 and 3 (12) would be greater than the sum of 7 a n d 3 (10). T h a t is,

If 9 > 7ther. 9 + 3 > 7 - r 3or 12 > 10

W h at a re the re p la c e m e n t for a, b, an d c in this illustration?T h e words “ same order” as used in the statem ent above implies th a t

if the left side of the inequality is larger than the right side a t the outset, then after addition takes place, the left side will still be larger than the right side.

Postulate 36 is helpful in simplifying inequalities of the form x — 7 > 4 . Thus,

* - 7 > 4 ■at — 7 + 7 > 4 + 7

x > 11

By exam ining the inequality x > 11, we are m ade aware of those num bers th a t are elements in the solution set of the original inequality x — 7 > 4 . T h a t is, any num ber greater than 11 will make the left side of this inequality greater than the right side.

P o stulate 37: If a > b

then a — c > b — c

This symbolic relation is often expressed as

“ If equals are subtracted from unequals, the differences will be unequal in the same order.”

As an illustration,9 > 7 '

9 - 3 > 7 - 3or 6 > 4

W hat replacem ents were used for a, b, and c?Postulate 37 is used to simplify the form of inequalities such as

x + 7 > 4, for

x + 7 > 4 x + 7 - 7 > 4 - 7

x > - 3 'P o s t u l a t e 38: If a > b

then - > - where c > 0c c

T his symbolic relation is often expressed as

“ I f unequals are divided by positive equals, the quotients will be unequal in the same order.”


As an illustration,1 5 > 6

5 > 2

W hat replacements were used for a, b, and c? W ere c negative in this illustration, how would our conclusion been, affected? W hat would happen were

c equal to zero? .This postulate is used to simplify the form of an inequality such as

i x > 20. Thus,4a: > 20

20 4 4

x > 5

W ith these assumptions as a foundation, we are prepared to graph the inequality x + Zy > 10. Let us first • consider the graph of the equalityx + 2y = 10. Each point on the line, such as P, satisfies the condition th a t

x be equal to 10 — 2y (1)

T o illustrate by using P, when y is replaced by 2, the value of * in equation(1) is 6, while the value of the abscissa of the point P in the graph is also 6 . By using Postulate 37 the form of the inequality

,r + 2y > 10

can be changed to* > 10 - 2y

T his implies that* m ust be greater than 10 — 2y (2) ■

W hen y was replaced by 2. we found th a t the value of IQ — Zy was 6 and, hence, the value cf x on the line was 6 . From our inform ation in (2), we see th a t x must now be greater th an 6 when y is equal to 2. W here will thesepoints fall? The points for which y is 2 and x is greater th an 6 will ap p ear •


on the dotted line in Figure 13-10. Thus, the x-coordinate of Q is greater than 6, for the point falls to the right of P.

In the same w ay, all points th a t fall to the “ right” of the line x -f- 27 = 10 will be such that the x-coordinate will be greater than the x-coordinate on the line. T he x-coordinate of a point on this line is equal to 10 — 2y, while the x-coordinate of a point to the “ right” of this line is greater than 10 — 2y. W e showed this to be the case for y = 2 . I t is true, however, for any replacem ent of y. Thus, for y ~ 71, the value of x on the line is

10 - 2y,

For a point to the right of (10 — 271, 71) the value of x will be greater than10 — 2y\, for the distance from this point to the 7-axis is greater than the distance from the point (10 — 271, 71) on the li.-.e to the 7-axis. This is illustra ted in Figure 13-11. In view of this, the graph of the inequality

x + 27 > 10

are the points in the half-plane to the “ right” of the line x + 27 = 10.

Figure 13-12.

Perhaps you m ay have been troubled by this question: “ H ad the inequality been solved for 7 ra th e r than x, in w hat m anner would the graph of th e inequality been effected?” L et us solve the inequality for 7 to see ju st w hat does occur.


x + 2> > 102)i > 10 - x

10 - xy > 0

Given

Postulate 37

Postulate 38

T he analysis, now, is similar to that used when the inequality was solved

for x. Thus, 7 is equal to1.0 ■ ; for all points on the line. Hence, for points

“ above the line,” y will be greater than10 Therefore, the g raph of

y >10 - x

will be the set of points in the half-plane “ above jh e line,” .

x + 27 = 10

Examination of the graph below will show th a t the graph of the solution set are points of the same half-plane as those when the equality had been

( in co m p le te g ra p h )

Figure 13-13.

solved for x. This, of course, is as it should be, for we would not w ant two different graphs representing the same inequality.

T o summarize, in graphing an inequality, the following steps should

1 be taken:(1) Remove the inequality sign and replace it w ith the equality sign.(2) G raph the equality.(3) If the inequality is solved for x, determ ine w hether the g raph of th e solution set will be the points in th e half-plane “ to th e right” o r “ to the left” of the line of equality.(4) If the inequality is solved for 7 , determ ine w hether the g raph of the solution set will be the points in the half-plane “ above” or “ below” the line of equality.

In the preceding unit we determined the intersection of th e solution sets of two equalities. T he same can be done were the sentences inequalities ra th e r th an equalities.

432

Illustration:COORDINATE GEOMETRY; THE GRAPH

D eterm in e the intersection o f the solution sets o f the open sen tences:

(1) 2.v + 3;y > 12(2) x - Ay > 4

M e t h o d : F ollo w in g t h e pattern outlined, w e w ill first graph th e eq u a lities 2x 4- Zy = 12 and x — 4y = 4.

Figure 13-14.Solving (1) for x , We obtain

2x > 12 - . 3y

. 12 - 3> .

T h is implies th a t the points of the graph will fall in the half-plane “ to the righ t” of the line 2* + Zy = 12 and, also, on the line itself.Solving (2) for y, we obtain

x — 4 > 4y

x ~ 4 — > >

Thus, the graph 9f the second inequality will be the points of the half-plane “ below” th e line x — Ay — 4. In solving for y, 4y was added to both sides of the inequality. This was done to make the coefficient of y a positive qu an tity so that Postulate 38 could be applied.

( in c o m p le te g ra p h )


The points in the region m arked A, being in both half-planes, will be points of the graphs of both inequalities. Hence, the points in the cross- hatcked, or doubly shaded region, are elements in the intersection of the solution sets of the two inequalities. Notice that whereas half of the line 2x + 3y — 12 is included in the intersection, none of the line x — Ay = 4 is included.

In this illustration, the first inequality was sojved for x, while the second was solved for y. Although both can be solved for the same variable, the graph of the intersection is m uch clearer and neater if one inequality is solved for one of the variables, while the other is solved for the o ther variable.

EXERCISES

1. Using Postulates 36, 37, and 38, determ ine the greatest or least replacement value for x so as to make each of the following open sentences true. Give two other values in the solution set of each of these inequalities,(a) 2x > 10 ' (b ) 24 < Zx(c) 5x — 1 > 29 (d) Ix — 36 < Ax(e) 9a: - 25 > l l x + 7 (f) 4(x - 3) < 5x + 2

2. G raph each of the following inequalities:(a) x < 10 (b ) y > — 5(c) x + y > 4 (d ) 2x — y < 6(e) y - Zx > 12 (f) Ay - x < 8(g) 2x - Zy > 6 (h ) 5x + 4;i < 20

3. By graphing, determine the intersection of the solution sets of the following pairs of inequalities:

4. Determ ine the intersection of each of the following systems of inequalities:fx > 0 f x < 0

(a) > 0 (b ) <7 < 0t * + > < 1 0 [x + 7 > — 4I* > 2 fx + 3 < 0 .

( c ) \ y < 5 ( d H > + 5 > 0

I* - > < 2 [ j - 2> > 0fx + y < 8 . fx + y > 0

(e) -UOx+ 7 > 10 (f) -{2* — 3/ < 6(* + 10)i > 1 0 |* + 2 y < 3


I Locus of Points

A very im portant topic in mathematics and one closely related to w hat we have been discussing is the topic known as the locus of points, f

D e f in it io n 74: T h e locus of points is the set of those points and only those points th a t satisfy certain given conditions.

Should the conditions that determ ine these points be described by an equation or an inequality, then the definition implies that the graph of this description will be the locus. Thus, let us say th a t the conditions describing the locus of points is

{(*,;■) | * = 3} (A)

This, you recall, was read as “ T he set of all points with coordinates (*, y) such that the abscissa of each of these points would always be 3.” The graph of this set of points is the line parallel to the 7-axis and 3 units to the right of it. Hence, the locus of points fulfilling the conditions specified in (A) is m erely the graph of the elements in (A).

Similarly, if the locus was described by the conditions

{(*,7) | 2x - iy < 12}

then the locus would be the points in the half-plane “ to the left” of the line 2x — 37 = 12.

Figure 13-16.

Actually, the definition of the locus of points calls for two requirements to be fulfilled:

(1) T h at every point that satisfies the given conditions be in the locus.(2) T hat every point in the locus should satisfy the given conditions.

Fortunately, this is exactly the relationship that exists between a graph, and the equality or inequality leading to that graph: every element in the solution set is a point of the g raph or locus, and every point of the graph

t Locus is a L a tin w ord m ean ing place; hence, locus of points m eans place of points.

LOCUS OF POINTS 435

or locus is an element in the solution set. Thu.., in the illustration above every ordered pair of values in the set

{ ( * , 7 ) | 2 x — 3 7 < 1 2 }

will lie in the half-plane to the “ left” of the line 2x — 3y - 12, and every point in this half-plane can be represented by a pa ir of values in this set.

In m any cases the description of the locus or set is not given by an equation or an inequality but by some narrative statem ent. Thus,

“ Find the locus or set.of points th a t are equidistant from the points (1, 0) and (5, 8).”

U nder these circumstances it is necessary for us to rewrite this description in terms of set symbols and then simplify the equality or inequality through

the use of our postulates.M e t h o d : O ur first objective is to draw a rough sketch based on the descrip-

------ *s

R (0,1) x

F ig u r e 13-17.

tion and allow the ordered pair (x, 7) to represent any point of the locus.In terms of this diagram the description of the set can be rewritten as

(1) {(*, y) \ d i = dt)Now, it is merely a m atter of simplifying this expression:

(2) {(X,y ) | V ( , - 0)* + (7 - l")2 = V ( x - 5)2 + (7 - 8)*}(3) {(x,7) | x2 + (7 - l )2 = (x - 5)2 + (7 - 8)*}(4) {(x,7) | x2 + 72 - 27 + 1 = x2 - lOx + 25 + y 2 - I 67 + 64}(5) {(x,7) j 10* - 27 + I 67 = 25 + 64 - 1}(6) {(x, 7) | 10x + 147 = 88}0 ) {(*,7) I 5x + ly = 44}

T hus, we have transformed the description for finding the elements of : the set from the narrative form to th a t of the simple equation 5x + ly = 44. T he graph of this equation is a line. Therefore, w hat we have, proved thus far is th a t all the points that are equidistant from the points (0, 1) and (5, 8) must be points of this line. However, it is possible th a t there m ay be some points of this line that are not equidistant from these two points. Hence, we must now show that every point of the line 5x + ly = .44 must be equid istant from ihe points (0, 1) and (5, 8).

Fortunately, we had proved in Theorem 80 th a t every element in the . solution set of 5x + 77 = 44 is on a line and every point on this line is an

436 COORDINATE GEOMETRY; THE GRAPH

element in the solution set of 5x + ly = 44. But by retracing our steps in the proof above, we find that the description Sx - f ly = 44 is but another form, or equivalent form, of the description

(2) V (" * -0 )* + (7 -T j* = V (* - _5)r +- ( y - 3YHence, the points on the line are also elements in the solution set of equation(2). However, this equation indicates that the distance from each point (x, y) on the line to (0, 1) is always equal to the distance from (x , y) to (5, 8), This is what we had hoped to show. In sum m ary, then, the locus or set of points that are equidistant from the point (0, 1) and (5, 8) are all the points on the line Sx + ly = 44.

Generally speaking, for the problems we will encounter on locus, once we have found the description leading to the graph, i t will not be necessary for us to prove that all points of this graph, fulfill the conditions of the problem . This will be so w ithout proof, for each step in our proofs are retraceable similar to w hat we had done in the illustration ju st completed. T h e only retraceable step that is questionable is when we had m oved from step 3 back to step 2 (see Proof, page 435). In doing this it was necessary to take the square root of each side of the equation. Thus, the poin t m ight be raised as to why the negative square root was not used. I t was simply because we .had agreed th a t distance would always be a positive num ber.

There are several things that are of interest about the line 5x ly — 44 in this illustration. By solving the equation for y and examining the coefficient of x, we see that the slope of this line is — 5/7 . T he slope of the segment US is 7 /5 . Thus, the locus turned out to be a line that is perpendicular to the line segment joining the two points (0, 1) and (5, 8). Furtherm ore, the coordinates of the m idpoint of JiS—(5/2, 9 /2 )—satisfy the description of the locus:

5x + ly = 44 5(5/2) + 7(9/2) JL 44

25/2 + 63/2 X 44 44 = 44

Therefore, the locus is r ; 't only the line perpendicular to US, but this line also turns out to be the bisector of RS.

T he description of a locus is sometimes given in terms of a com pound sentence. Thus,

“ Find the locus of points that are more than two units from the line x = 6 and also equidistant from the points (0, 1) and (5, 8) .”

Loci of this form are called compound loci.\ In reality they involve nothing m ore than the intersection of two sets: Symbolic descriptions of each of the sets are form ulated from the narrative descriptions. This is then followed by finding the graph of their intersection.

t P lu r a l o{ locv» is loci.

LOCUS OF POINTS437

M e t h o d : Starting with the first description,A = the set of points th a t are more than two units from the line x = 6

A = {(*,>) 1 x > 8 or x < 4}T h e description in set A cam e from the fact that the points had to be more than two units from the line x — 6. Since these points can lie on e ither side of x = 6, the description must include the fact th a t x m ay be e ither greater

than 8 or less than 4.T h e second description was analyzed in the preceding illustration and

led to the set B = {(x ,y ) \5 x + 7y = 44}

Hence, the problem resolves itself to one of determ ining the intersection t of set A w ith set B. Symbolically, this is expressed as A (~) B and is read as

(1) T h e intersection of A and B

or (2) A intersects B.T h e shaded region in the graph in Figure 13-18 represents those points

th a t a re more th an two units from the line x = 6. T h e line 5x 4- ly — 44

Figure 13-18.

represents those points th a t are equidistant from the points (0, 1) and (5, 8). H ence, th e points th a t are both two units away from th e line x — 6 and equidistant from (0, 1) and (5, 8) a re in th a t pa rt of the line 5x + ly = 44 th a t lies to the right of the line x = 8 and to the left of the line x — 4. T h e points of these sections of the line 5x + l y ~ 44 represent A C \ &•

EXERCISES1. Using set symbols, rewrite the description of each of th e following loci:

(a ) T h e locus of points that a re four units to th e righ t of th.e vertical axis.(b ) T he locus o f points that a re th ree units below the horizontal axis.

' (c) T he locus of points th a t a re four units to the right of the line x = 8 .(d ) T h e locus of points th a t are five units below the Hr*e> = 3.

t See page 11.

438 COORDINATE GEOMETRY: THE GRAPH(e ) T h e locus of points th a t are more than two units to the right of the

.y-axis.

(f) The locus of points that are more than seven units above the x-axis.(g) T he locus of points that are more than three units to the right of

the line x = 8.

(h ) T he locus of points that aie more than two units from the line v — 6.(i) T he locus of points that are more than five units from the line x = 3.

( j ) t T h e locus of points th a t ave less than three units from the line x = 5.2. D raw each locus in problem 1.

3. Using set symbols, rew rite the description of each of the following loci:(a ) T h e locus of points th a t a re a distance of five units from the origin.(b ) T he locus of points th a t are a distance of R units from the origin.(c) T he locus of points that are a distance of seven units from the

point (4, 6).

(d ) T he locus of points that are a distance of R units from the point (.h, k ) .

4. (a ) Using set symbols, express the locus of points that are equidistantfrom the points (0, 2) and (10, 4).

(b ) Draw this locus.

(c) Prove that this locus is the perpendicular bisector of the line segment• joining the two points.

5. Using set symbols, express the locus of points that are equidistant from • the two points in each of the problems below:(a) ( - 5 , 0) and (5, 0) (b) (0, - 7 ) and (0, 7)(c) (2, 5) and (12, 5) (d ) ( - 6 , - 3 ) and ( - 6, 9)(e) (1, 0) and (5, 2) (f) ( - 3, 4) and (5, 8)(g ) ( - 2, - 6) and ( 0 , - 1 4 ) (h ) ( - 1 , - 7 ) and (6, - 4 )(i) (a, b) and (2a, — 26)

6. D raw each locus in Problem 5.

7. (a ) Using set symbols, express the locus of points such that the differenceof the squares of the distances from the points A(Q, 0) and B(4, 5) is always 4.

(b ) By simplifying the description of the set in pa rt (a), show th a t this locus is a line.

(c) Prove th a t this locus is a line that is perpendicular to the line determ ined by tke points A and B.

(d ) G raph th e locus found in (b).

8. Using set symbols, express the locus of points such that the difference of the squares of the distances from any point of the locus to(a ) (3, 5) and (7, 5) is always 6 units.

LOCUS OF POINTS 439

(b ) ( — 4, 6) and ( — 4, —2) is always 2 units.(c) (0, 0) and ( — 5, - 8) is always 4 units.(d ) ( — 2, 4) and (3, 7) is always —3 units.

9. (a) By simplifying each of the descriptions in Problem 8, show that each of these loci is a line.

(b ) G raph each of the loci in Problem 8.

10. (a) Using set symbols, express the following locus as a problem in which

the intersection of two sets is to be fo u n d :

“ Find the locus of points that are two units to the right of the vertical axis and four units above the horizontal axis.”

(b ) Graph the intersection of the two sets found in part (a).

11. Draw the locus in each of the following problem s:(a) W hat is the locus of points that is m ore th an two units to the left

of the vertical axis and more than two units above the horizontal

axis?(b ) W hat is the locus of points that is more than four units to the right

of the vertical axis and on the line x — ly = 4?(c) W hat is the locus of points that is m ore than three units from the

horizontal axis and on the line x + y = 0?(d ) W hat is the locus of points that is two units from the x-axis and

equidistant from the points (0, 0) and (6, 0)?(e) W hat is the locus of points th a t is more than three units from the

line y = 7 and equidistant from the points ( — 2, 4) and (4, 6)?(f) W hat is the locus of points th a t is less than four units from the line

x = —8 and equidistant from the points ( — 1, 2) and (3, 8)?(g ) W hat is the locus of points that is equidistant from the points. (0, 0)

and (6, 0) and also equidistant from the points. (2, 4) and (4. 10)?

12. Prove that the locus of points equidistant from two fixed points is th e perpendicular bisector of the line segm ent join ing the two points. Let

(0, 0) and (2a, 2b) be the two fixed points.

13. (a ) W hat is the distance from the poin t (6, 4) to the _y-axis?(b ) W hat is the distance from the point <(x,y) to the ji-axis? .(c) Using set symbols, express the locus of points such that the distance

from each of these points to the^-axis is equal to its distance to the

point (6, 4).

14. Using set symbols, express the locus of points such that the distance from each of these points to the point (6, 0) is equal to its distance to

the line x ~ 2 .

440 COORDINATE GEOMETRY: THE GRAPH15. Using set symbols, express the locus of points such that the distance

from each of these points to the point (0, 4) is four times its distance to the jr-axis.

16. (a ) Using set symbols, express the locus of points such that the slopeof the line segment joining any of these points to the poin t (2 , 3) is 1/ 2.

(b ) By simplifying the description of this set, show that the locus will be a line.

(c) G raph the locus found in (b).

■ The Circle

Thus far we have considered only the properties of lines from an analytic standpoint. T here was, however, another figure th a t was exam ined briefly in synthetic geometry that , we would like to analyze a t this tim e in coordinate geometry. This figure is the circle. You m ay recall th a t by definition a circle is a set of points such th a t line segments draw n from all points of this set to a fixed point are congruent. H ad we preferred, the definition of a circle could have been given in the equivalent form of

A l t e r n a t i v e D e f in i t io n 36: A circle is the set or locus of points th a t are a fixed distance from a.fixed point.

E xam ination of this definition will reveal th a t it. is identically th e sam e as th e previous one. T he new definition, however, enables us to apply set symbolism to describe the set of points that lie on the circle.

Illustration;

Using set symbols, describe the circle wherein the fixed point is (2, 3) and the fixed distance is 7 units.

M e t h o d : In terms of the definition of a circle, this problem can be rew ritten as

THE CIRCLE 441

(1) This circle is the set of points such th a t each of these points is 7 units from the fixed point (2, 3).(2) C = { ( , , , ) | V ( , - 2)2 + (J- - 3)! = 7}(3) C — { ( x , y ) \ ( x - 2 )‘ .+ ( > - 3 ) * = 49}

In step. (2), the expression on the left side of the equality sign simply represents the distance from any point in the set to the point (2, 3). By the conditions of the problem this distance was always to be 7; hence, the equality between the radical and the 7. Step (3) was found by squaring both sides of the equality. T he equality in step (3) is called the equation of the circle for this problem. In exactly the sam e way we are now going to develop the equation of the circle where the fixed point may be any point in the coordinate plane and the fixed distance is any given distance.

T H E O R E M 81: T he equation of a circle whose center is (h, k) and whose radius is R is

(x - hy + (y - ky = r*

P R O O F

O n the basis of the definition m ade earlier we know th a t the fixed distance is the radius of the circle, while the fixed point is the

center o f the circle. Hence,

(1) This circle is the set of points such that each point is R units from the fixed point (h, k),(2) C - {{x,y) | V J T - ky + (y - k y = R}(3) c = {(*,>) | (* - *)* + (j. - k y = **}

Thus, points that are R units from (h, k) are elements in the solution set of

(* - h)1 + (y - k y = R* (1)W hat we have shown, thus far, is th a t each point o f the circle m ust

be an element in the solution set of equation (l)..Now it is necessary to show th a t each element in the solution set of equation (1) is a poin t on the circle


with center (h, k) and radius R. This can be done ra ther easily by simply reversing the steps in our proof. Thus,

(* - k y + ( y ~ k y = j?* (t>

V ( x - h )2 + (y - k y = R (2)

Equation (2) indicates th a t every element (x,y) in the solution set of (1)is a point whose distance from (h, k) is always R. But this is exactly w hat wem ean by a circle w ith cen ter (A, k) and radius R !

Y our attention m ust be called to one point of this proof. In finding the • square roots of both sides of the equality, the negative sign was not used. T his was avoided, for we had agreed that distance would always be considered as a positive num ber.

T he application of this theorem to problems is extremely simple.

Illustration:

Find the equation of the circle whose center is a t the point (2, —3) and whose radius is 5 units.

M e t h o d : This problem entails simply writing the general form of the equation of the circle. T his is the form found in T heorem 81. T h e letters k and k are then replaced by 2 and —3, and R by 5.

Thus, {x - h y + { y - k y = R 2(x - 2y + (y + 3)! = 25

Ju s t a word of caution: be careful to change the signs of the coordinates of the center when substituting them in the general form of the equation of the circle.

Should the center of the circle be at the origin, the equation of the circle will become

(x - h y + ( y - * )* .- R 2(* - 0)2 + 0- - 0) 2 = R>* * + > * = R 2 (General equation of a circle with center at origin.)

Ju s t as earlier we had shown that the graph of the equation of the form ax + by = c was a line, so we are prepared now to prove a com parable theorem relating to circles.

T H E O R E M 82: T h e g ra p h of every equation of the form

Ax* + ify2 + Bx + Cy + D = 0 (1) .

w ill be a circle.

A n a l y s is : To prove this, we will show that every point in the solution set of this equation is a fixed distance from a fixed point. This will imply, on th e basis of the reverse of the definition of a circle, th a t the graph of the solution set is a circle.

THE CIRCLE 443

PRO OFDividing both sides of equation (1) by A,

x 2 + y 2 + j x + j y + j = 0

Since B /A , C /A, and D /A are constants, we will replace each of them w ita

other constants so as to simplify our work.

Let B /A — 2p, CjA = 2q, D /A — s Then, x2 + y 2 + 2fix + 2 qy + s = 0

(,* + 2px + :?_ ) + O’2 + 2 qy + _?.)■ = - J

To determ ine the missing term, we complete the square of each tr inomial, remembering, of course, to add the same quantity to the right side

of the equality.

(.*2 + 2px -)- /* ) + (y! + 2 qy + q%) = p 1 + - s(* + p y + (y + q y = p 2 + ?2 — s

' / ( x + P ) , + {y + q)2 = + q2 - S (2)

Since the left side of the equation above represents the distance from the point (x ,y ) to the point ( —p, —q), this equation simply expresses the fact that each point (x, y) in the solution set is a fixed distance

V p 2 + q2 - s

from the fixed point ( — p, —q). Hence, the graph of the points in the solution set of equation (1) will be a circle.

Examine the right side of equation (2). If p 2 + q2 were equal to s, w hat would be the value of p1 + q1 — si Hence, w hat would the radius of thiscircle be? . U nder these conditions the only elem ent in the solution set willbe the center ( — p, —q). Thus, the circle will consist of only one point. A circle such as this is called a point circle. Similarly, if p 2 + q1 were less than s, the quantity under the radical would be i. Lgative. t In this event the radius of the circle will be an imaginary num ber. Circles such as these a re

called imaginary circles.

Illustration 1:Find the radius and the coordinates of the center of the circle whose

equation isx5 + f - 8x + 2y - 3 = 0

M e t h o d : Since the equation is of the form stated in Theorem 82, its graph will be a circle. By rewriting it in the general form shown in Theorem 81, we can easily determ ine th.e radius and the coordinates of the center,

t T h is sta tem en t is tru e only if t is a positive n u m b e r .

(x! - 8x + _? ) + (y* + 2y + _?_) = 3 (x* — 8x -f-16) + (y* + ly + 1) = 3 -f 16 - f ]

(x - 4)’ + ( y + 1)! = 20

Com paring this with the genera! iorm

(x - hY + (y - k y = R 2we can say that

coordinates of center: (4, —1); radius = V 2 0 = 2V 5

There is but one further m atter that we have to investigate. This is the situation that arises when we encounter inequalities with reference to circles.

Illustration 2:

Determine the locus of points th a t are less than 7 units from the point (3, 4).

M e t h o d : Rewriting the problem, we obtain

(1) C is the set of points such that the distance from each of these points to the point (3, 4) is less than 7 units.(2) C = {(x,y) | V (x — 3)« + (y - 4)* < 7}(3) C = {(*,>) | (x - 3)* + (y - 4)» < 49}

W ere the inequality sign in the third description replaced by an equality sign, the graph of this set would be the circle with center (3, 4) and radius 7. Hence, each point on the circle would be 7 units from the center (3, 4).

444 COORDINATE GEOMETRY; THE GRAPH

in

h

— 1 mFigure 13-21.

T h e conditions of this problem state th a t the points in the set m ust be less than 7 units from (3, 4). Hence, these points will lie within the circle. T hus, the region shaded in red will represent the locus of points th a t are less th an 7 units from the point (3, 4).

Similarly, were we required to determ ine the locus of points th a t w ere m ore than 7 units from the point (3, 4), these points would lie outside the circle.

THE CIRCLE 445

Illustration 3:Find the locus of points th a t are more than 4 units from the line x = 8

and less than 6 units from the point (8, 7).

M e t h o d : Using set symbols,

A = {(*, y) | x > 12 or x < 4}

B = {(*,>) I V (x - 8)2 + (y - i y < 6}

O ur problem is to find A f ] S . Rewriting B, we obtain

B = {(*>>) I (* ~ 8)2 + (> ~ 7)2 < 36}

Figure 13-22.

T he points to the right of x = 12 and to the left of x = 4 are in set A. The points within the circle are in set B. Hence, the intersection of A and B are points in the region shaded in red. T he points in '.Us region are more than 4 units from the line x = 8 and less than 6 units from the point (8, 7).

EXERCISES

1. Determ ine the equation of each of the following circles:(a ) Center (0 ,2 ) , radius 5 (b ) Center (3, 0), radius 4(c) Center (2, 5), radius 1 (d) Center ( — 3 ,1 ), radius 6

(e) Center (5, —3), radius V s (f) Center ( —1, —5), radius 2 v /32. F ind the equation of the circle

(a) Whose cen ter is the origin and passes through the point (6, 0).(b ) Whose center is the origin and passes through the point (0, —5).(c) Whose center is (6, 0) and passes through the origin.(u ) Whose center is (0, —8) and passes through the origin.(e ) Whose csn tcr is (4, 6) and passes through the point (4, 10).(f) Whose center is (2, 3) and passes through the point (-^4, 3);(g ) Whose center is (4, —5) and passes through the origin.(h ) Whose center is (3, 5) and passes through the, point ( — 1, 2).


3. Determine the center and radius of each of the following circles:(a) x* + y* - 6* - 16 = 0(b ) x ' 4- y 2 8y 4- 7 = 0(c) x1 ~r y 2 — 4x — 6y — 23 = 0(d ) x‘ + y* + iOx - 12y + 60 = 0(e) 2*2 + 2y2 - \6x + 12y - 48 = 0(f) 3x2 + 3 / - 6x + 487 - 105 = 0

4. Show th a t the point (2, — 5) lies on the circle (x — 3)2 + (^ + 4)2 = 2.

5. Show that the point ( — 3, 1) lies on the circle .*2 + f + Ax — IOji +12 = 0.

6. (a) Using set symbols, express the locus of points such th a t the sumof the squares of the distances from each of those points to the points . (0, 0) and (6, 0) is 36.

(b ) Show th a t this locus is a circle.(c) Show th a t the center of this circle is the m idpoint of the line

segment joining the two points.(d ) Explain how the locus would have been affected had the sum of

the squares of the distances been 18 ra th e r th an 36.(e ) Explain how th e locus would have been affected had th e sum of

the squares of the distances been less than 18 ra th e r than 36.(f) If the difference of the squares had been 36, ra ther than the sum •

of the squares, w hat would the locus have been?

7. (a ) Show th a t the locus of points such th a t the distance from each ofthese points to the point ( — 2, 0) is twice its distance to the point (2, 0) is a circle.

(b ) Find the center and radius of this ci:

8. Use the d iagram a t the rig h t toprove each of the following problems. The center of the circle is the origin. -----

(a) Prove that the radius drawn to the m idpoint of AB is perpendicular«-♦

to AB,

(b ) Prove th a t the perpendicular bisector of A B passes through the center of the circle.

9, (a) Express w ith set symbols and then draw the locus of points that are less than 5 units from the point (2 ,1 ).

A (0,-8)

TEST AND. REVIEW 447

(b ) W hat will be the locus of points th a t are m ore than 5 units from

the point (2, 1)?10. Draw the locus of points that are less than 10 units from the origin and

also on the line x = y.

11. Draw the locus of points that are more than 5 units to the right of the _y-axis and less than 10 units from the point (7, 2).

12. (a) Draw the locus of points that are 8 units from the origin and equid istant from the points (0 , —2) and (2 , 0).

(b) Draw the locus of points that are less than 8 units from the originand equidistant from the points (.0, —2) a.id (2, 0).

(c) D raw the locus of points that a re m ore th an 8 units from the originand equidistant from the points (0 , —2) and (2, 0).

13. (a) Draw the locus of points that are less than 2 units from the point(8, 10) and m ore than 6 units from the line 7 = 10.

(b ) D raw the locus of points th a t are more th an 2 units from the point (8, 10) and less than 6 units from the line y = 10.

14,* Draw the locus of points th a t are either less'than 2 units from the. point (8, 10) or m ore than 6 units from the line y ==10.

■ Test and Review1. How would you read each of the following expressions?

< « ).{ * !* + 4 - 7 }(b ) {(*,).) | > > 2 * + 3 }

2. Find two ordered pairs of numbers in {(jc, >) | 3x + y = 10}.

3. F ind two ordered pairs of numbers in the solution set of

2x ~ y > 4

4. D eterm ine the equation of each of the lines under the conditions given.(a) Passes through the points (2, 5) and ( 0 , - 4 ) .(b ) Passes through the point ( — 1 ,2 ) and has a slope of 2/3.(c) Passes through the origin and the point ( 4 , - 1 ) .(d ) Intersects the x-axis a t x = —3 and has a slope of — t.(e) Parallel to the^-axis and passes through the point ( 2 , - 1 ) .

5. A( — 3, 2), 5 (1 , —4), and C(7, 6) are the vertices of a triangle.

(a ) Find the equation of the m edian from A to BV.

(b) Find the equation of the perpendicular bisector of ~b C.

(c) Find the equation of the altitude from B to AU.

6. Is the following set of points collinear? Justify your answer.

{ ( - 2 , 5), (0, 2), ( - 6,


7. (a) Find the equation of the line th a t passes through ( —4, 2) and isparallel to the line 3* + y = 4.

(b ) Find the equation of the line that passes through the origin and is perpendicular to th e line r. —' 2y — 5 .

8. T he vertices of an isosceles triangle are *4(3, —5), B{20, 4), and C (l, 7) where Z B is the vertex angle. Prove th a t the altitude to the base is also the m edian to the base.

9. (a ) Find the intersection of the solution sets of the following two eq u ations : 5* — y = —11 and 2x + 5y = \ .

(b ) In terp re t the ordered pair of num bers th a t you found as your answer to part (a).

10. Will the graphs of the following two equations be parallel, intersect, or be coincident?

2x - Ay = 7 I x = 14y — 5

11. Prove that the medians of the triangle whose vertices are A (0, 4), B (6, 0), and C(2, 8) arc concurrent.

12. Prove that the diagonals of a rectangle are congruent.

13. Prove that the line segments joining a vertex of a square to the midpoints . of the opposite sides trisect the diagonal of the square that is no t drawn

from th a t vertex. L et (0, 0), (2a, 0), (2a, 2a), and (0, 2a) be the four vertices of the square:

14. Determ ine the element in the solution set of 3* + 12 < Ax th a t has the least value.

15. By graphing,- determ ine the intersection of the solution sets of the following pair of equations:

x > — 5 and y + Zx < 1

16. U sing set symbols, rewrite the description of each of the following loci:(a) T he locus of points that are 3 units to the left of the vertical axis.(b ) T h e locus of points th a t are m ore th an 5 units above the horizontal

axis.(c) T he locus of points that are m ore than 2 units from the line x = 7 .(d ) T he locus of points that are a distance of 2 units from the point (2, 6).(e ) T he locus of points that are less th an 4 units from the origin.(f) T he locus of points that are equidistant from the points ( — 5, 4)

and (7, 8).(g) T he locus of points such that the difference of the squares of its

distances from any point of th e locus to th e points ( —3, —2) and (1, 6) is always 3.


17. (a) Draw the locus of points that are 5 units from the vertical axis andequidistant from the points ( — 4, 1) and (6, —5).

(b) Draw the locus of points that are less than 5 units from the origin . and 2 units from the horizontal axis.

18. Determine the equation of the circle under each of the following con

ditions.(a) center (1, —4); radius 2.(b) center ( i, —3); passes through the origin.

19. Determine the center and radius of the circle whose equation is

x 1 + y l — Ax + 6y — 36 = 0

20. D raw the locus of points th a t are m ore than 5 units arid less than 10 units from the origin,

■ Try This For FunW hen light strikes a shiny surface, such as a m irro r, it is

reflected from that surface in a way so th a t the angle a t w hich it makes contact with the surface is congruent to the angle a t w hich i t leaves the

surface. In the drawing below, the light ray BA m ade contact w ith the m irror

forming the ZX. W hen it bounced off the mirror, the light ray BC form ed ' the Z 2 with the surface such that Z 2 = Z l.

This simple piece of information can help us find the height of an object whose base is inaccessible to us. Can you prove th a t the height of the flag pole below is

h -A Bx A C - D B

14The Circle

MANY O F T H E PR O PER TIES O F A C IR C L E CAN be developed far more easily through the processes of synthetic geometry ra th e r than coordinate geom etry. Hence, to formulate these principles'w V are going to return to the m ethods of proof used earlier in the course.

As usual, it will be necessary to define our terms before proceeding.

D efin itio n 75: A chord of a circle is a line segm ent w hose endpoin ts are tw o points of th e circle.

In Figure 14-1 the line segment CD is a chord of circle 0 , Segment A B is also a chord of circle O', however, it is a special chord since it passes through the center of the circle. W hat is its name?

450

THE CIRCLE 451

D e f i n it io n 76: A diam eter of a circle is a chord such that one of its pointsis the center of the circle. ......

D efin itio n 77: A central angle of a circle is an angle whose vertex is the center of the circle.

In Figure 14-2 /.E O F is a central angle of •'■he circle 0 . By having an understanding of the central angle at' our disposal, it is possible to define an arc of a circle.

D e f in it io n 78: A minor arc A B of circle 0 where A and B are points of that circle is the union of the points A and B and the points of the circle in the interior of central angle AOB.

In Figure 14-3 the points m arked in red are the points of the m inor arc AB of circle 0 . Each of these points are in the interior of /A O B and each is a point of circle 0 . In addition, points A and B are also points of m inor arc AB.

D e f in it io n 79: A m ajor arc A B of a circle 0 where A and B are points of that circle is the union of the points A and B and the points of the circle in the exterior of central angliA O B .

In Figure 14-3 the points of circle 0 that are draw n in black will be the points of m ajor arc AB. Each of these points is in the exterior of /.A O B and each is on circle 0 . Points A and B themselves are points of the arc, too.

D e f in it io n 80: A semicircle AB of a circle 0 where A and B are endpoints of a diam eter of this circle is the union of the points A and B and the

i—►points on the circle in the half plane on one side of AB.

All three of these terms—the minor arc, the m ajor atc,..and the semicircle— are referred to as arcs of a circle. T he symbol is usea to represent i

the word arc. Confusion immediately arises, for by-E F (read as arc EF). in Figure 14-2, do we m ean the minor arc or m ajor arc? Henceforth,‘"should

we refer to the .EF, we will be speaking of the m inor arc E F , not the ma jor arc. Should it be necessary to call attention to the m ajor arc EF, this will be^domTHy either referring to it as major arc E F or by nam ing another point

on the circle and calling the arc, EPF (see Figure 14-2). T here are times when more than two letters are employed to nam e a m inor arc also. T hus,

Figure 14-4, Figure 14-5. Figure 14-6.

452 THE CIRCLE

in Figure 14-1 AD m ay be called ACD, or CB may be called CDB. T h e letters that appear in the first and last positions are the names of the endpoints of the arc.

In Figure 14-4 A B is said to be the intercepted arc of / .A O B . In Fig

ure 14-5 CE is the intercepted arc nf ZC D E; and in Figure 14-6 GH is the intercepted arc of /F G H . Thus,

D e f in i t io n 81: A n a r c o f a c i r c le i n t e r c e p t e d b y a n a n g l e is a n a r c s u c h t h a t

e a c h s id e o f t h e a n g le c o n t a in s a t le a s t o n e e n d p o i n t o f t h e a r c a n d a l l

p o in t s o f t h e a r c o t h e r t h a n t h e e n d p o i n t s a r e in t h e i n t e r io r o f t h e

? ^ . a n g le .

In Figure 14-4 endpoint A of A B fell on side OA, while endpoint B felloil side OB. I t is possible for one endpoint to fall on both sides of the angle,

for in Figure 14-6 point G is a point of both GF and GH.A one-to-one correspondence exists between th e chords of a circle and

the minor arcs of th a t circle. Thus, in Figure 14-7 there is only one m inor

F ig u r e 14-7.

arc th a t bears the nam e CD and bu t one chord CD. Similarly, there is bu t

one A B and one chord whose endpoints are A and B. Hence, we refer to

T B as corresponding to. y lS ^an d , conversely,M B corresponds to ~AB. )" Tn order to maTce it possible to relate the m easure of an arc to thejn easure of the central angle th a t intercepted th a t arc, the following definition was devised :

D e f in i t io n 82: (1) T he measure of a m inor arc or a semicircle is the m easure of the central ang le th a t intercepts th a t arc.

(2) T he measure of a m ajor arc is 360 minus the m easure of the m inor arc having the same endpoints as the m ajor arc.

Thus, if in Figure 14-8 m /.A O B was 50, .the^measure of A B (written

as m AB)t would also be 50. In Figure 14-9, where central angle 1 is a straight

angle, the m ACB is 180. Similarly, m BD A is also 180. Hence, it appears th a t the arc consisting of the entire circle 0 will have a measure of 360.

THE CIRCLE 453

F ig u re 14-8. F ig u r e 14-9.

Since w e did not admit to the existence of angles greater than a straight angle in our work, it was necessary to define the measure of a m ajor airc as we did~ for no central angle less than a straight angle could intercept a

m ajor arc. Hence, to find in Figure 14-10 the m easure of ACB we would

first determine the measure of A B and subtract this value from 360, which is the measure of the entire circle. Therefore, if m /A O B . in Figure 14-10

is 70, then the m A B = 70 and m ACB = 360 — 70, or 290.Earlier it had been pointed out that m /A O B = 70 could be expressed

as the statement that /A O B was an angle.of 70 degrees where the degree was the unit of measure for the size of an angle. In the same way, if

/—S /—sm AB — 70, then A B is said to be an arc of 70 degrees. However, to distinguish the unit of measure for the angle from th a t of the arc, we call the former the angular degree, while the latter is the arc degree.

O ther than the fact that the angular degree and the arc degree are units of measure for two different geometric figures, there is ano ther very, very important difference between them. W hereas, by definition two angles having equal measures a re congruent, this relationship will generally no t be the case for arcs having equal measures! Consider Figure 14-11, I f /A O B

was an angle of 60 angular degrees, this would imply th a t both A B and <?B are arcs of 60 arc degrees. Obviously, to say th a t these arcs a re congruent

would stretch the imagination excessively! T h e statem ent th a t A B is an arc of 60 arc degrees implies merely th a t it is -jYtri or l i of its circle in size.

454 THE CIRCLE

In the same way CD is § of the size of its own circle. And although equal in the num ber of arc degrees in their measures, the two arcs are far from equal

in inches, or feet, or any linear unit. However, were both the m easure of A B and the m easure of RS in circle 0 equal to 60. it would seem th a t wc should say th a t A B = RS, for bcth are i the size of the same circle. First, however, it would be well to define congruent arcs before trying to prove this.

v' D efin it io n 83: Congruent circles are circles whose radii are congruent.

V D e f in i t io n 84: C ongruent arcs are arcs in the same or congruent circlesth a t have equal measures.

■V,

Now we are prepared to prove the relationship stated above.

T H E O R E M 83: If tw o c en tra l angles of a circle a re c o n g ru en t, th e n th e ir in te rc e p te d arcs a re congruen t.

Given: ZAO B ~ ZCOD in O 0 Concl.: A B ^ C D

Figure 14-12.


1. Z A O B ^ ZCOD in O 0 1. Given2. m lZA O B = m ZCOD 2. Def. of congruent angles

3., m A B = m Z A O B 3. Def. of the measure of an arc

4. pi CD =* m ZCOD 4. Same as 3

5. m A B = m CD 5. If two numbers are equal to twoequal numbers, then they are equalto each other.

6. A b =~ CD 6. Two arcs of equal measures are congruent arcs. (Reverse of definition ofcongruent arcs)

T H E O R E M 84: I f two arcs o f a circle a re congruen t, th en th e cen tra l ang les in te rce p tin g these arcs a re congruen t.

c

THE CIRCLE

Given: AB — CD in O 0 Concl.: /A O B = /.COD

455

PRO O F (The reasons will be left for yon to supply.)

1. A B = CD in O 0

2. m A B = m CD

3. m Z A O B = m AB

4. m ZCOD = m CD

5. m Z A O B — m ZCOD

6. Z A O B ^ ZCO D

TH E O R E M 85: If in a circle two chords a re co n g ru en t, th e ir co rre sponding arcs a re congruen t.

Given: AB SS CD in O 0

Concl.: T b ^ Gd


1. Let OA be the line through points 0 and A. T h e same

for OB, OC, and OD.

2. A S £ f f i (s)2. CM. = UC(s)

4. O B s t U l 5 CO'5. A 0 A B 9 1 & 0 C D6. Z A O B = ZCOD

7. A B = CD


2. Given3., T h e rsd ii of a circle a re congruent.

4. Same as 35. S.S.S.6. Def. of =. polygons

7. T heorem 83

456 THE CIRCLE

T H E O R E M 86: I f in a c irc le two arcs a re co n g ru en t, th e ir co rre sp o n d in g chords a re co n g ru en t.

P R O O F

T he proof of this theorem is very m uch the same as th a t for T heorem 85. I t will be left for you to do.

Theorem s 83 and 84 show the relationship that exists between central angles and their intercepted arcs. O n the o ther hand, Theorems 85 an d 86 point up how chords are related to their corresponding arcs. In sum m ary,

A. T o prove arcs of a B. T o prove chords of C. T o prove central a circle congruent:circle congruent:

(1) Prove arcs congruent

angles of a circle congruen t:

(1) Prove arcs congruent

(1) Prove chords congruent

or(2) Prove central

angles congruent

T h e definitions of such terms as “ bisector of an arc” or ‘‘m idpoint of an a rc” aTe similar to these definitions w ith reference to a line segment. In the work that follows, an understanding of these definitions is assumed.

Illustration:

Given: C m idpoint of BD in G O

Concl.: AD II OC

PR O O F (T he reasons will be left for you to supply,)'

1. C m idpoint of BD

2 . £ c s * 6 b3. Z l S / 24. m ZD O B = m Z 1 + m Z 2

5. UA S HD6 . Z A = Z D

7. m Z D O B — m Z A + m Z D8. m Z A + m Z D = m Z \ + m Z 29. m Z A + m £ II 3 + m ZZ

or 2m Z A = 2m Z l10. Z A ^ Z l

11. AD j| OC

THE CIRCLE

EXERCISES

A

t Given: /\A B C is isosceles

with AB = ~A^

Concl.: A B C ^ ACB

3 , G iven: / A = £ D

Concl .- .A B C & D C B

5 . G iven: AB == Concl.: Z B == Z C

Given: BAD = CD A

C oncl.: AB ■= CZ)

Given: OO w ith A B ~ CD

C oncl.: Z A — Z B

G iven: O 0 w ith C the m id

point of A B

Concl.: OC _L AB

458 THE CIRCLE

7 . Given: 0 0 with diam eter ABZC BA ^ Z D S A

C oncl.: CA =. DA

9. Given: O O with B B a diam.

A B II CO Concl.: C is the m idpoint

of AD.

11. Given: P m idpt, of A B in OO

Q m idpt. of CD

Concl.: A B II CD

G iven: Circles A and B with

AB the line drawn between the. centers

Cone!.: E m idpt. of CED

F m idpt. of CFD

8.

G iven: P m idpt. of APB

Q m idpt. of AQB

Concl.: PQ ± bi. A B

Given: 0 0 with AB || CD

P m idpt. o fA B

Concl.: Q m idpt. of CD

10.

12 .

THE CIRCLE 459

13. Given: AD = BCConcl.: {\ED C is isosceles.

15. Given: Circle 0C is the m idpoint

of AB.D is the midpoint

of I B .4—►

C oncl.: D lies on OC.

Given: OO with D the mid- 14.point of A BC is the m idpoint of

AB.4—>

Concl.: OD passes through C.(H int: See page 184.)

Given: Circles A and B P is the m idpoint

of CPD.Q is the m idpoint

of “CQD.

Concl.: QP passes through A.

16.

E1. I f the vertices of an equilateral triangle lie on a circle, they will divide

the circle into three congruent arcs.2. I f the vertices of a parallelogram He on a circle, then the diagonals

will be congruent.

460 THE CIRCLE

3. A diam eter of a circle bisects the circle, f4. T he line join ing the m idpoint of a chord to the m idpoint of its cor

responding arc is perpendicular to the chord.5. From the outer endpoint of a radius perpendicular segments are

draw n to two other radii. If the perpendicular segments are congruent; then the two arcs cut off by the three radii are congruent.

6. A central angle intercepts an arc on the circle. If from the m idpoint of the arc perpendicular segments are drawn to the sides of the angle, then these segments are congruent.

7. A radius perpendicular to a chord bisects the chord and its corresponding arc. f

8 . If the measures of two central angles of a circle are unequal, the meas- • ures of their intercepted arcs will be unequal. (H int: Use the indirect proof.)

9. If two chords of a circle are not congruent, their corresponding arcs will not be congruent.

10. If two arcs of a circle are not congruent, their corresponding chords will not be congruent.

11. If from a point on a circle two chords are drawn so as to make congruen t angles w ith the radius drawn to this point, then the arcs corresponding to the chords will be congruent.

12. T h e line joining the midpoint of a chord to the m idpoint of its corresponding arc passes through the center of the circle.

13. If two congruent chords intersect w ithin a circle, then the segments of one will be congruent to the corresponding segments of the other.

14. If two circles intersect, the perpendicular bisector of the chord th a t is common to the two circles passes through the centers of the two circles.

■ Chords Equ idistant from the Center of a CircleThus fa r we have but one m ethod for showing chords of a

circle to be congruent. A second method, to be presented in this unit, will depend upon our ability to show th a t the chords of the circle are equid istant from the center of the circle. As you recall, the distance from a point to a line is the measure of the perpendicular line segment drawn from th a t point to the line.J Thus, to say that point P is equidistant from lines a and b is to imply three things:

(1) P R ± a (2) PS l b (3) T K ^ P S

Similarly, were we asked to prove that P is equidistant from lines a and b, we would know that

t T hese problem s often ap p e a r as theorem s.} See p ag e 378.

CHORDS EQUIDISTANT FROM THE CENTER OF A CIRCLE 461

(1 ) PR Jl a a n d (2 ) PS 1 b

w h ile th e conclusion w ould be.

■ P R ^ T S

Figure 14-15.

T H E O R E M 87: If two chords are equidistant from the center of a circle, they are congruent.

G iven: G 0 w ith O E — OF

OE J. A B ...

OF L C D Concl.: T B ^ C D

F igure 14-16.

A n a l y s is ; By proving A O B E ~ A ODFt we can show B E == DF. In the sam e way A OAE can be shown congruent to A O ’CF, and hence, EA ~ FC. T h en by applying the addition postulate it will follow th a t A B = CD.


1. L et OB be the line through points 0 and B.

2. Sam e for OA, OD, and OC

3 . U E S Z U P (.1)

4. O E _L AB5. Z .0E B is a right angle.

7. AOFD is a right angle.

8 , 'O B ^ W D ( h )9. A O B E — A O D F

10. ~bE szEfIn the same way, A O A E can. be shown congruent to A OCF.

11. ~EA = FC

462 THE CIRCLE

The converse of Theorem 87 is also quite important. Its proof, however, is dependent upon the following statement.

TH EO R EM 88: A rad iiis p e rp en d icu la r to a chord bisects th e chord and its co rresp o n d in g arc.

PRO OF

Y o u w e r e a s k e d to p r o v e t h i s r e l a t i o n in P r o b l e m 7 ,

p a g e 4 6 0 ; h e n c e , i t s p r o o f w i l l n o t b e g i v e n n o w .

T H E O R E M 89: If two chords of a circle a re co n g ru en t, th ey a re eq u id istan t from th e cen ter of th a t circle.

Given: OO with/4.B S C£>

O E ± ? B

OF ± C D Concl.: m ^ O F

A n a l y s is : By using Theorem 8 8 and the postulate that halves of congruent segments are congruent, it can be shown that ~BE and E F are congruent. W ith this piece of information it will follow that A O £ E = A O D F , and, hence, OE = UF.

PRO OF (The reasons will be left for you to supply.)

4—V1. Lei OB be the line through 7. But, A B ^ U D

points 0 and B. 8. "BE £= U F (I)

2. Same for OD 9. /O E B is a right angle,10. /O F D is a right angle.

3. OE 1 AB *-► __ 11. OB = UZ) (h)

4. O E bisects AB. (Theorem 88) 12. A O B E — A O D F

5. OF 1 CD 13 . U E ^ U F

*-*6. OF bisects ?3).

All principles developed in this chapter wi'th reference to a single circle will hold as well for congruent circles. As an exercise, prove Theorem s 85, 86, 87, and 89 by using two congruent circles rather than the same circle.


EXERCISES

1. Given: OO with OC _L AE AB = D E

Concl.: A O B D is isosceles.

3 . Given: OO with AD _L BC C oncl.: A A B C is isosceles.

5. Given: OO with AO X BO <-» <-»CO 1 AB

Concl.: A OCB is isosceles.

D

Given: O 0 w ith OC _L AB

Concl.: CO bisects /A C B .

BCG iven: O 0 w ith BA

OD X A B

OE _L CB

Concl.: B ? bisects /A B C .

Given: ABCD is an isos. trap. 6.with A B = DC in OO.

C oncl.: AC and BD are equidistant from 0 .

464 THE CIRCLE

7 . G iven : O 0 with A B =

OD J_ AB

O E _L AC C oncl.: A A D E is isosceles.

A

9. G iven: O O with OA — OB

OA ± CD

O B 1 EF

C oncl.: 6 e S FD

11, Given: 0 0 with AC S BD

C o n c l.: A E — BF

G iven: Q 0 with E F = CD

OA ± CD

OB ± EF Concl.: Z O A B £ * ZODA

8 .

Given: OO with A B — DC 10. OE ± P A

OF 1 P D

C oncl.: PO bisects ZAPD.

Given: 0 0 with AB || CD

A B = ( IS P m idpt. of A S

Concl.: UP £* (TiJ

12.


13. Given: 0 0 with U E = D F

o i i c i <-> <->

OB 1 D F

Concl.: GO bisects ZCGD.

15. Given: 0 0 w ith/4C II D B

AOB is a diam eter.

Concl.: A S — D B (H int: Let

OP be the perpendicu-

lar from 0 to AC, and

OR be the perpendicu

lar from f? to D B .)

C

xyz------------ ,-----------_ |b°

G iven: Point 0 is the center 14. 'of both circles.Line I intersects the circles in points A,B, C, and D.

Concl.: A B ^ V D (H int:

Let OP be th e perpendicular from 0 t o !.)

Given: O O v /i th A B S lD C 16.*Concl.: P E ^ F U (H int:

Let OR be the perpendicular from 0

to PB, arid OQ be the perpendicular

from 0 to PC.)

466 THE CIRCLE

B1. The vertices of a triangle lie on a circle. If the sides of the triangle are

equidistant from the center of the circle, then the triangle is equilateral.2. If from opposite endpoints cf a diam eter chords are drawn so as to form

congruent angles with the diam eter, then the arcs corresponding to these chords will be congruent.

3. Two chords intersect within a circle. If the diam eter drawn to the point of intersection bisects the angle formed by the chords, then the chords are congruent. (H int: Prove that the chords are equidistant from the center of the circle.)

4. If two congruent chords are extended until they intersect outside the circle, then the extended segments will be congruent. Assume that the chords are not segments of parallel lines.

5. If two chords of a circle are not congruent, then they are not equidistant from the center of the circle. (H int: Use the indirect proof.)

6. If two chords of a circle are not equidistant from the center of a circle, then they are not congruent.

7.* I f a diam eter bisects two chords of a circle, then the chords are parallel.8.* If a diam eter bisects one of two parallel chords of a circle, then it bisects

the other also.9. * T he midpoints of congruent chords of a circle will lie on the same circle.

■ Tangents and SecantsT here are but two more lines that we need to consider in

connection with the properties of a circle.

D e f in it io n 85: A t a n g e n t t o a c i r c le is a l in e t h a t h a s b u t o n e p o in t i n

c o m m o n w i th t h e c i r c le .

D e f in it io n 86: A secant to a circle is a line that has two distinct points in common with a circle.


TANGENTS AND SECANTS 467

In Figure 14-18 the_ pointy P that is com rarm -tn .jê circle and the tangent is called th^ftomt of tangenqôr the(paint of contact.:Should we think of this in terms of our understanding of sets, we can consider the elements in set 0 as the points on circle 0 . Similarly, the elements in set I are the points on line I. Thus, if 0 f i ^—that is, the intersection of 0 and I—is but one element, then I is a tangent*to the circle. However, should 0 I be two elements as in Figure 14-19, then the line I will be a secant. W hat if the intersection of 0 and I be the null set; that is, contain no elements? This would imply that the line had no points in common with the circle and, hence, would have no connection to it.

D e f in it io n 87: Tangentîrcles are two circles th r t are tangent to the same line at the same point on that line.

The two circles in both Figures 14-20 and 14-21 are tangent circles. Should the line segment that joins the centers of the two circles intersect the common jtajigent lLne as it does in .F igure 14-20, then the circles are said to bejftangent externally. $.1 it does not, as in Figure 14-21, then the circles a r t,tangent internally. ;

Two circles can be tangent to the same line bu t not a t the same point on that line. This is the situation that exists in each of the drawings below. In Figure 14-22, there are four lines th a t are’scommon langenir*o circles A and B. The line A B joining the two centers is'called the ( ine of centers?-li the common tangents intersect line segment A B , as they do in Figure 14-22, they are said to be common internal tangents. If they do not intersect segment AB, such as I and m, they are common external tangents. .

468 THE CIRCLE

Notice that in the sequence of diagrams from Figure 14-22 through Figure 14-26, the center B is m ade to move along the line of centers and approach center A. As this occurs, the num ber of common tangents decrease from a m axim um of four to none a t all. In Figure 14-25 is the single common tangent a common internal tangent o r & common external tan -

> gent?

Exam ination of Figure 14-20, page 467, suggests that there m ust be some relationship of perpendicularity existing between the tangent to a circle and the radius drawn to the point of contact of that tangent. T o simplify the proofs of several theorems in connection with this relation, we will assume th e following:

\j P o s t u l a t e 39:. At a gi’--:n point on a circle there exists one and only one tangent to the circle.

T h e proof of each of the next three theorems expressing a relationship between tangents and radii is dependent upon the indirect, approach.

THEOREM 90; A lin e perpendicular to a radius at its outer endpoint is a tangent to the circle.

L

A Given: Circle 0 with P the ou ter end-O ) point of radius OP

J O P J L ly Cone!.: / is a tangent to the circle.

?Figure 14-27.

A n a l y s is : To show that I is a tangent to the circle, we must show th a t it has but one point in common with the circle. Since it already has point P in common with the circle, then our problem narrows to showing th a t I cannot have another point in common with the circle.


PRO O FBy the law of the excluded middle one of the following

statem ents is true and no o ther possibility exists:

P is the only point that I has in common with circle 0

or

P is not the only point that / has in common with circle 0 .

Let us accept the possibility that P is not the only point th a t I has in com-

mon with the circle; then let Q be another of these points. Also, let OQ be the line that passes through points 0 and Q. Since the radii of a circle

are congruent, O Q ~ O F . From the Given D ata we know th a t OP X hence, Z l and Z 2 are right angles and, therefore, are congruent. However, since UQ S— <57, then Z l = Z2. Thus, we have both Z l and Z l congruent to Z 2 and, therefore, /.X = Z3. However, we have a theorem th a t states th a t an exterior angle of a triangle m ust be greater than either of the remote interior angles. This would imply that m Z l > m Z l . Therefore, accepting the possibility th a t P is not the only point that I has in comm on w ith circle 0 led to the logical inconsistency of the tru th of Z l — Z l and Z l Z l . By the law of contradiction both cannot be true a t the same time. Since ZX Z l must be true, for it is the result of a theorem , then Z l ^ Z l must be false. Therefore the statem ent that P is not the only point th a t I has in common with circle 0 is also false. Hence, the stateihent th a t P i s the only.point that I has in common with 0 is true, for it is the only rem aining

possibility.

470 THE CIRCLE

T H E O R E M 91: A ra d iu s d ra w n to th e p o in t o f contact of a tan g e n t is p e rp e n d ic u la r to the tang en t.

Given: 0 0 with I tangent to the circle at point P

Concl.: OP ± I

F ig u r e 14-29.

PR O O F

By the law of the excluded m iddle one of the following statem ents is true and no o ther possibility exists:

OP JL I o r OP jL I

L et us accept the possibility th a t OP Y_l\ then at point P let m be the line

th a t is perpendicular to OP. By the theorem just proved, m must be tangent to the circle a t P, for if a line is perpendicular to a radius at its outer end

point, it is a tangent to the circle. However, we know from the Given D ata th a t I is also tangent to the circle at point P. Hence, there exists two tangents to circle 0 at point P, the tangents I and m. Postulate 39, however, states th a t there can exist one and only one tangent to a circle a t a given point

<—¥ jon th a t circle. Hence, accepting the possibility that OP / I led to the logical inconsistency of the tru th of a statem ent and its contradictory, By the law of contradiction both cannot be true at the same time. Since we have accepted the postulate as true, then the statem ent that I and m are both

tangents' to circle 0 at point P must be false, and, therefore, OP I is ■


also false. Hence, OP 1 I must be true, for it is the only remaining possibility.

T H E O R E M 92: A l i n e p e rp en d icu la r to a tan g en t a t its p o in t of contact w ith a circle passes th ro u g h th e c en te r of th e circle .

PR O O FBy applying the indirect proof and Theorem 91, this state

m ent can be proved. T he proof is left for you to do.

T he most im portant theorem concerning tangents to a circle is by far the very easiest to prove. Before doing so, though, there are several terms th a t will be clarified.

D efin it io n 88: A ta n g e n t segm ent from an ex te rn a l p o in t to a circle is th e line segm ent whose endpoints a re th e ex te rn a l p o in t and th e p o in t o f c o n tac t of th e tan g en t to the circle.

In Figure 14-31 the tangent segment from point P to the circle is the line segment PA on the line I where / is tangent to the circle a t point A. How m any tangent segments can be drawn to the circle from the point P? W hat do you think will be true concerning these two tangent segments? W hat do you believe can be said of the ray whose endpoint is P and that passes through 0? How will this ray be related to the chord that joins the two points of contact of the tangent segments from P?


D efin itio n 89: A secant segment from an external point to a circle is the line segment whose endpoints are the external point and the poirit furtherm ost from the external point that the secant has in common w ith the circle.

In Figure 14-32 the secant segment from point R to the circle is the line segment R b on the secant m. W hat can be said concerning the longest

. secant that can be d raw n from point R to the circle?

T H E O R E M 93: If two tan g en t segm ents a re d ra w n to a circle from a n ex te rn a l po in t, th en these segm ents a re congruen t.

472 THE CIRCLE

O iv PA tangent to GO at A

PB tangent to GO at B

Concl.: PA S FB

PR O O F (The reasons will be left for you to supply.)

1. L et OP be the line through 0 and P.

2. Same for OA and OB

3. (JA = UB (I)

4. U F ^ Z T F (A )

5. OA X AP (Theorem 91)

6. / OAP is a right angle.

7. OB 1 BP8. AO BP is a right angle.9. A O A P = A O B P

10. Fa s i FB

EXERCISES

1 , G iven: Circles A and B tan-

gent to PD at D

FU and FE are tangent segments.

Concl.: F V S lF E

Given: GO with PA and

PB tangent a t A and B respectively

C oncl.: OP bisects /.AOB.

TANGENTS AND SECANTS

3. Given: OO w ith PA and P B j G iven: Circles A a n d B tan- tangent a t A and B *-*respectively

Concl.: PO L A B

A

4.

473

gent to PD a t D

PE tangent to A at h.

P F tangent to B a t F

Concl.: F F S Z F E

5 . Given: CD common internal tangent to circles A and B a t C and D

C oncl.: / A =■ / B

7 . G iven: OO = O QI S a n d 'C S are com ' m on external tangent segments.

Concl.: A g£S5Z5

Given: PC and PD are tan gent to circles A 2nd B a t C, E, F, D respectively

Concl.: C E ^ D F

Given: PC and ~FD a re tangent 8 . segments.

A B tangen t a t E Concl.: PA + PB + AB =

PC + PD

474 THE CIRCLE

9. Given: Two circles with the same center, point 0 . (They are called ccn- centric circles.)<-»AB tangent to inner circle at C

Concl.: C is the m idpoint

of AB.

tangent to G O a t B, E, and C respectively.

Concl.: AD — A B + DC

to CD a t P «-»

PR 1 CD

C oncl.: PR passes through A and B.\ C

P

Given: Two concentric circles with center 0<—V <—►A B and CD tangent to inner circle a t E and F respectively

Concl.: AB ~ CD

G iven: T h e sides of quadri- 12. lateral ABCD are tangent to GO.

Concl.: AB + CD =AD + BC

Given: © A and B tangent 14. to CD a t PH

C oncl.: PA passes through B.


1 5 . G iven: T A and PC are tangent segments toGO.<-> <->EB 1 PA

E D I PC

Concl.: EB — ED

17.* Given: T he sides of A A B Care tangent to OO a t P, Q, and R.

A B S iA C <->

C oncl.: A lies on PO.

S P C

19.* Given: AD diam of 0 0

BC tangent to 0 a t P

AB and DC _L BC Concl.: AD = AB + DC

(H int: See Problem 12c, page 405.)

Given: O 0 with AC, AB, 16.*and BC tangents a t E, F , and D respectively.Z C is a right angle.

Concl.: AB = AC + BC —2 OD

G iven: BOC is a diam . 18.* in G O .

PA tangent to 0 0

AB || PO

Concl.: PC tangent to 0 0

Given: CD tangen t to cir- 20 . cles A and B. a t C and D

Concl.: AC'.BD — A E'.EB (H int: Prove A ACE ~ A BDE.).

476 THE CIRCLE

21. G iven: PB tangent to circles C and D a t B and A respectively.

Concl.: PC'.PD = BC :AD

G iven: PA and PB are tan- 22/ gent segments to circle 0 .

Concl.: (P A)2 = PO ■ PC

23. Using the diagram at the right, find the length of the tangent segment PA.

i24. Using the method illustrated above, find the length of the tangent

segment from the point given to each of the following circles:(a) (* - 2Y + (y - 3Y = 25, P (l, 6)(b) (* + 1)* + C - 2)» = 16, P(4, - 4 )(c) (* + 3 ) ' + 0 - + n * = 9 , i B( ~ 3 ,4 )(d ) x>~ + 1 0 ,P (5 ,7 )

" (e ) xJ + f + Zx + Ay - 20 = 0, P(10, 5)(f ) x* + / - <5* + Zy - 14 = 0, P(6, - 9 )

(g ) x1 — — 9 = 0, P(5, 4). (W hat can be said of point Pin this problem?)

25. (a ) W hat are the coordinates of the center of the circle, whoseequation is (x — 3)J + (y — 5)! = 25.

(b ) Show that the point (7, 2) is a point of this circle.(c) How will the tangent to the circle a t this point be related to the

radius to this point?

. (d ) W hat is the slope of the radius to the point (7, 2)?(e) W hat is the slope of the tangent to the circle a t the point (7, 2)?(f) W hat is the equation of the tangent to this circle a t the point

(7 ,2)?

THE SPHERE 477

(g) W hat is the equation of the tangent to this circle a t the point ( - 1, 2)?

UJ1. If two tangent segments are drawn to a circle from an external point,

then the line passing through that point and the center of the circle bisects the angle formed by the tangent segments.

2. If two circles are tangent externally, then the comm on in te rnal tan gent bisects-the common external tangent segment.

3. I f two circles are congruent, the line of centers bisects e ither common internal segment.

4. T he common internal tangent segments of two circles are congruent.5. I f two lines are tangent to a circle a t the endpoints of a diam eter,

then they are parallel.6. T he point of intersection of the line segment join ing the centers of

two noncongruent circles with a common internal tangent segment will divide the two segments into equal ratios.

7. T he common internal tangents of two noncongruent circles intersect a t a point such th a t the ratio of the segments of one is equal to the ratio of the corresponding segments of the other.

8. If the common internal tangents are drawn to two noncongruent circles, then the secant passing through the points of contact of the first circle is parallel to the secant passing through the points of contact of the second circle.

9. * If the sides of a triangle are tangent to a circle, then the perpendiculars to the sides at their points of contact with the circle will be concurrent; th a t is, meet a t a point.

10. * T he line of centers of two noncongruent circles will pass through the point of intersection of their common external tangent.

»*■

■ The Sphere

The circle in plane geometry has its counterpart in the sphere in space geometry. As we move through this unit, notice the sim ilarity th a t exists between the principles developed here and those developed earlier in this chapter,

478t h e C IR C L E

' f D e f in it io n 90: A sphere is a closed surface such th a t each point on this surface is a fixed distance from a fixed point, (Com pare this w ith the definition of a circle.)

As in the case of the circle, the fixed distar.cc is the rstHus oj the sfheie, while the fixed point is the center oj the sphere.

T H E O R E M 94: I f a p la n e a n d a sp h e re have m ore th a n one p o in t in comm on, these p o in ts w i l l l ie on a circle.

Given: Sphere 0

Plane a intersects 0 such that two of the points they have in common are A and B.

Concl.: All points th a t a and 0 have in common lie on a circle.

A n a l y s is : We will show th a t poin t B and point A are each the sam e fixed distance from P. Since they represent any two points that a and 0 have in common, then all points in com m on to the two will be th a t same fixed distance from P. Hence, by the reverse of the definition of a circle these points will lie on a circle.

PRO O F I STATEMENTS

1. Sphere 0 with point B, a point that plane a and sphere 0 have in common.

2. Let OP be the perpendicular from pojpt 0 to a.

3. Let PB be the line through

J3 and £ ; let OB be the line through 0 and B

4. OP X PB

5' /-OPB is a right angle.6 . A O P B is a right triangle.7. (s p y = (o b ) 1 - (o p y

In the same m anner, had any selected, it could be shown th a t

REASONS

2. From a given point not on a given plane there can be no m ore than one line perpendicular to th e given plane. (See Problem 2, page 288.)

3. There exists one and only one line through two points.

4. Def. of a line perpendicular to a a plane

5. Def. of perpendicular lines6. Rev. of def. of a right triangle7. T heo rem of Pythagoras

other point A com m on to a and 0 been

ip-r

THE SPHERE 4798. (AP)1 = (OA)1 - (OP)19. OA = OB

10. .■.(AP)2 = (OB)1- ( O P ) 1

8. Same as 79. Def. of a sphere


Thus, each point that a and 0 have in common can be shown to be a

distance of V (O B )1 — (OP)1 from P. Hence, by the reverse of the definition of a circle the points common to the two fall on a circle.

If we analyze the previous proof, we will notice th a t point P is the center of the circle on which the points of intersection lie. Thus, by this proof we have shown also that

TH EO R EM 95: I f a p lane intersects a sphere , th e p e rp en d icu la r d raw n from the center of the sphere to th e p lan e w ill pass th ro u g h the center of th e circle on w h ich the po in ts o f in tersection lie.

Now let us turn to tangency as related to a sphere.

D e f in it io n 91: A tangent plane to a sphere is a plane th a t has one and only one point in common with the sphere. (Com pare this definition with th a t of a line tangent to a circle.)

As before, the point that the tangent plane has in common with the sphere is called the point oj tangency, or the point oj contact. Since a sphere is considered as a set of points, 0 , and a plane as a set of points, p, then the definition of a tangent plane implies that 0 C) p will have but one element, the point of tangency. Similarly, if 0 and p have more than one point in • common, then according to Theorem 94 0 p will be a set of points th a t lie on a circle.

P o s t u l a t e 40: At a given point on a sphere there exists one and only one plane that is tangent to the sphere. (Com pact this w ith Postulate 39.)

T H E O R E M 96: A p lan e p e rp en d icu la r to a ra d iu s o f a sp h ere a t its o u ter endpo in t is tan g en t to th e sphere .

G iven: Sphere O w ith P the outer endpoint of radius OP.

OP ± aConcl.: Plane a is tangent to the sphere.

Figure 14-35.

A n a l y s is : T he analysis and proof of this theorem arc almost identical to that used to prove Theorem 90. The difference arises only in the fact that, the word sphere replaces the word circle, and plane replaces line.

480 THE CIRCLE

PR O O F

Bv the law of the excluded m iddle one of the following statem ents io true and no o ther possibility exists:

P is the only point that a has in common with sphere 0 .or

P is not the only point that a has in common with sphere 0 .

Let us accept the possibility that P is not the only point that a has in common4—>

with 0 ; then let Q be another of these points. Also, let OQ be the line th a t passes through 0 and Q. T he two intersecting lines OP and OQ determ ine

a plane. This plane intersects plane a in the line I. OP _!_ I by definition of a line perpendicular to a plane. Hence, Z l and Z 2 are right angles. T h ere

fore, Z l == Z2. But UF = OQ by definition of a sphere. Hence, Z 3 = Z 2.

In view of this, Z l •= Z3. However, we have a theorem that an exterior angle of a triangle must be greater than either of the remote interior angles. This would imply that m Z l > m Z3. Therefore, accepting the possibility th a t P is no t the only point a has in common with sphere 0 led to the logical inconsistency of the tru th of Z l — Z3 and Z l 3= Z3. By the law of contradiction both cannot be true at the same time. Since Z l Z 3 m ust be true, for it is the' result of a theorem, then Z l = Z3 must be false. Therefore, the statem ent that P is not the only point that a has in common with sphere 0 is also false. Hence, the statem ent th a t P is the only point th a t a has in common with sphere 0 is true, for it is the only rem aining possibility.

T h e proofs of the next two theorems are very m uch the same as those for Theorem s 91 and 92. They will be left for you to do.

T H E O R E M 97: A rad iu s d raw n to the p o in t o f contact of a p la n e ta n gen t to a sp h ere is p e rp e n d ic u la r to th e p lane.

T H E O R E M 98: I f a p lan e is tangen t to a sp h e re , th en a line p e rp en d ic u la r to th e p lan e at th e p o in t o f contact w ill pass th ro u g h th e cen ter of the sphere.

THE SPHERE

EXERCISES

481

A1. Given: 0 is the center of the sphere.

C is the m idpoint of AB.

C oncl.: OC JL AB

2. Use the same diagram as in Problem 1. Given:. 0 is the center of the sphere.

OC JL A B

C oncl.: C is the m idpoint of AB.

3. G iven: 0 is the center of the sphere.P is the center of the circle.__ _ «->

Concl.: AB — A 5 (H int: D raw PB

and PC.)

4. G iven : 0 is the center of the sphere.A and B are the centers of two circles of the sphere.UA £* UB

Concl.: 0 /1 = O B (H int: See definition of congruent circles.)

5. Use the same diagram as in Problem 4.G iven : 0 is the center of the sphere.

A and B are the centers of two circles of the sphere.

Concl.: 0 A & OB

480 THE CIRCLE

P R O O F

Bv the law of the excluded m iddle one of the following statem ents ij true and no other possibility exists:

P is the only point that a has in common with sphere 0 .or

P is not the only point that a has in common with sphere 0 .

Let us accept the possibility that P is not the only point that a has in commoni—y

with 0 ; then let Q be another of these points. Also, let OQ be the line th a t passes through 0 and Q. T he two intersecting lines OP and OQ determ ine

a plane. This plane intersects plane a in the line I. OP _L I by definition of a line perpendicular to a plane. Hence, Z l and Z 2 are right angles. T h ere

fore, Z l == Z 2. But UP — 0 $ by definition of a sphere. Hence, Z 3 = Z2.

In view of this, Z l — Z3. However, we have a theorem that an exterior angle of a triangle must be greater than either of the remote interior angles. This would imply th a t m Z l > m Z3. Therefore, accepting the possibility th a t P is no t the only point a has in common with sphere 0 led to the logical inconsistency of the tru th of Z l S= Z 3 and Z l 9= Z3. By the law of contradiction both cannot be true a t the same time. Since Z l 3£ Z 3 m ust be true, for it is the result of a theorem, then Z l = Z3 must be false. Therefore, the statem ent that P is not the only point th a t a has in common with sphere 0 is also false. Hence, the statem ent th a t P is the only point th a t a has in common with sphere 0 is true, for it is the only rem aining possibility.

T h e proofs of the next two theorems are very m uch the same as those for Theorem s 91 and 92. They will be left for you to do.

T H E O R E M 97: A rad iu s d raw n to the p o in t o f contact of a p lan e ta n g en t to a sph ere is p e rp en d icu la r to th e p lane .

T H E O R E M 98: I f a p lan e is tangen t to a sp h ere , th en a lin e p e rp en d ic u la r to th e p lan e a t th e p o in t of contact w ill pass th ro u g h th e cen ter of the sphere.

THE SPHERE

EXERCISES

481

A

1. Given: 0 is the center of the sphere.

C is the m idpoint of AB.

C oncl.: OC 1 AB

2. Use the same diagram as in Problem 1. Given:. 0 is the center of the sphere.

OC ± AB

C oncl.: C is the m idpoint of AB.

3. G iven: 0 is the center of the sphere.P is the center of the circle.__ _,

Concl.: AB S A 5 (H int: D raw PB

and PC.)

4. G iven: 0 is the center of the sphere.A and B are the centers of two circles of the sphere. U A ^ U B

Concl.: 0 A = O B (H int: See definition of congruent circles.)

5. Use the same diagram as in Problem 4.G iven : 0 is the center of the sphere.

A and B are the centers of two circles of the sphere. O A ^ O B

Concl.: 0 A ~ U B

482 THE CIRCLE

1. If two chords are equidistant from the center of a sphere, then the two chords are congruent. (H int: See proof of Theorem 87.)

2. If two chords of a sphere are congruent, then they are equidistant from th e center of the sphere. (H in t: Use Problem A-2 and see the proof of Theorem 89.)

3. Tangent segments from an external point to a sphere are congruent.4. If a line is drawn from th e center of a sphere perpendicular to a tangent

plane, then it passes t 1,rough the point of tangency. (H int: Use the in direct proof.)

■ The Relation Between Angles and ArcsEarlier in this chapter we discussed th e relation th a t ex

isted between the measure of a central angle and that of its intercepted arc. In reality, the measure of every angle whose sides intercept arcs on a circle is in some way related to the measures of these arcs.

Q uite often in this section we will speak of points lying “ within a circle” o r “ outside a circle.” Although all of us have an intuitive understanding of these terms, it would be best were they defined formally.

D e f in it i o n 92: T he interior of a circle is the set of points such that i f a line was drawn through any given point of the set it would intersect the circle in two distinct points where the given point would be between . the points of intersection.

D e f in i t io n 93: T he exterior of a circle is the set of points of the plane that are neither interior points or points of the circle.

Henceforth, whenever the phrase “ inside a circle” or “ within a circle” is used, i t will refer to the set of points of the interior of a circle. Similarly, the phrase “ outside a circle” refers to the points of the exterior of the circle.

T he.vertex of an angle may lie in any one of four different positions relative to the circle: either (1) a t the center, (2) w ithin the circle, (3) on

.<*>

THE RELATION BETWEEN ANGLES AND ARCS 483

the circle, or (4) outside o f th e circle. T h e relation between the angle and its intercepted arcs will depend upon which of the four positions the vertex of the angle takes. Should the vertex fall at the cen ter of the circle, then, being a central angle, its measure will be equal to the measure of its in tercepted arc. This we know from Definition 82. I t is to the angles whose vertices fall in the remaining three positions th a t we turn o u r attention.

In the Figure 14-38 Z A B C is said to be an inscribed, angle, and it is said

to be inscribed in ABC. The endpoints o f the arc, A and C, are points o f the sides of the angle, while the vertex of the angle, point B, is a point th a t lies somewhere on the arc “ between” the endpoints A and C. Form ally, it is defined as

D efin itio n 94: An inscribed angle in an arc is an angle such th a t each endpoint of the arc is a point on each side of the angle, while the vertex of the angle is a point on the arc not coincident w ith an endpoint.

T H E O R E M 99: T h e m easure of a n in sc rib ed a n g le is equal to o n e -h a lf th e m easure o f its in te rc e p te d aro.

Given: Inscribed Z B

C oncl.: m Z B = $ m AC

Figure 14-39.

A n a l y s is : Having no more than the measure of a central angle upon which, to base our proof, v/e are forced to consider the,proof of this theorem under three different circumstances. T he first of these, obviously, m ust be in term s of a central angle. Each of the others will be in term s of the conclusions draw n in the first proof.

THE CIRCLE

G iven: Inscribed A B with side BA passing through center 0

Concl.: m A B = j m AC

Figure 14-40.

| P R O O F | STATEMENTS REASONS4—►

1. L et OC be the line through 1. Why?points 0 and C.

2. m A AO C = m AC 2. T he measure of a central angle is* equal to the measure of its in te r

cepted arc.3. m A AO C - m A B + m AC 3. The m easure of an exterior angle of

a triangle is equal to the sum of themeasures of the remote interior an gles.

4. m A B + m A C = m AC 4. Why?5 . U B ^ O U 5. Why?6. A B S i A C 6. Why?

7. m A B + m A B — m AC 7. Why?or, ■ 2 m A B = m AC

8. m A B = J m AC 8. Halves of equals are equal.

Case 2

G iven: Inscribed A A B C w ith center 0 in the interior of A A B C

Concl.: m A A B C = \ m AC



1. Let BO be the line through 1. Why?points B and 0.

2. m A A B D = \ m AD 2. Case 1

3. m AD BC = \m f> C 3. Case 1

4. m AABC = \ m AC 4. Addition postulate

Case 3

Given: Inscribed A A B C w ith center 0 in the exterior of A A B C

Concl.: m A A B C = ^ m AC

Figure 14-42.


1. Let BO be the line through 1. Why?points B and 0.

2. m A D B C = | m DC 2 . Case 1

3. m A D B A = | m.DA 3. Case 1

4. m A A B C — | m AC 4. Subtraction postulate

THEOREM 100: An angle inscribed in a semicircle i s a righ t angle.

A n a l y s is : The proof of this theorem follows directly from th e previous theorem . If an angle is inscribed in a semicircle, the arc it in tercepts will be a semicircle. Since the measure of a scmicircle is 180, the m easure of the inscribed angle intercepting this arc will be 90 and, hence, a right angle.

T he proof of Theorem 100 makes possible the proof of another theorem concerning the relation of the measure of an angle to. the m easure pf its intercepted arc. Once again the vertex falls on the circle. Now, however, oniy one of the sides of the angle is a chord of ihe circle. T he other side is a ray th a t is tangent to the circle a t the endpoint of the ray. Before tu rn ing the page, try to draw this angle.

T H E O R E M 101: T h e m easure o f an angle form ed b y a tan g e n t a n d a chord is equal to o n e-h alf the m easure of its in te rc e p te d arc.

0

<—>Given: 0 0 with BC tangent to 0 at

point B

Concl.: m /.A B C = \ m AB

486 THE CIRCLE

B C

Figure 14-43.

A n a l y s i s : T o a r r i v e a t t h e c o n c lu s io n , i t is n e c e s s a r y t o r e l a t e /A B C to

an inscribed angle that intercepts AB. This is done by drawing BO and introducing / D into the problem . O u r objective will be to prove / D — /A B C .

P R O O F STATEMENTS REASONS4—>

1. L et BO be the line through 1. Why?

points B and 0 ; let AD bethe line through A and D.

2. D A B is a semicircle. 2. Rev. of def. of a semicircle.3. /D A B is a right angle. 3. An angle inscribed in a semicircle

is a right angle.4. / D is complementary 4. The acute angles of a right triangle

to Z l. are complementary. (See Theorem53.)

<-♦5. BC is a tangent to O 0 at 5. Given

point B.<-+ <-+

6. D B 1 B.C 6. Why?7. /D B C is a right angle. 7. Why?8. / A B C is com plem entary . 8. Why?

to Z l ,9. / A B C = Z D 9. Why?

10. m / D = i m A B 10. T he measure of an inscribed angleis equal to one-half the measure ofits intercepted arc.

11. m /A B C = h m T B 11. Why?


We can turn our attention now to those situations in which the vertex of an angle falls either outside the circle or inside the circle, bu t not a t the

center.

TH EO R EM 102: I f the vertex o fa n ang le falls w ith in a circ le , th e measu re of the angle is equal to o n e-h a lf the sum of the measures of the two arcs in te rce p ted b y th e sides o f th e ang le and th a t of its vertical angle.

/ A B C whose vertex lies w ithin circle 0m / A B C — ^(m AC + m ED) ■

G iven:

Concl.

Figure 14-44.


1. Let CD be the line through 1. Why?

points C and D.2. T he m easure of an exterior angle of a

triangle is equal to the sum of the measures of the rem ote interior an

gles.

2. m /A B C = m / D + m / C -

3. m / D = $ m AC 3. Why?

4. m / C = J m ED 4. Why?

5. m /A B C — | m AC + h m &D 5. Substitution postulate

or,

m /A B C = \{m AC + m ED)

T H E O R E M 103: I f th e vertex of an a n g le falls ou tside a c irc le a n d th e sides o f th e an g le in te rce p t arcs on th e c irc le , th e n th e m easure of the ang le rs eq u al to o n e-h a lf th e d iffe ren ce of th e m easures of the arcs in te rc e p te d b y th e sides of th e ang le .

A n a l y s is : There are three different cases under which the vertex can lie . outside the circle. Although the proofs are very m uch the same, each will •

be examined separately.

488

Case 1

THE CIRCLE

Giver.: PA and T E are secant segments to circle 0■

Concl.: m Z P = %(m AB — m DC)

P R O O F | STATEMENTS

1. L et AC be the line through points A and C.

2. m Z P + m Z A = m Z 1

3. m Z P = m Z \ - m Z A

4. m Z \ = J m A B

5. m Z A = \ m DC

6. m Z P = \ m A B - \ m D C or, m Z P = J(m A B - m DC)

REASONS □1. Why?

2 . Theorem on the exterior angle of a triangle.

3. If equals arc subtracted from equals, the differences are equal.

4. Why?

5. Why?

6. Substitution postulate.

Case 2

Given: ~PA is a tangent segment to OO.Y B is a secant segment to OO.

C oncl,: m Z P = J(m — m AC)

Figure 14-46,

P R O O F (The reasons are left for you to supply.)

1. L et A C be the line through points A and C.

2. m Z P + m Z A = m Z l -3. m Z P = m Z l — m Z A

4. m Z l = J m AB

5 . m Z A = ? m A C

6. m Z P = \ m T B - \ r n X d

or, m Z P — \{rn A B — m 4C)


Case 3

G iven: PA and T B a re tangent segments to circle 0 .

Concl.: m Z P = i(m AC B — tn AB)

P R O O F

T he proof is left for you to complete.

In terms of the angles we have examined, we can sum m arize the relationship between the measure of an angle and the measures of its intercepted arcs as follows:

(1) If the vertex'of the angle lies a t the center of the circle'.T he measure of the angle is equal to the m easure of its intercepted arc.

(2) If the vertex of the angle lies within the circle bu t no t a t the center: The measure of the angle is equal to one-half the sum of the measures of its intercepted arc and th a t of its vertical angle.

(3) If the vertex of the angle lies on the circle:T he measure of the angle is equal to one-half the measure of its intercepted arc.

(4) If the vertex of the angle lies outside the circle:The measure of the angle is equal to one-half the difference of the measures of its intercepted arcs.

EXERCISES

1. Given: m DC = 100, m AD = 70,

m A B — 60

Find the measure of each of the following angles:(a) Z D AC (b ) ZB C D .(c) Z B A C(d ) ZA D C (e ) Z A C B ( f ) Z A B C + Z A D C

2. Given: Circle 0 with rr. AC = 80 <—►

D E tangent to 0 0 a t C

490THE CIRCLE

Find the measure of each of the following angles (a ) AAOC (b ) Z A B C (c) Z B A C(e ) ZE C B (f) ZO C E (g) ZOCD

3. G iven: m AB = 120, m D C — 80► •

50, FG tangent

(d) ZA C B (h ) Z ^ C #

TO Z.&4C > a t C

F CFind the measure of each of the following angles:(a ) Z B D C (b ) ZA C D (c) ZPCF(d ) ZACG (e ) Z ^jS tf (f) Z P

4. Use the diagram a t the righ t to answer each of the following questions:(a ) m Z P = 30, m DC = 40,

m <4B = ?

(b ) ot Z P — 55, 70,

m A B — ?

(c) m Z P = 50, m .4.B = 140,

m DC = ?

(d ) m Z P = 20, m A B = 70, m DC = ?

1^) m .Z l = 65, m 4.B = 85, m DC = ?

(f) m Z l = 55, m £ c = 30, m A B = ?

(g ) ct Z l = 60, m ^ 5 = 80, m Z P = ?

(h ) m Z l . = 75, m D& = 65, m Z P = ?

(i) m Z P = 25, w .DC = 3 0 , m Z l = ?

( j ) m Z P = 35, m A B = 100, m Z l = ?(k) m Z il = 20, m Z l = 7 0 , m Z P = ?

(1)* m Z l = 80, m Z P = 50, m A B = ?

5. Given: m ZC = 85, m /ID = 100,/~N

m CD = 110, PE tangent to circle at A

Find the value of each of the following:

APPLICATIONS OF THE THEOREMS 491

(b ) m Z A B C + m ZA D C

E

(a) m Z P

Given: m BC = 85, m CD = 120, m Z l = 70

E is the m idpoint of BA. Find the measure of Z P.

7. Prove: If the vertices of a quadrilateral lie on a circle, then the opposite angles are supplementary. (This statem ent often appears as a theorem.)

8. The measure of the angle formed by two tangents to a cjrcle from an external point is 40. Find the measure of each of the arcs of the circle whose endpoints are the points of tangency.

9. T he two tangents from an external point to .a .circle are perpendicular to each other. Show that the major arc whose endpoints are the points of tangency is three times as large as the m inor arc with these same endpoints.

10. Using the diagram at the right where A B is a diameter, prove that Z P ZQ.

■ Applications of the Theorems on A ngle MeasurementT he theorems that were proved in the preceding section

have wide application to m uch of the work we developed earlier in the course; in particular, the section on similarity. Before we investigate illustrations of this, there is one theorem yet to be presented that will help simplify the proofs of m any problems.

T H E O R E M 104: I f in scribed angles o f a c irc le in te rc e p t th e same a rc , th e n they a re co n g ru en t.

492 THE CIRCLE

Given: Inscribed A P, Q, and R in te rcept AB.

Concl.: Z P = ZQ =. Z R

1 P R O O F ] STATEMENTS REASONS

1. Inscribed A P, Q, and R 1. Givenintercept AB.

2. m Z P = \ m AB, 2 . Why?tn Z Q = i m AB,

tn Z R = J m A B3. Z P ~ Z Q ^ Z R 3. Rev. of def. of congruent angles

Illustration 1:

Given: In A ABC, AD is an a ltitude\o~SC.A E is a diameter.

Concl.: A B 'A C = A D 'A E

A n a l y s is : By proving A A B D ~ A A E C it will follow that A B '.A E = A H ’.AC. O u r conclusion will then be an im m ediate consequence of this proportion.

PR O O F STATEMENTS

1. AD is altitude to BC.

2. AD ± BC3. Z A D B is a right angle.

4. A E is a diameter.5. ZA C E is a right angle.

REASONS

1. Given

2. Def. of an altitude3. Def. of perpendicular lines4. Given

5. An angle inscribed in a semicircle is a righ t angle.

APPLICATIONS OF THE THEOREMS 493

6. Z A D B = ZACE 1. Z B == Z E

8 . A A B D ~ A A E C

9. A B'.AE = AD'.AC 10. A B-A C = A D -A E

Illustration 2:

6 . Why?7. I f inscribed angles of a circle in ter

cept the same arc, then they are congruent.

8. A.A. theorem on sim ilarity9. Def. of similar polygons

10. T h e product of the m eans of a p roportion is equal to the product nf the extremes.

G iven: Circles R and S tangent to B C a t P

CD tangent to S a t D

BA tangent to R a t A

Concl.: A B || CD

A n a l y s is .: By showing Z D to be congruent to Z A , A B will be parallel

to CD.

PRO O F STATEMENTS

1. Circles R and S tangent to

BC a t P

2. CD tangent to S a t D.

BA tangent to R a t A

3. m ZCPD = \m P D

m Z D = \m P D

4. Z D S * ZCPD

5. tn Z B P A = i m p A

m Z A — $ m PA6. Z A £* ZB P A7. But, ZCPD S ZB P A8. Hence, Z D Z b Z A

9. A B \\ CD

REASONS

1. Given

2. Given

3. T h e m easure of an angle form ed by a tangent and a chord is equal to one- half the m easure of its intercepted

arc.

4. Why? .

5. Sam e as 3

6. Why?7. Why?8 . Why?

9. Why?

i

1

494

EXERCISES

THE CIRCLE

1 . Given: Circle 0Concl.: AE'.ED = B E:E C

G iven: PB and PC are secant segments.

Concl.: PA'.PD = PC'.PB

3 . Given: ED _L ADO O with A B a d iam eter

Concl,: AC'.ED = A B '.B E

5 , Given: BD bisects ZABC . Concl.: A B -D C =

A E -D B

G iven: PA = F E Concl.: ~AB = U S (H in t:

Draw AC and DB.]

Given: A l i £ i B U £ i ( J B

Concl.: PB and PC trisect ZAPD.

2 .

c

T"

APPLICATIONS OF THE THEOREMSmm *7 . Given: CA tangent to circle

to AM is the m idpoint

of AB.

M S L A C

M R L A B

Concl.: M R = M S (H int:4.—)

Draw AM .)

9. Given: PA is a tangent segment.

Concl.: A B:C A = PA:PC

11. Given: AD ^ AE

D F = EG Concl.: A A B C is isos.

G iven: A A B C is isosceles with A B = AC.AC is a d iam eter of G O .D is the m idpoint of

BC. (H in t: Draw

AD.)

Concl.

8.495

G iven: C is the m idpoint 1 0 .

of AD.Concl.: A B'.EA — BC:AC

Given: OO w ith AC _L EG 12. Concl.: A E -A D = AG-A.F

13. Given: PQ tangent to circle a t C

R S i! PQ Concl.: CA'.CB = CE:CD

496

15. G iven: Circles P and Q tan-4—

gent to AC at A

A E intersects O P a t D and OQ at E.4—>

BD tangent to P a t i>

CE tangent to Q at E

C oncl.: D B II EC

17. Given: OO with tangentsegment PA

Concl.: (AC)' = PC-CB

A P

G iven: Circles P and Q tan- 14.gent to A F a t A

Cor.ci.: B C W D E

THE CIRCLE

D

G iven: Circles R and 5 tan- 16.gent to E F a t P

Concl.: C D :A B = PD'.PB

Given: T hree intersecting 18.* circles

4—►

C oncl.: E P passes through F.(H int: Use indirect

APPLICATIONS OF THE THEOREMS*19. Given: S T tangent segment

to circle F

I)A tangent segment to circle E

Concl.: (AB)1 = B C -BD

21.’

497*

Given: PED comm on external tangent to © A and B a t E and D

PAB is the line of centers.

C oncl.: CD II F E (H in t:

Draw BD and AE.)

2 2 / G iven : © R and 5 tangent to / a t P4—►

CD tangent to O R at E

Concl.: PE bisects ZCPD. (H int: Prove / I ^ Z2.)

G iven: A is the m idpoint 2 0 .

of D E.Concl.: A B -A G —

A C -AF

G iven: AC is a diam eter 23. of O E , .

is a diam eter of OF.A B is the common chord to © £ and F.

C oncl.: C, B, and D are collinear. (H int: Draw

CB and BD ; then prove th a t ZC B D is a straight angle.)

498 THE CIRCLE

i

f

fI

i-

<■

|vI

B1. If the vertices of an isosceles triangle lie on a circle, then the tangent at

the vertex of the vertex angle bisects the exterior angle a t this point.2. If from any point on a circle a perpendicular line segm ent is dropped

to a diameter, then '! e square of the measure of the perpendicular is equal to the product of the measures of the segments of the diam eter.

3. I f parallel lines intersect a circle, they will cut off congruent arcs on the circle.* T here are three cases that exist; you are to prove all three.

(a ) Case 1: The parallel lines are secants.(b) Case 2: T he parallel lines are tangents.(c) Case 3: One of the parallel lines is a tangent, while the o ther is a

secant.

4.. I f two chords intersect w ithin a circle such that one bisects the other, then the square of the measure of one of the segments of the bisectedchord is equal to the product of the measures of the segments of the otherchord.

5. A line segment is drawn through the point of tangency of two externally tangent circles term inating a t points of each of the circles. If tangents are drawn to each of the circles at these points, then the tangents are parallel.

■ Chords, Tangent Segments, and Secant SegmentsSeveral of the problems in the preceding exercises have

wide application in num erical situations. In view of this they will be established as theorems, and illustrations of when they can be applied will be presented.

T H E O R E M 105: I f two chords in tersect w ith in a c ircle , th e p ro d u c t of the m easures of the segm ents of one w ill be equal to the p ro d u c t o f th e m easures of the segm ents o f th e o ther.

Given: Chords A B and TJD intersect at E.

Concl.: A E -E B = CE-ED

Figure 14-51.

CHORDS, TANGENT SEGMENTS, AND SECANT SEGMENTS 499

1 PROOF (The reasons will be left for you to supply.)

1. Let AD be the line through 4. / A ED ZC E B> points A and D. 5. t \A E D ~ ACEB '■■■

; 4—►2. Let CB be the line through 6. AE-.CE = E D :E B

points C and B. 3. Z C S ZA

7. .-. A E -E B = CE-ED

T H E O R E M 106: If a tan g en t segm ent an d a secant segm en t a re d ra w n to a circle from an ex ternal p o in t, th e n th e sq u a re o f th e m easure of th e tan g en t segm ent is eq u al to th e p ro d u c t o f the m easures of the secant segm ent a n d its ex te rn a l portion .

Given: T angent segm ent PA Secant Segm ent PB

Concl.: {PAY = PB-PC

A n a l y s i s : Note that the external portion of a secant segment is that segment between the external point and the first point of intersection with the circle. In Figure 14-52 it is the segment PC. The internal portion is the segment between the points of intersection of the secant with the circle. By proving A PAC ~ A PBA the conclusion will follow.


1. Let AC be the line through 5. m ZP B A = J m A cpoints A and C. 6. ZPAC ^ ZP B A

2. Let A B be the line through points A and B.

3. Z P ^ Z P

7. & P A C ~ A P B A i8. P A \P B = PC:PA9. A (P/1)2 = P B-P C

4. m ZPAC II 3 &

500 THE CIRCLE

'1

T H E O R E M 107: If two secant segm ents a re d raw n to a c irc le from an ex te rn a l po int, th en the p ro d u c t o f th e m easures o f one of these segm ents an d its ex te rn a l p o rtion is eq u al to the p ro d u c t o f the m easures of th e o ttie r w ith it,« exte rn a l portion.

Given: PB and PC arc secant segments. Concl.: PB-PA = PC-PD

PRO OF (Tne reasons are left for you to supply.)

1. L et BD be the line through points B and D.

2. L et CA be the line through points C and A.

3. A P S H A P

4.. A B S AC5. A P B D ~ A P C A6. PB'.PC = PD'.PA7. P B -P A = PC-PD

J

H ad the tangent segment in Figure 14-53 been drawn from point P to the circle, then (PR)1 = P B -P A ; also, (P R)1 = PC-PD. In fact,

R

Figure 14-54.

(P R )1 would be equal to the product of any secant segment from point P with its external portion. Since each of these products is equal to (P R )1, we say that from a fixed point the product of the measure of a secant segment with the measure of its external portion is constant. Notice that as the secant seg-

Figure 14-55.


m ent rotates counterclockwise about the point P, its length increases until it reaches a certain position and then it begins to decrease. W hat is that position? How do you account for the fact th a t although the secant segment grows larger, its product with its external segm ent remains the same?

PA = 6, AB = 4, PD = 5 : CD = ?

Figure 14-56.

M e t h o d : I t is advisable to label all six segments in Figure 14-56 before applying the theorem needed; in this case it is Theorem 107.

PB-PA = PC-PD 10-6 = (x + 5) '5

60 = 5* + 25 35 = 5 x .

7 = x or CD

Illustration 2:

In a circle whose diam eter is 10 inches, how far from the center is a chord th a t is 8 inches long?

M e t h o d :

Figure 14-57.

Using Theorem 105: C E-ED = A E -E B (W hy is J E ^ E B l )

. (5 + x)(5 - x) = 4-4 25 — *! = 16 25 - 16 = x‘‘

9 = x13 = x

Illustration 1:

G iven: T o Find

502 THE CIRCLE

Illustration 3:

Two secant segments are drawn to a circle from an external point. If the secant segments are congruent, then their internal portions are congruent.

Given-. P B ^ P C C oncl.: A B — DC

Figure 14-58.


1. P B-P A =2. PB = PC3. .’. PA =4. .-. A B =

PC-PD

PDDC . "

1. Theorem 1072. Given3. Division postulate4. Subtraction postulate

(Step 3 from step 2.)

EXERCISES

A

1. Use the figure a t the right for each of the following problems:(a) CE = 9, ED = 4, A E = 12, E B = ?(b ) A E = 12, E B = 4, ED = 3, CE = ?(c) CD = 18, ED = 6, EB = 8, A E = ?(d ) 4 B = 10, A E = 6, ED - 2, CD = ?(e) CD = 16, £ D = 6, £ B = 3, A B = ?(f). A B = 16, C £ = 32, AE = £B ,

ED = ?(g ) t = 3, £ B = 2, CD = 7, CE = ?(h ) f CD = 15, CE = 12, = 13,

E B = ?

t T h e solution o f a q u a d ra tic equation by factoring is requ ired for this problem .


2. Use the figure at the right for each of the following problem s:

(d ) PA = 8, PC = C B ,PC = ? ( f) PA = 4 ,P C = CB, CB = ? ( h ) t PA = 8, BC = 30, PC = ?

G iven: PA is a tangent segment.

PB is a secant segment.

(a) PA = 8, PB = 32, PC = ?(c) PC = 3,C B = 24, P /i = ?(e) PA = 6,PC = 4 ,C B = ?(g ) t P.-4 = 2, BC = 3, PB = ?

3. Use the figure at the right for each of the following problem s:(a) PA = 5, PB = 8, PD = 4,

PC = ?(b ) P<4 = 6, AB = 4, PD ■= 5,

CD = ?(c) CD = 8, PD = 2, PA = 4,

PB = ?(d ) PC = 16, CD = 13. PB =

A B - ?(e )f PB = 6, PA = 4, CD = 5,

PD = ?( f ) t PC = 12, PD = DC, A B = 1,

PB = ?

4. In the diagram at the right, PQ is an arc of a circular race track where M

is the m idpoint of PQ and R is the

midpoint of PQ. If PQ = 120 (feet) and M R — 10 (feet), w hat is the diam eter of the track?

5. A circular play area is being laid out in a field. T he points P, R, and Q have been found such that PS = 24 (feet), PQ = 66 (feet), and RS = 12 (feet). If T is to be another point on the circle, how far from R will it lie?

o. Two secant segments arc drawn to a circle from an external point. . T he length of one of these secants is 9 inchcs, while its internal portion

t T h e solution of a q u ad ra tic eq u a tio n by factoring is requ ired for th is prob lem .

504 THE CIRCLE

is 1 inch in length. If the circle bisects the o ther secant segment, how long is it?

7. T o a circle whose diam eter is 20 inches a tangent segm ent is draw n from a point that is 20 inches from the center of the circle. How long is the tangent segment?

8. T he diameter of the earth is approxim ately 8,000 miles. An astronaut is a t a point 100 miles above the surface of the earth. Assuming th a t he can see th a t far, w hat is the farthest point on earth he would be able to see from his position?

9. A surveyor places his transit along the line tangent to the circle a t point A such that PA = 200 (feet). He locates another point B on the circle and finds PB = 80 (feet). If a third

point, C, on the circle lies along PB, how far from point B will it be?

10. If in the figure at the right AG = CG and PF = 4 , FD = 8, PE = 3, AC — 12, then w ha.'is the measure

of EG?

B C

11, A circular archway is to be built such th a t the distance betw een the endpoints of the arch is to be 14 times as long, as the height of the arch-. If the radius of the circle of this arch is 20 feet, how high is the arch?

1.2. A tangent segment and a secant segment are draw n to a circle from an external point. T he measure of the tangent segment is twice th a t of. the external portion of the secant segment. W hat is the measure of the tangent segment if the measure of the internal portion of the secant segment is nine?

P


13.* The measure of a chord of a circle is 14. A point is selected on the chord so that its distance from one end of the chord is 8, while its distance from the center of the circle is 4. W hat is the measure of the chord that is perpendicular to the radius that passes through this point? (This chord is the shortest chord that can be draw n through this point.)

B

1 . Given: Chords AB and CD intersect at E.AE - ED

Concl.: A B = CD

A _____

3 . G iven: © O and R tangent

to PQ a t Q Concl.: PB-PA =

PD-PC

G iven: Secant segments

PB and PC PB = PC

Concl.: AB = DC

G iven: PD common secant segment to © 0 and R

7 1 tangent segment to 0

FB tangent segment to R

Concl.: PA = PB

508THE CIRCLE

15.* Given: © 0 and R w ith secants

PABC and P D EF

Concl.: AD II CF (H int: Prove A P A D ~ A FCF.)

16. Given: Sphere 0 with secant

segments PB and PD intersecting sphere at A and C

Concl.: PB-PA = jPD-PC |

Given: Sphere 0

PA tangent segment

PD secant segment intersecting sphere at C and D

Concl.: {PAY = PD -PC

17.

Test and Review

Find the measure of each of the following:

(a ) Z A D B (b ) Z E A B

(d ) Z BAD (e ) ZA D C(g ) Z E A L (h ) Z B A F

(c) BC

(f) AC(i) m Z A B C + m ZA D C

r -

TEST AND REVIEW 509

2. Use the diagram at the right to answer each of the following questions:

(a ) m BC = 100, m AD = 40, rn Z P = ?

(b ) m BC = 120, m AD = 30, m Z B E C = ?

(c) m A B + m C D = 160, m ZA E D = ?(d ) m ZB D C = 75, m Z A B D = 25, m Z P = ?

(e) m Z P — 30, m AD - 100, m BC = ?

(f) m ZB E C = 115, m ZP = 75, m AD = ?

3. Given: D is the midpoint of CE.

m CD = 60 m ZB F C = 85 m Z D B A = 70

Find the measure of Z P.

4. In a circle a t the right CD — 10, D E = 4 , B E = 12. Find P<4. :

6. In the circle a t the right PA is a. tangent segment whose measure is 12.

If the measure of PB is 8, then what is the measure of BC?

7. T h e measures of the radii of two concentric circles are 10 and 8 respec-

510 THE CIRCLE

10.

tively. W hat is the m easure of a chord of the larger circle that is tangent to th e sm aller circle?T h e m easures of the radii of two circles th a t are tangent externallyare 8 and 3 respectively. W hat is the distance between the points oftangency of one of their common external tangents?T h e measures of the diameters of two circlcs that are tangent internallyare 18 and 8 respectively. W hat is the length of the tangent segmentfrom the center of the larger circle to the smaller circle?

___ pIn the diagram at the right PE is atangent segment. If PA = 4, AB = 6,w hat is PE?

11. A point is 4 units from the center of a. circle. A secant, segment is drawn from that point to the circle such that the measure of the external portion is 3 while the internal portion is 1. W hat is the radius of the circle?

12. T h e m easure of an angle formed by two tangents to a circle is 90. If the radius of the circlc is 8, how far is the point from the center of the circle?

B

1 . G iven : G 0 with C the

m idpoint of AB

CD J .O A

CE 1 OB

Concl.:

C


Given: GO with diam eter A B extended to C

DC 1 AC Concl.: AB-AC =

AE-AD

TEST AND REVIEW

3. Given: Secant segments and PD

„ , PA _ PF Concl.: p c pD

5, Given: AD = EBC is the midpoint of BD,

Concl.: M S ®

7 . G iven: © A and B

AB bi. CD and EF.

Concl.: CD || .EF

Given: GO with T 2 andT B tangent segments

~BC is a diam eter.

Concl.: ACWPO (H int:Prove M to be the

m idpoint of AB.)

Given: A B II CD

EF passes through center of G O .

OF

Concl.: A B & CD

Given: GO Concl.: AE-EB =

(CO)2 - (OE)1 (H int: CE = CO - f OE E 5 = CO - OE)

512

9. Given: AB _L BC

DC 1 BC

Concl.:

11, Given: Q 0 w ith EA and ~UB tangent segments

Concl.: (A B ) J = A E -B C

Given: Circie 0 with 1 0 .

A S ^ V D

Concl.: A E £= D E (H int:Draw perpendiculars from 0 and prove triangles congruent.)

THE CIRCLE

12.*

8

Given: AB and CD are common external tangents.

HG is the common internal tangent.

Concl.:

cProve each of the following statem ents:

1. I f the endpoints of two perpendicular diameters are joined in order, the ,

quadrilateral formed will be a square.2. T h e perpendicular bisector of a chord passes through the center of the

circle.

17

TRY THIS FOR FUN 5133. T he diagonals of an equilateral pentagon inscribed within a circle are

congruent. (The vertices of the pentagon are points of the circle.)4. If from the endpoints of a diam eter perpendiculars are drawn to any

secant of the circle, then the points of intersection of the perpendiculars with the secant will be equidistant from the center of the circle.

5. If two chords intersect within a sphere, then the product of the measures of the segments of one is equal to the product of the measures of the segments of the other.

I Try This For FunA question that often arises in the minds of students is the

possibility that when a chord of a circle is doubled in measure will its distance from the center become only half as great? Should a chord be selected at random , then the answer is a most emphatic “ N o !” However, if we make our choice a bit more carefully, there do exist chords in every circle for which this is true. These are the chords whose measures are the same as their distances from the center of the circle. Can you prove that this is so for these chords?

Given: OE — AB, OE J_ AB

CD = 2AB, OF JL. CD C oncl.: OF = iO E

(H int: Let OE = 2a.)

/

/

/ \ // \ /

15 V

LocusSynthetic Geometry

O U R FIR S T E N C O U N T ER W IT H T H E T E R M locus was in connection with the study of analytic geometry. We are going to take another look a t this area of our work; now, however, it will be from a synthetic point of view rather than an analytic.

By definition, the locus of points was the set of points and only those points that satisfied certain given conditions. Thus, if the locus is a line, the definition implies that two properties will ho ld :

(1) Every point on the line satisfies the given conditions.(2) Every point th a t satisfies the given conditions lies on the line.

LOCUS: SYNTHETIC GEOMETRY 515

Before plunging into the formal study of locus, let us exam ine several problems from an intuitive point of view. T h a t is, let us determ ine what to us seems to be the set of points that fulfill the requirem ents w ithout handicapping ourselves with the necessity of trying to .justify our answer.

Illustration:

W hat is the locus of points that are five inches from a given line?

A n s w e r : The locus o r set of points that are five inches from a given line are two lines, one on each side of the given line, parallel to it and five inches from it. (See Figure 15-1.)

In answering the question in this illustration, we have, in reality, made two assertions, neither of which we have attem pted to prove'.

(1) All points on a and b are five inches from /.(2) All points that are five inches from / lie on either line a or line b.

EXERCISES

Discuss informally the answer to each of the following problems:

1. W hat is the locus of points in a plane that are 10 inches from a givenpoint? j

2. W hat is the locus of points in space that are 10 inches from a given point?

3. W hat is the set of points in a plane that are equidistant from two parallel lines?

4. In regard only to the points in your classroom,(a) W hat is the set of points that are 7 feet from the front wall?(b ) W hat is the set of points that are 8 feet from the wall containing

the windows?(c) W hat is the set of points that are both 7 feet from the front wall

and 8 feet from the wall containing the windows?(d ) W hat is the locus of points that are equidistant from the ceiling

and floor of the room?

5. W hat is the locus of points in space that are 7 feet from a given plane?How does this problem differ from Problem 4 (a)?

6. W hat is the locus of pdints in a plane that are more than 6 inches from a given point?

7. W hat is the iocus of points in space that are less than 10 inches from a given point?

516 LOCUS: SYNTHETIC GEOMETRY

8. W hat is the set of points outside a circle in a plane th a t arc 10 inches from the circle? (The distance from a point to a circle is the external portion of the secant segment drawn from that point through the center of the circle.)

9. W hat is the set of points in space inside a sphere of radius 25 inches that are 10 inches from the sphere?

10. W hen in a plane a coin rolls along one side of a line, w hat is the locus of the center of the coin?

11. A dime is rolled around a half-dollar; what is the locus of th e center of the dime?

12. Two points A and B are 10 inches apart. Consider your answer to each of the following questions in terms of coplaner points only:(a ) W hat is the locus of points that are 8 inches from A and 8 inches

from B?(b ) W hat is the locus of points that are 5 inches from .4 and 5 inches

from 5?(c) W hat is the locus of points that are 2 inches from A and 2 inches -

from £?(d ) W hat is the locus of points that are less than 8 inches from A and

less than 8 inches from B ?(e) W hat is the locus of points that are less than 8 inches from A and

8 inches from B ?

13. Two parallel lines I and m are 6 inches apart. Point P is a point on /. Consider your answer to each of the following questions in terms of coplaner points only:(a) W hat is the set of points equidistant from

from P?(b ) W hat is the set of points equidistant from

from P?(c) W hat is the set of points equidistant from

from P?(d ) W hat is the set of points equidistant from

4 inches from P?(e) W hat is the set of points equidistant from

3 inches from. P?

14. W hat is the locus of points that are on a given plane and 5 feet from a second given plane?

I and m and 5 inches

I and m and 3 inches

/ and m and 2 inches

I and m and less than

I and m and less than

THEOREM, CONVERSE, INVERSE, AND CONTRAPOSITIVE 517

B Theorem, Converse, Inverse, and ContrapositiveEarlier in our work it was pointed ou t th a t in order to

justify a set of points as fulfilling the conditions set up in the problem , i t is

necessary to prove that(1) Each point in the set fulfilled the requirements of the problem.

and(2) Each point that fulfilled the requirements was an element in the set.

In analytic geometry we were extremely fortunate, for there the description of the set of points was given by an equation. Hence, it immediately followed that all points on the graph of this equation satisfied this equation, while all the points that satisfied the equation fell on its graph. This is the relation between a graph and its equation. Now, however, we are faced with the problem of not only having to name what we believe the locus to be but also justifying this by proving the two properties stated above.

T o illustrate, let us say that we are required to determine the set of points that are equidistant from points A and B in Figure 15-2. An intelligent

A m

• 6

Figure 15-2.

guess would be that the set of points equidistant from A and B is the perpendicular bisector of ~AH. To justify our conviction, we are faced with having to prov-e two things. The first is that

(1 ) Every point equidistant from A and B is a point on the perpendicular

bisector of ~AB.

This, however, is not enough, for our guess implied more than that which ! is stated in (1). By (1) we have merely shown that those points that are equidistant from A and B have to be points on the line that is the perpendicular bisector of ~AB. This perpendicular bisector contains many points, and although we have shown that some of its points are equidistant from A and B , it may have many others that are not equidistant from A and B. Thus, points P, Q, and R that we know to be equidistant from A and B lie

on the perpendicular bisector of A B . Points W and S that are also on the

perpendicular bisector of A B , however, may not be equidistant from A and 5 . Hence, the second part of our proof consists in showing that


(2) Every point on the perpendicular bisector of AB is equidistant from points A and B.

Figure 15-3.s-

E xam ination of statem ents (1) and (2) reveals th a t they are simply a statem ent and its converse. Thus, justifying a set of points as fulfilling the requirem ents of a given problem necessitates proving a theorem and its converse.

Although the term converse had been used a num ber of times earlier in the text, only a ra th e r weak description had been m ade of the word. I t h ad been pointed out th a t the converse of a theorem was a statem ent in w hich th e “ given d a ta ” and the “ conclusion” of the original theorem had been interchanged. But w hat if the given da ta contained m any pieces of inform ation, and, in addition, we were asked to draw several conclusions? Do we interchange all of the given data with all the conclusions o r with only part? In reality, should we interchange any piece of given data with any one of the conclusions, a converse statem ent will follow. Thus, an original statem ent m ay have m any converses depending on the num ber of pieces of given data and the num ber of conclusions. None or perhaps even all of these converses m ay be true. T he num ber that are true will depend on how m any can be proved to be so.

N orm ally, th e given d a ta is listed in term s of angle bisectors, or perpendicular lines, or isosceles triangles, while the conclusion is considered in term s of parallel lines, o r congruent triangles, or similar triangles. T o make our discussion as general as possible, we will consider the given data as, “ Item A ,” “ Item B,” and "Item C,” while the conclusion will be called “ Item D ,” T hus, the statem ent of a theorem in term s of these items will be

“ If ‘Item A,’ ‘Item B,’ and ‘Item C’ are true, then ‘Item D ’ is true.”

O r, even m ore concisely, this, can be stated as

“ If A, B, and C are true, then D is true.”

In terms of these symbols, a converse of the above statement will be


“ If A, B, and D .are true, then C is tru e .”

Can you write a t least two other converses to the original statem ent?M any of the theorems in the past and, even m ore im portan t, those

concerned with locus consist of merely one piece of given d a ta and one conclusion. It is to this type of statement that we now turn our attention. Its form will be that of the conditional statement with which we are fam iliar:

If p then q.Thus, in the statement

(1) “ If two angles are right angles, then the two angles are congruent” p is the antecedent (Two angles are right angles) while q is the consequent (The two angles a re congruent).

Early in the year we proved the truth of statem ent (1). Now we are interested in knowing what effect the tru th of this statem ent has on statements such as

(2) “ If two angles are congruent, then the two angles are right angles.”(3) “ If two angles are not right angles, then the two angles a re no t congruent.”(4) “ I f two angles are not congruent, then the two angles a re not right angles.”

Statem ents (2), (3), and (4) are called respectively the converse, inverse, and contrapositive to statem ent (1). Notice that the antecedent and consequent of the inverse are the contradictory statements to the antecedent and consequent of the original statem ent (1).

In general, our present objective is to investigate the relation th a t exists between the following four conditional statem ents:

(1) I f p then g. (original statement)(2) If q then p. (converse of original statem ent)(3) If then (inverse of original statem ent) ;(4) If ~ y then ~/>. (contrapositive of original statem ent)

T h e ideas we are trying to develop can be expressed quite clearly through the use of the Venn diagram. In Figure 15-5 we have shaded those elements that are members of the set p. In Figure 15-6 the region shaded

t Recall that is the contradictory statement to q and is read, as “not q."


Figure 15-5. F igure 15-6.

contains those elements th a t are in set q\ in Figure 15-7 it was the elements in set ~ / i , while in Figure 15-8 it was those in ~ y . Each of these figures

•vq

Figure 15-7. ■=* Figure 15-8.

expresses diagramatically th e truth of th e original statem ent

“ I f p then

Now, based on the tru th of this statem ent, w hat will these diagram s tell us of the tru th of the rem aining three statements?

W e will first examine“ If q then />.”

Notice in Figure 15-6 th a t although some of the elements of q a re contained in p, such as element A, not all the elements of j are in p. In particular, elem ent B is not in p. Thus, it appears that we cannot say that

“ If q then p" must be true on the basis of the tru th of

“ I f p then q."

This could also have been seen by examining the two statem ents:

(1) I f two angles are right angles, then the two angles are congruent.(2 ) If two angles are congruent, th en the two angles are righ t angles.

Although we had proved the tru th of th e first, the second m ay n o t be true, for if two angles are congruent, they m ay be straight angles ra th e r than right angles or, in fact, they m ay be any two angles having equal measures ra th e r than right angles,

Examination of Figure 15-7 will give us a clue as to the tru th of the

inverse

(3) “ If then ”In Figure 15-7 wc see that some of the elements that a re no t in ~/>, such as B, are in q. However, if statement 3 is to be true, the.' all the elements th a t are not in p cannot be in q. Thus, the tru th of the original statem ent does notimply the tru th of its inverse.

Figure 15-8 points out th a t all the elements th a t are not in q are also not in p. Thus, apparently, the tru th of

(1) “ I f p then q"

implies also the truth of its contrapositive

(4) “ If then.In the same way, were a Venn diagram to be draw n such as Figure 15-9

cFl

THEOREM, CONVERSE, INVERSE, AN D CONTRAPOSITIVE 521

Figure 15-9.

it would be possible to demonstrate th a t the tru th of the statem ent

“ If then ~ p "

w ould lead to the tru th of the statem ent“U p then q."

In Figure 15-9 all elements in p must be outside of circle . By being outside of. circle (~ p ), they are also outside of circle (~ ? ) - Hence, the elem ents of p a re also elements of q. T hus, since Figure 15-9 implies the tru th of

“ If thenit also implies the truth of

“ If p then q.”

Two statem ents of the form

“ If p th e n ? ” and “ If ~ q then

are called equivalent statements. Equivalent statem ents are statem ents th a t are either both true or both false at the same time. I t is not possible for one to be true while the o ther is false.

R eturning again to the conditional statem ent

(1) “ If two angles are right angles, then the two angies are congruent"

we see th a t its contrapositive

522 LOCUS: SYNTHETIC GEOMETRY(4) If two angles are not congruent, then the two angles are not right angles.”

will be true since the original statem ent was true.B ut w hat of the converse and the inverse cf a statement. Aiihough the

tru th cf neither follows from the tru th of the original statement, can anything be said of the relationship th a t oxists between them? Oddly enough, they, too, are equivalent statements. T he truth of

“ I f q theii p " (converse)

gives rise to the Venn diagram in Figure 15-10. An element that is not in p

F I

Figure 15-10.

w ould lie outside circle (p) and, hence, could no t bp an element of q. Thus, the tru th of the converse leads to the tru th of the inverse:

“ If then

Similarly, the tru th of

“ If thenwill imply the tru th of

“ I f q then p "D raw a Venn diagram to show why this would be so.

Based on this analysis, we will assume that

P o s t u l a t e 41: T h e statem ents

“ If p then q” and “ If ~ q then ~ p ”

are equivalent statements.

Postulate 41 points out no t only the equivalence of a statem ent and its contrapositive but also the equivalence of the converse and the inverse of a statem ent. In reality, the converse and the inverse of a statem ent can be considered as a statem ent and its contrapositive relative to each other, for

“ If q then p " and “ If ~ p then

are bu t a statem ent and its contrapositive.Knowing these relations is very valuable, for there are times when it

m ay be exceedingly difficult to prove a statem ent. However, proving its contrapositive m ay turn out to be a simple m atter. Hence, the contrapositive is proved and the tru th of the original statem ent immediately follows from Postulate 41.


EXERCISES

1. Given the fact that each of the following statem ents is true, write its converse, inverse, and contrapositive. State which of your statem ents is true without need for proof.(a ) If it rains tomorrow, then I shall stay at home.(b ) If the test is not difficult, then I shall receive a passing grade.(c) If I buy the book, then it is not expensive.(d ) If at least 100 people do not purchase tickets, the show will no t go on.(e) If two triangles are congruent, then an angie of one triangle is con

gruent to its corresponding angle in the other triangle.

2. Rewrite each of the following statements as conditional statem ents; then write the contrapositive of each.(a) Any citizen is eligible to run for Congress.(b ) A ll Golden Brand candies are good to eat.(c) All of Jo h n ’s shirts are m ade from cotton..(d ) A book th a t is written in Sanskrit does not sell in H arb o r City.(e) A person who does not enjoy Shakespearean plays will not enjoy

Hamlet.(f) A person who eats excessively will not lose weight.(g) Two tangent segments to a circle from an external point are con

gruent.

3. (a ) W rite the converse to the following statem ent: If a polygon is a tr iangle, then it has three sides.

(b ) W hy is the converse of this statem ent true w ithout need of proof?

4. (a ) Using your postulates, prove the statem ent th a t “ I f 2x = 10, thenx = 5.”

(b ) In view of the fact that you have proved this statem ent to be true, w hat o ther statem ent will be true w ithout need of proof?

(c) W hat is the converse of the statem ent in pa rt (a)? Prove th a t the converse is also true.

(d ) In view of the fact that you have proved the converse o f the sta tem ent in (a) to be true, what other statem ent will be true w ithout need of proof?

5. (a) W rite the converse, inverse, and contrapositive to the statem ent“ If Z A and Z B are straight angles, then Z A = Z B ."

(b ) In view of the theorems we have proved, which of the statem ents th a t you wrote in answer to part (a) are true?

6. W rite the converse and inverse of each of the following statem ents:

(a) Any point on the perpendicular bisector of a line segm ent is equidistant from the endpoints of the line segment.

(b ) If a point is on the bisector of an angie, then it is equidistant from the sides of the angle.

(c) All points on a circle are a fixed distance from the center of the circie.

7• Justify why the converse, inverse, and contrapositive of any definition are true statements w ithout the necessity of having to prove them to be.

HI Locus Theorems

There are basically only five theorems upon w hich the gTeat bulk of the locus problems in synthetic geometry are based. Justification of each of these theorems, as you know, involves the proof of both a statem ent and its converse. For a few of these theorems it will be far easier to prove the inverse ra ther than the converse. From the information in the preceding unit, however, the tru th of one will imply the tru th of the other.

T H E O R E M 108: T h e locus o f po in ts eq u id istan t from two fixed po in ts is th e p e rp e n d ic u la r b isector o f th e l in e segm ent jo in in g th e two points. (First theorem on locus)

Part A—Proof of StatementIf a point is on the perpendicular bisector of the line seg

m ent joining two fixed points, then it is equidistant from these two points. (See Theorem 17.)

524 LOCUS-. SYNTHETIC GEOMETRY

Figure 15-11.


1. PQ L A B 5. I R & 1 U } (j)2. / PRB and /P R A are right angles. .(*)3. /P R B S i Z PRA (a) 7. & P R A & & P R B

4. PQ bisects AB. 8. T Z S i T B

525

Part B—Proof of ConverseI f a point is equidistant from two fixed points, then it lies

-r k.i» joining these two points.

LOCUS THEOREMS

Given: PA S P SP is any po in t equidistant fromA and B.

Concl.: P is on th e X bisector of ~SB.

Figure 15-12--P R O O F (The reasons will be left for you to supply.)

1 - A O B J ~ k P R B5. A PRA S* A P R S

6 .

7 . .- .P R is the X bisector of

J B .

I . H s H (A)

2. L et P R be the perpendicular that< ’>

exists from P to AB.3. /P R A and /.P R B are right angles.

4 . P R ^ P j i (0R ath er than prove the converse as we had in P a rt B, i t would have

been possible for us to have proved the inverse to justify th e second half of the proof of this locus theorem. As a point of inform ation, this will be

done now.

Alternate Proof of Part B—Proof of inverseIf a point is not on the perpendicular bisector of the line

segment jo in ing two fixed points, then it is not equidistant from these two

points.

♦-» __Given: R S _L bisector of A B

Point P does not lie on RS.

Concl.: T A — T B


P R O O F

By the law of the excluded m iddle one of the following sta tem ents m ust be true and no o ther possibility exists t

Ta ^ T S or

__ _ <-»L et us accept the possibility th a t PA = PB. L e t PS be the line th a t exists through points P and S. By th e S.iS.S. congruency theorem it is possible to show th a t A P S A ~ A PSB. Hence, /P S A S /P S B , and, therefore,

PS A. AB. T his would imply th a t /P S A is a right angle. The Given D ata,

however, informs us that RS X AB ; hence, /R S A is a right angle. T herefore, /P S A S /R S A . However, by our postulate that the whole is greater th an any of its parts, it follows th a t /P S A ^ /R S A . Therefore, accepting

the possibility th a t PA £= PB led to the logical inconsistency of the tru th of both /P S A == /R S A and /P S A /R S A . By the law of contradiction both cannot be tru e a t the same time. Since /P S A 3= /R S A is true, for it is the result of a postulate, then /P S A ~ /R S A must be false, and so

PA ~ P B m ust also be false. Hence, P AZ£ PB is true, for it is the only rem aining possibility.

THEOREM 109: T he locus of points equidistant from two intersecting lines is the two lines that are the bisectors o f the angles formed by these lines. (Second theorem on locus)

Part A—Proof of Statement■ ■ ■ ■ ■ ■ If a point is on the bisector of one of the angles formed by two intersecting lines, then it is equidistant from these lines.

Given: S T is the bisector of an angle formed by a and b.

P is any point on ST .

PQ l b , PR l aConcl.:

LOCUS THEOREMS 527

1 PROOF 1 (The reasons will be left for you to supply.)

1. S T bisects /Q S R . 5. /P Q S and / P RS are right angles.2. /Q S P ^ /R S P («) 6. /P Q S = / P R S (a)

3. P S ^ T S ( s ) i . a p q s ^ a p r s

4. PQ 1 b, PR l a8. T Q ^ P R

Part B—Proof of Inverse■ ■ ■ ■ ■ ■ ■ ■ If a point is not on the bisector of one of the angles formed by two intersecting lines, then it is not equidistant from these lines.

G iven: S T is the bisector of an angle formed by a and b.

P is any point th a t is not on S T .

PQ l b , PR l a Concl.: P Q m F R

F ig u r e 15-15.

P R O O F

By the law of the excluded middle one of the following statem ents m ust be true and no o ther possibility exists:

T Q ^ P K or F Q g P T i

L et us accept the possibility th a t TQ == PR. Let PS be the line through points P and S. By the hypotenuse-leg congruency theorem it is possible to show A P Q S == A P R S . Hence, /P S Q = /P S R , and, therefore, m /P S Q —

i m /R S Q . T he Given Data, however, informs us th a t S T is the bisector of /Q S R ; therefore m /T S Q = J m /R S Q . Hence, m / PSQ — m / TSQ . However, by our postulate th a t the whole is greater than any of its parts,

m /P S Q 7* m /T S Q . Therefore, accepting the possibility th a t PQ ~ TFR led to th e logical inconsistency of the tru th of both m /P S Q = m / T S Q anc. m /P S Q 7 m /T S Q . By the law of contradiction both cannot be true, a : th e sam e time. Since m /P S Q m / T S Q is true, for i t is the result o f a

postulate, then m /.PSQ = m /T S Q m ust be false, and, hence, PQ ?= PP-

is false. Therefore, TQ =£ 'PR is true, for it is the only rem aining possibility.

T H E O R E M 110: T h e locus of points a t a fixed d istance from a g iven lin e is tw o lines p a ra lle l to th e g iven lin e a t th e fixed d istance from it. (Third theorem on locus)

Part A—Proof of Statement

■ ■ ■ B B If two lines are parallel to a given line such th a t two points, one oil each line, are each a distance d from the given line, then all points on these lines are a distance d from the given line.


b II /, a II /P on line J a t a distance d from I Q is any other poin t on b.

PS 1 I, QR 1 1 QR = d

P R O O F (The reasons will be left for you to supply.) 1

l . i l l / 5. PQRS is a parallelogram .

2 . PS 1 1 6. QR = PS7. But PS ~ d

3 . QR 1 1

4, PS II QR

8. .'. QR*= d

T h e same proof would apply for points on line a.

Part B—Proof of Inverse■ ■ ■ ■ ■ ■ I If two lines are parallel to a given line such th a t two points, one on each line, are each a distance d from the given line, then all points no t on these lines are not a distance d from the given line.

Given: HI /, a II /P on line b a t a distance d from / Q is any point no t on either a or b,

K 1 U Q R . 1 1 Concl.: QR ^ d

Figure 15-17.

LOCUS THEOREMS 529

PR O O FThe proof is indirect as in T heorem 109. I t will be left for

you to complete.T H E O R E M 111: T he locus of points e q u id is ta n t from two g iv en p a r

allel lines is the p e rp e n d ic u la r b isector o f th e segm ent th a t is the com m on p e rp e n d ic u la r to these tw o lines.(Fourth theorem on locus)

Part A—Proof of StatementIf a point is on the perpendicular bisector of the line seg

m ent that is the common perpendicular to two parallel lines, then it is equidistant from the two lines.

G iven : a II I, b II /

R S 1 a, R S 1 b

Q is any o ther point on I.C oncl.: Q is equidistant from a and b.

Figure 15-18.

A n a ly sis: To prove that <2 is equidistant from a and b, it is necessary to show th a t the perpendiculars from Q to a and b are congruent. In order to

4—>

arrive a t this, we allow Q T to be the perpendicular from Q to a and then4—► 4-**

extend Q T to intersect b. After proving th a t Q W is perpendicular to b, we

will show th a t Q T S QW.

P R O O F | STATEMENTS REASONS |

1 . a II M l/ 1. Given4—►

2. L et Q T be the perpendicu 2. T here exists one and only one per

lar from Q to a. pendicular .from a given point to agiven line.

3. Extend Q T until it inter 3. A line can be extended as far as de

sects b a t W. sired.

4. Q T l b 4. I f a iine is perpendicular to one oftwo parallel lines, it is also perpendicular to the other.

5. F R ^ T S 5. Given

6.

530

If three or more parallel lines cu t off congruent segments on one transversal, they cut off congruent segments on every transversal.

LOCUS: SYNTHETIC GEOMETRY

Part 8— Proof of Inversem a m R M If a point is not on the perpendicular bisector of the line segm ent th a t is the common perpendicular to two parallel lines, then it is not equidistant from the two lines.

G iven: a II /, b II I

RS l a , R S L b

Q is any point not on /.Concl.: Q is no t equidistant from a

and 6.

P R O O F

The proof is indirect as in Theorem 109. I t will be left foryou to complete.

THEOREM 112: The locus of points that are a fixed distance from a fixed point is a circle. (Fifth theorem on locus)

P R O O F

This statem ent should not, strictly speaking, be called a theorem , for i t is simply the alternative definition of a circle th a t had been given on page 440.

W hen you are confronted with a locus problem, it would be best to follow the patte rn suggested below:

(1 ) D raw a diagram in w hich you have placed any two points of th e set or locus.

(2) Prove some relation about these two points th a t falls under one of the five locus theorems.

LOCUS THEOREMS 531

(3) W rite the proof in paragraph form similar to th a t used for the indirect

proof.Several illustrations will be given to present a clearer picture of this m ethod

of approach.

Illustration 1:W hat is the locus of the midpoints of congruent chords of a circle?

Given: "AB S "UD in circle 0P is the m idpoin t of AB.Q is th e m idpoint of CD.

Concl.: T o find the locus of the m idpoints of congruent chords of circle 0

> 4—>By allowing OP and OQ to be the lines th rough points O

and P and through 0 and Q, we can show th a t OP 1 . A B an d OQ L CD.

Hence, OP S OQ since congruent chords of a circle are equidistant from the center. Thus, the m idpoints of congruent chords of a circle are a fixed distance from the center of the circle. Hence, their locus is a circle, for the set of points th a t are a fixed distance from a fixed po in t is a circle (fifth

theorem on locus). ..

Illustration 2:In a right triangle with a fixed hypotenuse w hat is the locus of the

vertex of the right angle?

Given: R igh t A A P B w ith Z P th e righ t angleR ight A A Q B w ith Z Q the right angle

Concl.: T o find the locus of the vertices of the right angle of the right

triangle w here A B is the hypotenuse

Figure 15-20.

PRO OF


PR O O F

L et M be the m idpoint of A B and let P M and Q M be the lines through / ’ and M and through Q and M . By Problem 10, page 274, we

know that in A APB M P ~ M A since the median to the hypotenuse is con

gruent to the segments of the hypotenuse. Similarly, in & A Q B M Q ~ M A.

Hence, M P = M Q. Thus, the vertices of these right triangles are a fixed distance from point M . Hence, their locus is a circle, for the set of points th a t are a fixed distance from a fixed point is a circle (fifth theorem on locus). Incidentally, since A and B lie on this circle and also on the hypotenuse, they will have to be excluded from the locus. Can you explain why?

EXERCISES

1. W hat is the locus of the centers of circles that are tangent to two parallel lines?

2. W hat is the locus of the centers of congruent circles that are tangent externally to a given circle?

3. W hat is the locus of the centers of congruent circles that are tangent to a given line?

4. W hat is the locusof the midpoints of the radii of a circle?5. W hat is the locus of points that are equidistant from the sides of an

angle?

6. W hat is the locus of the vertices of the vertex angles of isosceles triangles drawn on the same base?

7. Each of the radii of a circle is extended its own length. W hat is the locus of the endpoints of these extended segments?

8. W hat is the locus of points that are always a fixed distance from a fixed circle? (See Problem 8, page 516, for the distance from a point to a circle.)

9. .W hat is the locus of the outer endpoint, not the point of tangency, of congruent tangent segments that are drawn to a circle?

COMPOUND LOCI IN SYNTHETIC GEOMETRY 533

10. W hat is the locus of the centers of circles th a t have a given line segment

as a common chord?11. W hat is the locus of the centers of circles th a t are tangent to bo th sides

of a given angle?12. In the diagram a t the right what is •

the locus of the midpoints of all linesegments whose endpoints are P and some point on /?

1

13. W hat is the locus of points th a t a re equidistant from two opposite sides

of a rectangle?14. W hat is the locus of the midpoints of line segments whose endpoints

lie in a pair of given parallel lines?15.* Prove the theorem in space geom etry th a t the locus of points th a t are

equidistant from two fixed points is the plane th a t is the perpendicular bisector of the line segment join ing these two points. Use the diagram below and the proof of Theorem 108 as a model.

16. W rite the statements of the theorems in space geometry th a t are comparable to Theorems 109, 110, 111, and 112. Do no t prove these statem ents.

■ Compound Loci in Synthetic GeometryIn our study of the intersection of two sets in analytic

geometry, we found that this simply implied th a t we. w ere searching for those elements th a t were common to the two sets.f These elements were either points in the coordinate plane or ordered pairs of values that represented these points. Now. however, the intersection of two sets, or loci, can only be points th a t are common to the two sets, for in synthetic geometry we do

t See page 4 t9.

not consider the correspondence between points and ordered pairs of num bers.

W here previously the proof of a problem was shown in terms of the intersection of the solution sets of equalities or inequalities, now it m ust bc given in terms of two of the five locus theorems. Several examples will be given to illustrate this point.

Illustration 1:

W hat is the locus of points that are equidistant from the points A and B and from the points C and D ?

534 LOCUS; SYNTHETIC GEOMETRY

P R O O FT he set of points equidistant from points A and B is the

perpendicular bisector of AB, the line I. T he set of points equidistant from

points C and D is the perpendicular bisector of VD, the line m. To find the points that are equidistant from the points A and B and also from the points C and £>, it-is necessary to find I Pi that is, the intersection of I and m. This is point P. Hence, P is equidistant from the points A and B and also equidistant from the points C and D.

U nfortunately the proof is not yet completed at this stage, for the points A, B, C, and D may take positions other than shewn in the diagram above. W ere the line segments A B and CD parallel, then I would be parallel to m. Hence, / pi m would be the null, or empty, set; that is, there would be no points that are common to the two loci. See Figure 15-23.

E-oc0

A C D B

Figure 15-23. Figure 15-14. Figure 15-25.


Finally, if A, B, C, and D were so situated th a t th e perpendicular

bisector of A B was also the perpendicular bisector of CD, then / P | m would be all the points on I or all the points on m. See Figures 15-24 and 15-25.

Illustration 2:W hat is the locus of points that are a fixed distance from a fixed point

and equidistant from the sides of a given angle?

Figure 15-26.

PROOFT he set of points th a t are a fixed distance from the fixed

point is a circle with the fixed point (/4) as the center and the fixed distance(d) as the radius. T he set of points equidistant from the sides of the angle

is the bisector (/) of the angle (/.B C D ). T he set o f points th a t is a fixed>

distance from A and equidistant from ZBC D is I f~) A, or points P and Q. T he placement of the fixed point and the given angle m ay be such that the

possibilities shown in Figure 15-27 will exist.

i n a = {p} I n A = 0 (em pty set)

Figure 15-27.

You may have noticed in our study of coordinate geometry th a t although we proved the concurrency of the m edians o f any trian g le ,! we

t See Problem 20, page 392.


avoided doing the same for either the perpendicular bisectors of the sides of a triangle or the angle bisectors of a triangle. This omission was not an oversight, but ra ther it was deliberate, since the proofs of the concurrency of these lines are far simpler and far more elegant in syr.thctic geometry than in coordinate geometry. T he locus theorems provide us with the tools that make this so.

T H E O R E M 113: T h e p e rp e n d ic u la r bisectors of th e sides c f a tr ia n g le a re c o n cu rren t a t a p o in t th a t is eq u id is tan t from the vertices o f th e tr ia n g le .

G iven: I is the X bisector of ~KC.m is the _L bisector of A B . n is the X bisector of ~BV.

Concl.: n passes through P (the point of intersection of I and m).

6

P R O O F

T he set of points that are equidistant from A and B is the

line m, the perpendicular bisector of AB. Similarly, the set of points equidistant from A and C is I. Hence, P, the intersection of I and m, is equidistant from A, B, and C. In particular, it is equidistant from B and C and, therefore, must be a point on n, for the locus of points equidistant from two fixed points is the perpendicular bisector of the line segment joining these two points.

T he proof of this theorem depends upon the fact th a t I and m can be ■ shown to intersect a t some p o in t P. By using the indirect m ethod of proof, it is possible to show that if we accept the possibility that I does no t inter-

sect m, then AC would be perpendicular to m, for if a line is perpendicular to one (0 of two parallel lines (/ and m), then it is perpendicular to the other

also. Hence, AC would have to be parallel to AB as two lines perpendicular to the same line are parallel. This, however, is in contradiction to the Given D ata that A S and AC have the point A in common. Thus, I must intersect m, for the possibility th a t I does not intersect m led to contradictory statements.

I t m ay be interesting to you to compare the proof above with the one you gave for Problem 18, page 190.


TH EO REM 114: T h e bisectors of th e angles o f a t r ia n g le a re co n cu rren t a t a p o i n t th a t is eq u id is tan t fro m th e sides o f th e

triang le.

4Given: i bisects Z A .

m bisects Z B .

n bisects ZC .

Concl.: n passes th rough P (the point of./7. —►

intersection of I and m).

PRO OF—► —■*>

T he locus of points th a t a re equidistant from AC an d A B

is the ray I, the bisector of Z A . Similarly, the locus of points equidistant —* —► “f, .

from BC and BA is m. Hence, P, the intersection of I and m, is the point4—* > 4—►

that is equidistant from AB, CA, and CB. In particular, it is equidistant—> ——* "4

from CA and CB and, therefore, must be a point on n, for the locus of points equidistant from the sides of an angle is the bisector of the angle.

As In Theorem 113 the proof of this theorem depends upon our ability

to show th a t ( and m intersect a t some point P. This can be shown ra th e r easily by the application of Pasch’s Axiom.

EXERCISES

In each of the following problems you are to find the locus in terms of the conditions given. In addition, you are to draw diagram s showing the other possibilities that may arise. Specifically, follow the p a tte rn

presented in Illustration 2, page 535.1. W hat is the locus of points that are equidistant from two parallel lines

and lie on a third line?2. W hat is the locus of points that are a fixed distance.from a fixed point

and lie on a given line?3. W hat is the locus o f points that are equidistant from the sides of an

angle and lie on a given line?

4. W hat is the locus of points that are a fixed distance from a fixed line and lie on a given circle?

5. W 'hat is the. locus of points th a t are equidistant from two fixed points and lie on a given circle?

6. W hat is the locus of points that are a fixed distance, <A, from p o in t A and a fixed distance, di, from point B ?

7. W hat is the locus of points th a t are a fixed distance from the point of intersection of two intersecting lines and lie on these two lines?

8. W hat is the locus of points that are equidistant from two parallel lines and are a t a fixed distance from a given point th a t lies on one of these lines?

9. W hat is the locus of points th a t are equidistant from two fixed points and are a t a given distance from a given circle bu t lie outside the circle?

10. W hat is the locus of points that are equidistant from the points A and B and also from the points A and C? T h a t is, w hat is the locus of points- equidistant from three given points, A, B, and C?

11. W h at is the locus of points th a t are equidistant from two intersecting lines and are a t a given distance from their po in t of intersection?

12. W hat is the locus of points th a t are a fixed distance from p o in t A and equidistant from points A and B?

13. * Using the diagram a t the right, prove that the bisectors of ZEAC,ZD C A, and Z A B C are concurrent.

538 LOCUS; SYNTHETIC GEOMETRY

14.* Using the diagram a t the right, prove that the altitudes of a triangle

«-► .

ft

c

pDiscuss each of the following problems informally; no

proof is required.

1. W hat is the locus of points that are less than a given distance from a fixed iine and that lie on a second fixed line?

are concurrent. (Hint! Let QR be the line through A th a t is parallel to

BC. Same for PQ and PR, Then use Theorem 113.)

STRAIGHTEDGE AND COMPASS CONSTRUCTIONS 539

2. W hat is the locus of points that are greater than a given distance from a fixed line and that are equidistant from two fixed points? (See similar problem in Coordinate Geometry, pages 436-437.)

S. W hat is the locus of points that are less than a given distance from a given point and lie on a given line?

4. A'f is the midpoint of line segment AB. W hat is the locus of points that are not equidistant from A and B and are less than a given distance from M?

5. W hat is the locus of points that are equidistant from two parallel lines and less than a given distance from a given line?

6. W hat is the iocus of points that are greater th an a distance, di, from point A and also less than a distance of d% from point A? (Assume di to be less than di.)

7. In space geometry w hat is the locus of points th a t are a fixed distance from a fixed point and lie on a given plane?

8. In space geometry what is the locus of points th a t are less than a fixed distance from a fixed point and lie on a given plane?

9. In space geometry, w hat is the locus of points th a t are a given distance, di, from a fixed point and a t a given distance, di, from a fixed plane?

10. In space geometry w hat is the locus of points th a t are less than a distance, dh from a fixed point and a t a given distance, di, from a fixed plane?

11. In space geometry what is the locus of points th a t arc equidistant from two fixed points (see Problem 15 page 533) and equidistant from two parallel planes?

12. In space geometry what is the locus of points that a re equidistant from two fixed points and less than a given distance from a given plane?

■ Straightedge and Compass ConstructionsClosely allied to the work on loci is the topic of construc

tion, for it enables us actually to draw the locus by using certain instruments. I t m ay have occurred to you that the num ber of curves considered in this course has been ra ther limited. In fact, they have been lim ited to but two: the linef and the circle. No other curve was exam ined, not because of lack of space but ra ther because of the definition of plane geom etry as agreed upon by Greek m athematicians:

Plane geometry is that branch of m athem atics in which the only instruments permissible are the straightedge and compass.

As you well know, the compass is the instrvment used for drawing circles or arcs of circles. Th< straightedge, on the other hand, is the instrument for drawing lines,

We arc assuming that a line can be considered as a special type of curve.


T he straightedge is very m uch the same as the instrum ent you have come to know as the “ ruler.” However, unlike the ruler the straightedge has no markings on it. Hence, i: cannot be used to measure the lengths o i line segments.

A question such as “ Can the bisector of an angle be constructed?” is largely meaningless until '.ve are told what instruments we are perm itted to use. If none are available, then by all means no construction is possible! However, in our work we will accept the Greek definition of geom etry and consider that the only instruments that are both available and permissible are the straightedge and compass. Thus, the above question should be interpreted as “ Can the bisector of an angle be constructed by using straightedge ?.nd compass only?” T h e answer to this question and others like it will be the subject of this unit.

In ar, tarlier topic t we discussed the existence of certain lines or angles. T he proofs of these “ existence” statem ents were based on three postulates:

(1) A line may be extended as far as desired in either direction.(2) There exists one and only one line through two points.(3) At a given point on a given line there exists an angle whose vertex is the given point and one of whose sides is the given line such th a t this angle is congruent to any given angle.

In this unit we will reprove not only a few of the existence statem ents examined previously but also m any more that we were unable to prove at th a t time. Now, however, our theorems will be phrased in the language of construction.

You may have wondered how it will be possible to construct congruent line segments without the aid of the markings on a ruler. T he next theorem should answer that question.

T H E O R E M 115: At a g iven p o in t on a given lin e a l in e segm en t cong ru e n t xo a g iven lin e segm ent can b e constructed . (How would this theorem have been worded as an existence statement?)

F ig u r e 15-30.

Given: Line segment A BLine I containing point P

T o Construct: A line segment on / a t P th a t is congruent to AB

M e t h o d :

(1) Using A as the center and a radius congruent to AB, draw an arc of a circle.

t See page 255.

( i ) W ith P as the center and the same radius, draw an arc of a circle

intersecting ( a t Q.(3) Then, P Q ^ l B


I P R O O F ! STATEMENTS REASONS j

1. O A S I O P1. Reverse of the definition of congruent

circles

2. P Q = A %2. Definition of congruent circles

TH E O R E M 116: At a g iven p o in t on a g iv en lin e a n an g le co n g ru en t to a given an g le can b e constructed .

G iven: Z BLine I containing point P

T o C onstruct: An angle a t P, one of whose sides is I and that is congruent to Z B .

M e t h o d :

(1) Using B as the center and any convenient radius, draw an a rc in te rsecting the sides of Z B at A and C.(2) Using P as the center and the sam e radius as in step 1, draw an arc

intersecting I a t E.(3) W ith C as the center and S T as a radius draw an arc.(4) W ith E as the center and UA as a radius draw an arc intersecting th e

previous arc a t D.

(5) Using the straightedge,' draw PD.(€) T hen Z D P E ^~ ZA B C

PRO O F | STATEMENTS REASONS |

1. Draw AC and DE. 1. W hy possible?

2 . T A ^ T D (s )2. Def. of congruent circles .

3. W ^ . T E { s ) 3. Same as 2

4. £ 4 = * 1 5 0) 4. Same as 2

5. A A B C S A DPE 5. Why?

6. Z A B C £ ZD P E 6. Why?

542 LOCUS: SYNTHETIC GEOMETRV

T H E O R E M 117: T h e p e rp en d icu la r bisector of a l in e segm ent can be constructed .

A < -

V '

Given: Line segnjent AB To C onstruct: T he perpendicular bisec

tor of AB

4 ?

Figure 15-32.

M e t h o d :

(1) Using A as a center and a radius whose measure is greater than one- half of AB, draw arcs above and below AB,(2) Using 5 as a center and the same radius as in step 1, draw arcs above and below AH, intersecting the previous arcs in P and Q.

<—>(3) Use the straightedge to draw PQ.

_(4) T hen, PQ is the perpendicular bisector of AB.

P R O O F STATEMENTS REASONS j

1. D raw PA, PB, QA, and QB. 1. W hy possible?

2. TA & P B 2. Def. of congruent circles

3. ($A — ($B 3. Same as 2

4. PQ is the perpendicular bi 4. If two points are each equidistantsector of AB. from the endpoints of a line segment,

then the line joining them will be the perpendicular bisector of the line segment.

T H E O R E M 118: T h e bisector of an an g le can b e constructed.

G iven: /.ATo Construct: T he bisector of /.A

Figure 15-33.


M e t h o d :

(1) Using A as a center and any convenient radius, draw an arc intersecting the sides of Z A at- B and C.-(2) Using B as a center and a radius whose m easure is greater than one- half of BC, draw an arc.(3) Using C as a center and the same radius as used in step 2, draw an arc intersecting the one in step 2 a t P.

(4) Use the straight edge to draw AP.

(5) Then, AP bisects ZBAC .

PROOF


THEOREM 119: At a g iv en p o in t on a g iven lin e a lin e p e rp en d icu la r to th is lin e can b e constructed .

G iv en :' Line / w ith point P on I T o C onstruct: A line perpendicular to I

at P

Figure 15-34.

M e t h o d :

(1) Using P as a center and any convenient radius, draw arcs intersecting / at A and B.(2) Using A as a center and a radius whose measure is greater than AP, draw an arc above /.(3) Using B as a center and the same radius as used in step 2, draw an arc intersecting the arc in step 2 at Q.

(4) Use the straightedge to draw QP.

(5) Then, QP J_ t

PRO OF

The proof will be left for you to do.

544 LOCUS: SYNTHETIC GEOMETRYTHEOREM 120: From a g iv en point not on a given lin e a lin e perpen

dicular to the g iven lin e can be constructed.

Given: Line I with P not on I T o C onstruct: A line from P perpendic

u lar to I

M e t h o d :

(1) Using P as a center and a radius, whose measure is greater than the distance from P to I, draw arcs intersecting I a t A and B.(2 ) Using A as a center and a radius whose measure is greater th an one- half of AB, draw an arc below I.(3) Using B as a center and the same radius as used in step 2, draw an arc intersecting the arc in step 2 a t Q.

V(4) Use the straightedge to draw PQ.

(5) Then, PQ X I

PR O O F


Illustration;

Construct an angle of 45°.

A n a l y s is : By using either Theorem 117, 119, or 120 a right angle can be constructed. Theorem 118 is th en used to bisect th a t angle, thus obtaining an angle of 45°.

T o avoid the necessity of having to write out the steps in your m ethod of attack, simply draw the arcs of your circles large enough so th a t it is possible to determine what points had been used as the centers of the circles. Should this be done, an exam ination of your drawing is all that is necessary to understand the m ethotl you used to draw the diagram . C an you tell


from the arcs in Figure 15-36 whether T heorem 117, 119, or 120 was used to construct the right angle?

Z A B C = 45

EXERCISES

1. (a) In Theorem 117, step 1, why was it necessary to use a radius whosemeasure was greater than one-half of AB?

(b ) In Theorem 119, step 2, why was it necessary to use a radius whose measure was greater than AP?

(c) In Theorem 120, step 1, what represents the distance from P to I? Why must the measure of the radius be greater th an this distance?

2. Construct a line segment equal to the-sum of two given line segments.3. Construct an angle equal to the difference of two given angles.4. Divide a given line segment into four congruent segments.5. Draw an obtuse angle and divide it into four congruent angles.

6. (a) Construct an angle of 135°.(b) Construct an angle of 22\°.(c) Construct an angle of 67

7. I f a and b are the measures of two given line segments where a > b, construct a line segment whose measure is equal to {(a — b).

8. If Z A and Z B are two given angles, construct an angle whose measure is equal to \(m Z A + m Z B ).

9. D raw a triangle similar to the one a t the right. Construct the m edian from

C to AB in this triangle.

6 C

10. Draw a triangle sim ilar to the one in Problem 9. Construct the line *

that is perpendicular to AC a t point P. ■


11. D raw a triangle sim ilar to the one in Problem 9. Construct the altitude

from B to A€-12. Construct the angle bisectors of a triangle and find their point of con

currency.

13. (a ) Construct the perpendicular bisectors of the sides of an acute ■triangle. Do the same for a right triangle and for an obtuse triangle,

(b ) W hat conclusion can be drawn concerning the point of concurrency in each case?

14. (a) Construct the altitudes io the sides of an acute triangle. Do thesame for a right triangle and for an obtuse triangle. ■

(b ) W hat conclusion can be drawn concerning the point of concurrency in each case?

15. C onstruct th e locus of points that are equidistant from two intersecting lines.

16. Construct the locus of points that are equidistant from two fixed points.17. Construct the locus of points that are equidistant from two fixed points

and a t a given distance from a given point.18. C onstruct the locus of points that are equidistant from two fixed points

and equidistant from the sides of a given angle.

■ More About Construction with Straightedge and CompassT he great bulk of the problems on construction in plane

geometry are based on the six construction theorems developed in the p receding unit. T he key to determining how a particular construction can be done is to sketch the figure in completed form. An inspection of this drawing will lead to those construction theorems th a t m ust be applied. Application of this m ethod will be m ade to the construction theorems in this unit.

T H E O R E M 121: T h ro u g h a given p o in t no t on a g iven lin e a l in e can b e constructed paralle l to th e g iven line .

Given: Line I and point P not on I

T o Construct: A linethrough P parallel to /

A n a ly sis: By inspecting the completed sketch we realize that were Z l £= Z2, the'lines would be parallel, Hence, the problem simplifies to the point of m erely constructing an angle congruent to a given angle.

CONSTRUCTION WITH STRAIGHTEDGE AND COMPASS 547

M ethod :

(1) Through P draw a line intersecting / a t Q.(2) On this line at point P construct an angle congruent to ZQ(3) Then, m 11 /


1. Z2 was constructed congruent to Z l .

2. / II m

1. An angle congruent to a given angle can be constructed.

2. If two lines are cut by a transversal such that the corresponding angles are congruent, then the lines are. pa rallel.

Before developing the next two construction theorems, several definitions are necessary.

D efin itio n 95: An inscribed polygon in a circle is a polygon whose vertices are points of the circle.

D efinition 96: A circumscribed polygon about a circle is a polygon whose sides are tangent to the circle.

ABC D E is an inscribed polygon, or FG H IJ is a circumscribed polygon, or circle 0 is said to be circumscribed circle P is said to be inscribed withinabout ABCDE. FG H IJ..,

Ju s t as we refer to points that fall on the same line as being collinear points and points that are elements of the same plane as coplaner points, so, too, do we have a name for points that are members of the same circle. These points are called concyclic points T here are a num ber of interesting theorems that can be proved to show when points are concyclic—the most im portant of these being the one in which the vertices of a quadrilateral are concyclic if its opposite angles are supplementary. I t is possible to word the following theorem as, the vertices of a triangle an concyclic.

548 LOCUS.- SYNTHETIC GEOMETRYT H E O R E M 122: A circle can be c ircum scribed about a g iven trian g le .

is circum scribed abou t A A B C

A n a l y s i s : By inspecting th e completed sketch we realize th a t the center of the circle is equidistant from the three vertices of the triangle. This is a direct application of the theorem that the perpendicular bisectors of the sides of a triangle are concurrent at a point th a t is equidistant from the three vertices. T he radius of the circle is then merely the line segm ent from the po in t of concurrency to any one of the. vertices.

M et h o d :

(1) Construct the perpendicular bisector of ~BC.

(2) Construct the perpendicular bisector of AC. This will intersect the line in step 1 at some point 0.

(3) Using 0 as a center and OB as a radius, draw the circle.

’ P R O O F

By using the indirect m ethod of proof it is possible to show th a t points A and C m ust lie on the circle.

T H E O R E M 123: A circ le can b e in sc rib ed w ith in a g iven tr ia n g le .

Given: A ABC To Construct:

A circle th a t is inscribed within A A B C

-A n a ly sis : By inspecting the completed sketch we realize th a t the center of t h e circle is equidistant from the three sides of the triangle. This is a direct a p p lica tio n of the theorem th a t the bisectors of the angles of a triangle are c o n c u rre n t a t a point th a t is equidistant from the sides of the triangle.

T he radius of the circle, as seen in the sketch, is the length of th e perpendicular segment from the center to any one of the sides.

M e t h o d :

(1) Construct the bisector of /.ABC.(2) Construct the bisector of /A C B . This will intersect the line in step 1

at some point I. i—►(3) Construct the perpendicular from I to BC.

(4) Using I B as a radius, draw the circle.

PRO OFT he proof rests upon the need to show th a t both A B and

AC are tangent to the circle. I t will be left for you to do.

The remaining construction theorems are concerned w ith the ability to construct triangles when you are given certain parts of these triangles. In reality, they are but the five congruence statem ents couched in the language of construction theorems.

The need for a simple way to refer to the line segments and angles connected w ith a triangle gave rise to the following sym bols:

CONSTRUCTION WITH STRAIGHTEDGE A N D COMPASS 549

A

(1) For the angles of a triangle the capital letters A, B, and C are used.(2) The side opposite / A is called side a; similarly, b lies opposite / B ,and c opposite /C .(3) T he altitude to side a is while th a t to side b is At, etc.(4) The angle bisector of / A is while that of / B is <b, etc.(5) T he median to side a is m„, while th a t to side b is m», etc.

Thus, to be given a, b, and C would imply that we knew two sides of a triangle and also the angle included between these sides. Similarly, if the given data were In, and c, it would imply th a t line segments ware, given' for the altitude and angle bisector to one side of a triangie; in addition, one of the sides of the triangle was given; however, it was not the side to which the altitude and bisector were drawn.

THEOREM 124: A triangle can be constructed if g iven two sides and the angle included between these two sides (a, b, C).

G iven; a i


To Construct: A triangle with the above elements

sk e tch

5v...JL

M e t h o d :

(1) O n line I select point C.(2) W ith C as one endpoint construct a line segment congruent to b.(3) At C construct an angle congruent to ZC.(4) Wi.th C as one endpoint construct a line segment on m th a t is congruent to a.(5) D raw the line segment joining the endpoints of a and b.

T H E O R E M 125: A triangle can be constructed if given two angles and the side included between these two angles (B, C, a).

G iven: ____ o

To Construct: A triangle with the above elements

sk e tch


M e t h o d :

(1) On line I select point B.(2) A t B c o n s t r u c t a l i n e s e g m e n t c o n g r u e n t t o a.(3) A t B c o n s t r u c t a n a n g l e c o n g r u e n t t o ZB .(4) At C construct an angle congruent to ZC.(5) The intersection of m and n will be the third vertex of the triangle.

A n a l y s is : This construction is possible only if Z B and Z C are so given that m Z B + m Z C < 180. W ere the m Z B + m ZC either equal to or greater than 180, there would exist a contradiction with the theorem that the sum of the measures of all the angles of a triangle must be 180. In thissituation the sum of the measures of only two of them would be either 180or greater than 180. This can not be so.

T H E O R E M 126: A triangle can be constructed i f given three sides (a, i>, c).

G iven: _____ 2____________ I

T o Construct: A triangle w ith the above elements

M ethod :

(1) On line I select point P.(2) At P construct a line segment congruent to a.(3) Using P as a center and c as a radius, draw an arc above line I.(4) Using Q as a center and i as a radius, draw an arc to intersect the arc in step 2 at some point R.

(5) Draw RP and RQ.(6) A RPQ is the required triangle.

A n a l y s is : This construction is possible only if a, b, and c are so given th a t • the sum of any two of these segments is greater than the third. Exam ination of the Sketch reveals the fact that any side of the triangle must be smaller than the sum of the other two sides since the shortest path between two vertices of the triangle is the line segment joining them.

552 LOCUS: SYNTHETIC GEOMETRYT H E O R E M 127: A. tr ia n g le can b e constructed if g iv en two ang les a n d

a side opposite one of them (/I, B, a).

T o Construct: A triangle with the above element

A n a ly sis ; By examining the Sketch we realize th a t the th ird angle of the triangle can be found by subtracting the sum of m ZA and m Z B from a straight angle. T o save time this construction is done a t the third vertex of th e triangle.

This construction is possible only if m Z A + m Z B < 180.M e t h o d :

(1) On line I select point P.(2) At point P construct a line segment congruent to a.(3) At point P construct an angle equal to the sum of angles A and B.(4) At B construct an angle congruent to Z B .(5) The intersection of m and n will be the third vertex of the triangle.

T H E O R E M 128: A r ig h t tr ia n g le can be constructed i f g iv en th e hy po tenuse a n d one of the legs.

Given: ---------- :----- -------------1

CONSTRUCTION WITH STRAIGHTEDGE AN D COMPASS 553

M e t h o d :

(1) At point P on line m construct a perpendicular to m.(2) With P as one endpoint construct a line segm ent congruent to I on the perpendicular constructed in step 1.(3) Using Q as a center and h as a radius, draw an arc intersecting m at R.

(4) Draw QR.(5) Then A Q P R is the required triangle.

A n a l y s is : This construction is possible only if A > !. C an you justify why

this should be so?In applying the five construction theorems just proved to problem

situations, it is best to draw a sketch of the completed figure. O ne of the triangles in this drawing will be such that it can be constructed by using one of these five theorems. M ark this triangle in red and under it write the letters of the construction theorem you p i ; t o use. O nce you have completed the construction of this triangle, the rem aining vertices of the figure you have been asked to construct can usually be found ra ther easily.

Illustration 1:Construct an isosceles triangle given a leg and the a ltitude to the base.

Given: ______ 1________ j

To C onstruct: An isosceles triangle with the above elements

F ig u r e 15-48.

A n a l y s is : Exam ination of the Sketch revealed th a t the triangle a t the right in this sketch can be constructed under the hypotenuse:leg construction theorem. After tins triangle was constructed, the th ird vertex was found by using ^ as the center of a circle and / as the radius. An arc was draw n intersecting m at point R. T h is is th e third vertex of the isosceles triangle.

Construct a triangle giveu m„, a, and b . _______

Given:

554 LOCUS: SYNTHETIC GEOMETRYIllustration 2:

A triangle with the above elements

Figure 15-49.

A n a l y s is : Exam ination of the Sketch revealed th a t th e triangle a t the right in this Sketch can be constructed under the S.S.S. construction theorem, where b, ma, and \a are the three sides. W ith this triangle completed, there was no need to find the third vertex, P, for it was already there as the leftmost point o f segment a.

EXERCISES t

A _

1. Construct an isosceles triangle under the conditions listed.(a ) Given a leg and the vertex angle.(b) Given a leg and the base.(c) Given the base and a base angle.(d ) Given the base and the median to the base.(e) Given a leg and the bisector of the vertex angle.(f) Given the altitude to the base and one of the base angles.(g )* Given the base and the altitude to one of the legs.

sketch

S.S.S.

POK a the required triangle X

t N o p roo f is req u ired for any of the problem s in this g roup of exercises.


a, b, hc U, A, B

2. Construct a right triangle under the conditions listed.(a) Given the legs.(b) Given a leg and the acute angle adjacent to that leg.(c) Given a leg and the acute angle opposite that leg.(d) Given the hypotenuse and one of the acute angles.(e) Given one leg and the median to the other leg.

3. Construct an equilateral triangle under the conditions listed.(a) Given one side.(b) Given an altitude. (Hint: Find a 60° angle by constructing any

equilateral triangle.)(c) Given a bisector of one of the angles.

4. Construct a triangle under the conditions listed.(a) a, hc, c (b) c, hi, B (c)(d) a, b, mb (e) b, U, A (f)(g) ha, ma, a (h) Ac, tc, C (i) hb, tb, c( j) A, B, ha (k) ha, ta, B (1) A, C, hb(m) <1,, B, hc (a) ht, ntt, A (o) * m„, mb, c

5. Construct a parallelogram under the conditions listed.(a) Given two sides and an angle.(b) Given the diagonals and an angle included between them.(c) Given the diagonals and one side.(d ) Given sides a and b and the altitude to side b. (An altitude of a

parallelogram is the common perpendicular to a pair of parallel• sides.)

6. Construct a square(a) Given a side.(b ) Given a diagonal.

7. Construct a rhombus(a) Given a side and an angle.(b ) Given a side and a diagonal.(c) Given the diagonals.(d ) Given an angle and the diagonal to the vertex of that angle.

8. Construct a quadrilateral given sides a, b, c, and d, and also the angle between sides a and b.

9. Draw a circle similar to' the one at the right and construct a tangent to the circle at point P. (H int: See Theorem VO.)

i t

556 LOCUS: SYNTHETIC GEOMETRY10. Construct a chord through a point within a circle such that this point

will be the m idpoint of the chord. (H int: See Theorem 88.)11. D raw a diagram sim ilar to the one

a t the right and construct a tangent to circle 0 that is parallel to line I.

u

/12. Construct a circ le that has a radius

a and is tangent to c irc le 0 a t po in t P.

13. Construct a circle through three noncollinear points. (H int: See T heorem 122.)

14. (a ) Using the markings on the d iagram below as a guide, divide a linesegment into three congruent line segments.

(b ) W hat theorem m ust be used to justify the conclusion that

a p s z p q ^ T Z b

15.* Construct an isosceles triangle given its perim eter and the altitude to the base. (Hint: Construct the altitude a t the midpoint of the perim e ter; then analyze your sketch.)

16.* C onstruct a right triangle given the hypotenuse and the altitude to the hypotenuse.

T1. Construct the locus of points tha t are a g iven distance, d, from a fixed line.2. Construct the locus of points tha t are equidistant from two p ara lle l lines.

TEST AND REVIEW 557

3. Construct the locus of points that are equidistant from two fixed points and also equidistant from two parallel lines.

4. Construct the locus of points that are a given distance, a, from a fixedpoint and also a given distance, b, from a fixed line.

5. Construct the locus of points that are equidistant from two intersectinglines and also equidistant from two parallel lines.

| Test and Review

X1. Given the fact that each of the following statem ents is true, write its

converse, inverse, and contrapositive. Indicate which of your statements is true without need of proof,(a) If Jo h n goes to the movie tonight, then I shall go.(b) If a subject is not difficult, then Bill does not enjoy it.

<-* ___ —*(c) If A B bisects CD, A B will not bisect ZA .

2. State as a conditional statem ent the contrapositive of

“A person who does not like m athem atics should not plan to be an engineer.”

3. (a) Give the definition of a right a;ngle as a conditional statem ent.(b) W rite the inverse of the definition of a right angle.

4. (a) If the converse and the inverse of a statem ent are true, will this implythat the original statem ent is also true? Justify your answer.

(b) If the contrapositive of a statem ent is true, does this imply that the original statement is also true? Justify your answer.

5. W rite three converse statem ents with reference to the statem ent

“U P A S Z P B an d <£? $ B , th en PQ J_ A B ."

L UAnswer each of the following questions w ithout formal

proof:

1. W hat is the locus of the centers of circles th a t are tangent to a given line a t a given point of th a t line5

2. W hat is the iocus of the centers of congruent circles th a t are tangent intem aliy to a given circle?

3. W hat is the locus of the centers of all congruent circles th a t pass through a fixed point?


4. W hat is the locus of the midpoints of all line segments whose endpoints are one of the vertices of a triangle and any point of the side opposite th a t vertex?

5. W hat is the locus of the points of intersection of the diagonals of a rhom bus that have a fixed segment as a side?

6. W hat is the locus of the m idpoint M

of A B where points A and B are always points of each side of right

angle C and m A B is fixed?

7. W hat is the locus of the centers of circles that are tangent to a given circle at a given point of that circle?

8. W hat is the locus of points that are equidistant from the centers of two congruent intersecting circles?

9. W hat is the locus of points that are equidistant from the sides of an angle and lie on another given angle?

10. W hat is the locus of points that are equidistant from the sides of an angle and also equidistant from two parallel lines?W hat is the locus of points that are a fixed distance from a given point and equidistant from the endpoints of a given line segment?W hat is the locus of points that are equidistant from the vertices of a rectangle?

13. W hat is the locus of points that are equidistant from the vertices of an isosceles trapezoid?

14. W hat is the locus of points that are equidistant from the sides of a rhombus?

11.

12.

0Using straightedge and compass only, complete the con

struction in each of the following problems:

1. Construct a triangle under the conditions given.

(a) a, c, ma (b) ht, tt, a

2. Construct a rectangle given a base and a diagonal.3. Construct a rhom bus given one of its angles and the diagonal to the

vertex of its next consecutive angle.4. Construct an isosceles trapezoid given the lower base, a leg, and an

upper base angle.

TRY THESE FOR FUN 559

5. Construct a rectangle given a side and one of the angles formed by the diagonals.

6. Divide a line segment into five congruent segments.7. Using a straightedge and compass, j

find the point on I that is equidistant ___ *8from points A and B.

8. Construct the locus of points that are equidistant from the sides of an angle and at a fixed distance from a fixed lin e .'

9. Through a given point construct a line that intersects a given line in an angle of a given measure.

10. Through a given point construct a line that will cut off congruent segments on the sides of a given angle.

Try These For Fun

If the point of concurrency of the altitudes of a triangle is found and also the point of concurrency of the perpendicular bisectors of the sides, then the distance from the latter to a side is one-half the distance of the opposite vertex to the point of concurrency of the altitudes.

Can you prove this?

Given: Any triangle ABC

OM is the _L bisector of BC.

ON is the J. bisector of AC.AD and B E are altitudes intersecting a t F.

Concl.: A F = 2 OM

Suggestion: Let circle 0 be the circle circumscribed about A ABC.

Let CP be th e ‘line through 0 . Draw PA and PB.


BI t would have been possible for us to have proved the rela

tionships between a theorem., its converse, its inverse, and its contrapositive had we resorted tc “ tru th tables.” This is a device that logicians use to show the equivalence, or nonequivalence, of statements. By this scheme of things, “ tru th values” are assigned to any statem ent; that is, a statem ent is said to be either “T ru e” or “ False.” Thus, the statem ent

p\ “ I shall pass this test”

is either a true statem ent or a false statem ent. We express this by saying th a t the tru th values of p a re T or F , and -write this as

P

TF

Figure 1

Were we now to examine a second statem ent,

q: “ I shall receive a passing mark for this term ”

the tru th values of this statem ent would also be

TF

Figure 2

Similarly, as we recall, i f p represents “ I shall pass this test,” then (read as not p) was interpreted as “ I shall noi pass this test.” And it is possible for us to draw up a set of tru th values for based on the tru th values of p. T o illustrate, if p is true, then ~ p would certainly have to be false; and when p is false, must be true. This idea is expressed in a much, m uch more compact form by establishing a “ tru th table," W hat is the truth

Figure 3

table relating q and ~q ?Further, we m ight be interested in examining all the different combina-'

tions of tru th values th a t exist between two distinct statements. Thus, they


m ay both be true a t the same tim e; they may both be false a t the same tim e; or finally, one may be true when the other is false. Again resorting to tru th tables, the picture would be

p ?

T TT FF TF F

Figure 4

On the first line we note that when p is true, q is also tru e ; on the second appears the possibility of p being true white q is false, and so on. These are the only combinations th a t exist between the tru th values of p and q.

As a final step we would like to determ ine the tru th values that should bc assigned to the conditional statem ent

“ If p then q”

O r as wc had learned earlier, this statement can be expressed symbolically as

/> -* ?Thus, if both p and q are true, would it be advisable to assert that “ If p then q” is also true? Or, if both p and q are false, w hat tru th value should be assigned to “ If f> then q"? Let us examine some definite statem ent before we try to commit ourselves. For instance, Miss M artin said to Joe,

“ If you pass this test, then you will pass for the term ”

where the statements p and q are

p: “ You pass this test” q: “ You will pass for-the term ”

How should Joe interpret the teacher’s honesty under each of the following conditions?

A. (1) Joe passes the test, [p is true.)(2) Jo e passes for the term, (q is true.)

He should certainly agree that Miss M artin told the tru th , for this is exactly w hat she said would happen.

B. ( I ) Jo e passes the test, (p is true.)(2) Jo e does not pass for the term, (q is false.)

l t seems that Joe would be justified in his reaction that the teacher had not told the tru th . 7'hus, if p is true ar.d q is false, then “ If p then q" is a false statement.

C. (1) Jo e does not pass the test, {p is false.)(2) Jo e passes for the term, (q is true.)


Joe would joyously accept the passing grade and point out to ail who would care to listen th a t the teacher said nothing about w hat would happen in the event h r failed the test. Hence, since she is not telling a falsehood, she m ust be telling the t r u th ! Thus, ifp is false and q is true, then “ I f p then q" is true.

D. (1) Jo e does not pass the test, (p is false.)(2) Jo e does not pass for the term. (q is false.)

Jo e is a sad young m an indeed, for by his analysis in C, he has equally well justified that the teacher is still telling the tru th in D for she had m ade no m ention of the consequences should he fail the test. Thus, if p is false and q is false, then “ If p then q" is true.

T he tru th table for “ If p then q" is accepted as

Figure 5

Hence, the tru th value of p —> q is false only when p is true and q is false.T o lend m eaning to the background we have established, we first

define equivalentstatements as statements that have the same tru th values. H ence, the question of the relation of a statem ent to its contrapositive m erely reduces to, “ Do these statements have the same tru th values?” If so, then the tru th of one implies the truth of the o ther or the falsity of one implies the falsity of the other. We have already determ ined the tru th values of p —> q (see Figure 5), and now let us do the same with its contrapositive

(1) (2) (3) (4) (5)

Figure 6

Columns (3) and (4) were baseu on the analysis given to arrive at Figure 3. T o determ ine column (5), we examined columns (3) and (4) only, rem em bering th a t —» ~i> is a false statem ent a t no other time but when ~ y is true and ~ p is false. Now, should we compare the truth values of p —* q with we note that they are identically the same. T h a tis, when p and q are both true, then p -> q and ~ y —* are also true;


similarly, when p is true and q is false, then p —* q is false and so, too, is false; and so on. Thus, p —■> q and its contrapositive

are equivalent statements.

(1) Can you prove that a statem ent and its converse are not equivalent statements? T h a t is, p > q and q —* p do n o t have the same tru th values.(2) Can you prove that a statem ent and its inverse are not equivalentstatements? T h a t is, p —> q and do not have the same tru thvalues.(3) C an you prove that the converse and the inverse of a statem ent areequivalent statements? T h a t is, q —> p and —> ~q are equivalentstatem ents.

16Inequalities

T H U S FA R O U R STUDY O F SY N T H E T IC G E O M - etry has been concerned prim arily with methods for showing conditions under which quantities will be either equal or congruent to one another. On those rare occasions when we have discussed the inequality of the measures of line segments or angles, it has been, to a large extent, only w hether they are unequal, not which of the two is the larger or which the smaller.

In this un it we plan to present six theorems th a t will enable us to prove w hich of two line segments has the larger measure or which of two angles has the larger measure. T he postulates necessary for the proofs of these theorems were presented earlier in our work. They arc

P o stulate 3 0 : Given any two num bers a and b, one and only one of these three relations m ust be tru e : a > b, a = b, a < b. (Existence and Uniqueness of Order)

P ostulate 3 1 : Given any three numbers a, b, and c where a > b and b > e, then a > c. (Transitivity of O rder)T he first of the theorems on inequalities is actually a form of the inverse

of the theorem on the base angles of an isosceles triangle. I t does far more for us, however, than m erely justify that if the measures of two sides of a triangle are unequal, the measures of the angles opposite them will be un-

564

INEQUALITIES 565

equal. The theorem will specifically tell us which of the measures of the two angles is the larger.

Even in terms of the very little work that we have developed thus far in this chapter you may have noticed how cumbersome it was to word some of our statements. In order to greatly simplify the language to be used throughout this chapter, we will agree upon the follow ing:

(1) Whenever we refer to the fact that one angle is larger than another, it will imply that the measure of the first angle is larger th an the measure of the second angle.(2) W henever we refer to the fact that one line segm ent is larger than another, it will imply th a t the measure of the first line segm ent is larger than the measure of the second line segment.

THEOREM 129: If two sides of a triangle are unequal, the angles opposite them are unequal and the angle opposite the greater side is the greater angle.

Given: A A B C w ith A B > AC Concl.: m Z \ > m Z l


1. A ABC with A B > AC

2. Let AC be extended to point P such that AP = AB.

♦ 43. Let BP be the line through

points B and P.4. m ZA > m Z P

5. m Z A B P = m Z P

6. m ZX > m Z A B P7. But m Z A B P > m Z l

8. .'. m Z \ > m Z 2

1. Given

2. A line can be extended as far as desired in either direction.

3. T here exists one and only one line through twr> points.

4. T he measure of an exterior angle of a triangle is greater than the measure of either of the rem ote interior angles.

5. Theorem on the base angles of an isosceles triangle.

6. Substitution postulate7. The whole is greater than any of its

parts. .8. Postulate on transitivity of order

(Postulate 31)

566 INEQUALITIES

The converse of Theorem 129 is also true. Its proof, as we have often found to be the case with converse theorems, is by the indirect approach, Nov/, however., we will have to apply the assumption on the existence and uniqueness of order between two quantities rather than the law of the excluded middle. We are forced to take this position, for we are concerned w ith more than just the fact that two quantities are unequal but with which of these is the greater.

T H E O R E M 130: I f two angles of a tr ia n g le a re u n e q u a l, th e sides opposite them a re u n eq u a l a n d th e side opposite th e g re a te r ang le is th e g re a te r side.

Figure 16-2.

P R O O F

By the postulate on the existence and uniqueness of order one of the following statem ents must be true and no o ther possibility exists:

(X ).A B > AC, (2) AB = AC, or (3) A B < AC

O rder (1) will be shown to be true by proving that both order (2) and order (3) lead to contradictory statements.

Part 1

Let i<s accept the possibility that A B = AC. If this is so, then m ZC *» m Z B , since if two sides of a triangle are congruent, the angles opposite them are congruent. T he Given Data, however, states th a t m Z C > m Z B . Hence, accepting the possibility that A B .= AC led to the logical inconsistency of the truth of both m Z C > m Z B and m Z C > m Z B . By the law of contradiction both cannot be true at the same time. Since m Z C > m Z B must be true by virtue of the Given D ata, then m Z C > m Z B m ust be false, and so, therefore, is A B = AC false.

Part 2■ ■ ■ ■ ■ ■ ■ Let us now accept the possibility that A B < AC. If this is so, then m ZC < m Z B by Theorem 129. But the Given D ata states th a t m Z C > m Z B . Hence, the possibility that A B < AC led again to the logical inconsistency of the tru th of both m Z C > m Z B and m Z C > m Z B .

INEQUALITIES 567

Since m ZC > m Z B must be true by virtue of the Given Data, thenm ZC > m Z B must be fake, and so, therefore, is AB < AC false. .

Thus, we can conclude that A B > AC, for it is the. only remaining possibility.

T he next two theorems on congruency bear a resemblance to two of the congruency statements.

TH E O R E M 131: I f two sides of one tr ia n g le a re co n g ru en t respec tive ly to two sides o f a second tr ia n g le b u t th e in c lu d e d ang le of the first tr ia n g le is g re a te r th an the in c lu d e d ang le of th e second tr ia n g le , th e n th e th ird side of th e first trian g le is g re a te r th an th e th ird side of th e second trian g le .

Given: A A B C and D EF A B ^ M A U ^ W m Z B A C > m Z D

Concl.: B C > EF


1. m Z B A C > m Z D 1. Given

2. W ithin Z B A C construct AQ so 2. An a:..^le congruent to a giventhat Z Q A B £* ZD . (a) angle can be constructed.

3. Extend /4Q to point Q so that 3. A line can be extended as far a

A Q = D F (s). (The proof does desired in either direction.no t depend on whether Q fallsin the interior or exterior ofA A B C .)

4. L et AP be the bisector of ZQAC. 4. W hy possible?

5. Let QP be the line through 5. W hy possible?points P and Q.

6. I b s M ( i ) 6. Given7. A A B Q Z S A D EF 7. S.A.S.

8. m & D F 8. See statem ent 3.

9. A C .z z D F 9. Given

10. A § & AC (s) 10; Why?

568 INEQUALITIES

11. ZQ A P 3? ZCAP (a)12. A F ^ A ? ( s )13. A A Q P ^ A A C P 1415. In A BPQ, BP + PQ > BQ

16. But BQ = EF17. And PQ = PC18. .'. BP + P C > EF19. But, BP + PC=> BC20. Hence, BC > EF

Figure 16-3.

U . W h y ?

12. Why?13. S.14. Why?15. T he shortest path between two

points is the line segm ent jo in ing the two points.

16. Why? (See step 7.)17. Why? (See step 14.)18. Substitution postulate19. Def. of the sum of two segments20. Same as 18

THEOREM 132: If two sides o f one triangle are congruent respectively to two sides o f a second triangle but the third side o f the first triangle is greater than the third side of the second triangle, then the angle opposite the third side o f the first triangle is greater than the angle opposite the third side of the second triangle.

Given: A A BC and D EF with A n __ __A B S ^ D E A C ^ D F BC > EF

C oncl.: m ZA > m Z DC E

Figure 16-4.

PR O O F

By the postulate on the existence and uniqueness of order one of the following statements m ust be true and no other possibility exists:

(1) m Z A > m ZD , (2) m Z A — m ZD , or (3) m ZA < m Z D

O rder (1) will be shown to be true by proving that both order (2) and order (3) lead to contradictory statem ents.

Part 1■ ■ ■ ■ ■ ■ ■ I Let us accept the possibility th a t m ZA = m Z D , From the

Given D ata A B ~ B E and AC S DF. Hence, by the S.A.S. congruence

INEQUALITIES 569

postulate it follows that A A B C £= A D E F . Therefore, BC = EF. However, the Given D ata states that BC > EF, Hence, accepting the possibility that m Z A = m Z D led to the logical inconsistency of the tru th of both BC > EF and BC > EF. By the law of contradiction both cannot be true at the same time. Since BC > EF is true by virtue of the Given D ata, then BC > EF must be false and, therefore, so must m ZA = m Z D be false.

Part 2

■ ■ ■ ■ ■ ■ ■ Let us now accept the possibility that m Z A < m ZD . If

this is so, and also A B S D E and AC = DF, then by T heorem 131 BC < EF. Again, however, the Given D ata states th a t BC > EF. Hence, accepting the possibility that m Z A < m Z D led to the logical inconsistency of the tru th of both BC > E F and BC > EF. By the law of contradiction both cannot be true at the same time. Since BC > E F is true by virtue of the Given Data, then BC > E F must be false and, therefore, so must m ZA < m Z D be false.

Hence, m Z A > m Z D m ust be true, for it is the only remaining possibility.

T he last two theorems to be proved on inequalities provides us with a tool for com paring the measures of chords of a circle. T heir proof, however, is dependent upon a statem ent that we have not had.

THEOREM 133: If unequals are subtracted from equals, the differences w ill be unequal in the reverse order.

Given: a = b c > d

C oncl.: a — c < b — d


1. d < c2. b = a3. d — b < c — a

4. a — c < b — d

1. Given2. Given3. If equals are subtracted from un

equals, the differences will be un equal in the same order. (Postulate37)

4. If equals are added to unequals, the sums will be unequal In the same order. (Both a — c and b — d were added to both sides of the inequality in step 3.)

570 INEQUALITIESA simple num erical illustration of this theorem is

10 = 10 7 > 2

Hence 10 — 7 < 10 — 2or 3 < 8

W hereas 7 was greater than 2 with the greater quantity , 7, appearing on the left, after subtracting each of them from 10 and 10, the differences were such that the smaller quantity, 3, now appears on the left. Thus,

' 3 < 8

T H E O R E M 134: If two chords of a circle a re u n e q u a l, th en th e g re a te r chord is th e sm aller distance from th e cen ter.

Given:

Concl.

«-vOD L A B in OO

OE L BCAB > BCOD < OE

A n a l y s is : Although the chords do not have to have an endpoint in com m on, for simplicity of proof they were drawn in th a t m anner. In addition, this theorem will hold whether the chords appear in the same circle, as above, o r in congruent circles.

PROOF STATEMENTS REASONS4—¥

1. Let D E be the line through points D and E.

1. W hy possible?

2. OD L A B , OE ± BC 2. Given

3. OD bisects A B \ 3. A radius perpendicular to a chord<-> _OE bisects BU. bisects the chord.

4. But A B > BC 4. Given5. D B > B E 5. If unequals are divided by equals,

the quotients will be unequal in the same order.

6. tn Z 2 > tn Z l 6. If two sides of a triangle are u n equal, the angles opposite them are unequal, and the angle opposite the greater side is the greater angle.

INEQUALITIES 571

7. Z O D B and ZO EB are right angles.

8. m /.O D B = tn ZO EB9. tn ZO E D < m ZO D E

10. OD < OE

7. Def. of perpendicular lines

8. Why?9. If unequals (m Z2 and m Z l ) are

subtracted from equals (m Z O D B and m ZO EE ), the differences will be unequal in the reverse order.

10. Converse of reason 6

T H E O R E M 135: I f two chords of a c irc le a re un eq u al, th en the ch o rd th a t is the g re a te r d istance from th e cen ter is th e sm aller chord.

G iven:

Concl.:

OE L AB

OF L C D OE < OF A B > CD

PROOFBy the postulate on the existence and uniqueness of order

one of the following statements must be true and no other possibility exists:

(1) A B > CD (2) A B = CD (3) AB < CD

O rder (1) will be shown to be true by proving that both order (2) and order (3) lead to contradictory statements.

Part 1

Let us accept the possibility that AB — CD. From the theorem that congruent chords of a circle are equidistant from the center of the circle, it follows that OE = OF. However, the Given Data states that O E < OF. Hence, accepting the possibility that AB = CD .led to the logical inconsistency of the tru th ol both OE = OF and OE ?±.OF. By the law of contradiction, both can not be true a t the same time. Since OE < OF by virtue of the C-iven Data, then OE = OF must be false and, therefore, so must A B — CD be false.

Part 2(The proof of this part is left for you to do.)

572 INEQUALITIES

Illustration:

Given: A A B C is equilateral. C oncl.: BD > AD

A n a l y s i s : T o s h o w t h a t BD > AD, i t is n e c e s s a r y t o p r o v e t h a t /.D A B > / B .


1. A A B C is equilateral.2. m /C A B = m / B

3. m /D A B > m /C A B

4. tn /D A B > m / B5. BD > AD

1. Given2. An equilateral triangle is equiangu

lar. (See Problem 20, page 152.)3. The whole is greater than any of its

parts.4. Substitution postulate5. If two angles of a triangle are un

equal, the sides opposite them are unequal, and the side opposite the greater angle is the greater side.

EXERCISES

1. Given : A A C B is a right triangle with /A C B the right angle.

Concl.: A B > AC AB > BC

Given: / A B C is an obtuse angle.

Concl.: AC > AB A C > CB

2.

3 . Given: A ABC is isosceles with A B — AC.

BA extended to D C oncl.: BD > DC

INEQUALITIES

D

5. Given: A A B C is isosceles with A B = AC.

A B extended to D Concl.: tn /A C D > tn / D

"7, Given: A B Jl CD BD > BC

Concl.;. AD > AC (H int:

Construct B E S BC)

Given: A A B C is isosceles 4. with A B = AC.

CD draw n within /A C B

Concl.: DC > D B

573

A

Given: A B i . B D 6<

Concl.: AD > AC (H in t:Prove th a t / \ is obtuse.)

Given: AC > A B 8«

BD bisects /A B C .

CD bisects /A C B .Concl.: D C > DR

574 INEQUALITIES

9. Given: A ABC is isosceles with A B ~ AC. m Z l > m Z2

Concl.: BD > DC

11. Given: A M is m edian to BC. AC > AB

Concl.: m ZA M C >771 Z A M B

13. Given: ABCD is a parallelogram.77i Z B A D >771 Z 4 Z > C

C oncl.: BD >

Given: A ABC is isosceles 10. with A B ^ ~AC.DC > BD

C onci.: 77i Z D AC >77i Z D A B

Given: M is the m idpoint 12. of ~BC.

i?Z> s UE D M > E M

C oncl.: AC > A B

Given: D B = EC 14.DC > B E

C oncl.: AC > A B

<—15. Given: AO intersects 0 0 a t B.

Concl.: AC > AB (H int: Use Postulate 19.)

INEQUALITIES

17. Given: A A B C is isosceles with AB S AC.—►

BP intersects

A c at R.C oncl.: PB > PC

19. G iven: © 0 with m Z2 > m Z l

C oncl.: 77i DC > m

Given: Point P is any point in the interior of the triangle.

C oncl.: at 4- y + z >+ b -f- c) (H int:

Use Postulate 19.)

16.

575

Given: A A P C is isosceles 18. with A B S AC.4—AC extended to D

Concl.: BD > CD (H int:Prove m Z l > 77i ZZ.)

G iven: M is jhe m idpoint 20/ of BC in 0 0 .

OA J . D E Concl.: D E > B C

576 INEQUALITIES

21. G iven: O 0C oncl.: m / P C S > tn / P B C

(H im : Draw OC.)

23. If light, being reflected from the m irror surface BC, travels from point A to poin t D, it will take the pa th APD. Show th a t this pa th is shorter than any other path , such as AQD.

Given: GO with AC > AD 22. Concl.: m /.D A B > m /C A B

(H int: Mo other lines are needed.)

Given: PQ J . plane a

R Q > QS Concl.: P R > P S (H int:

Use same m ethod as in problem 7.)

24.

31. T h e sum of; two sides vf a triangle is greater than the third side .t2. If a triangle is no t isosceles, then a m edian to any side is greater than the

a ltitude to th a t side.3. (a ) How would you express Problem 19 in Group A in the form of the

statem ent of a theorem?(b ) W rite the converse of your answer to Problem 3(a) and prove it.

4. Using the information in Problem 3(b), prove th a t if two m inor arcs of a circle are unequal, then the chord corresponding to the larger arc is greater than the chord corresponding to the smaller arc.

5. W rite the converse of Problem 4 and prove it.

t T h is p rob lem often appears as a theorem .

TEST AND REVIEW577

6.* A point is in the interior of a circle bu t not a t the center. T he smallest chord that can be draw n through this point is the one th a t is perpendicular to the radius through the point, (H int: Use Postulate 32.)

J | Test and Review


41. Given: A B and CD intersect a t E.

Concl.: A B + CD > AD + CB (H in t: See Problem 1,

page 576.)

3. Given: D C > A B |

B D bisects /A B C .

Concl.: DC > D A (H in t: Prove

tn / \ > m / 2 . )

Given: AC > A B 2 .D is any point of BC.

C oncl.: AC > AD

Given: OO w ith CD > EF 4.A U & U S

OA J . E F

O B I CD Concl.: m /A C O > m /B C O

578 INEQUALITIES

5. Given: ABCD is a rhombus. BC > AC

C orel.: m Z D AD > m ZAD C

Given: ABCD is a paral- 6.lelograro.m /.B A D > rr. /C 'B A

Concl.: B E > A E


1. T h e sum of the diagonals of a quadrilateral is less than the sum of the sides.

2. If a triangle is not isosceles, then the angle bisector of any angle of the triangle is greater than the altitude from that vertex.

3. If two oblique lines and a perpendicular are drawn to a plane from an external point, the greater oblique line will intersect the plane a t a greater distance from the foot of the perpendicular than the smaller oblique line. (H int: Use the indirect proof and apply Problem 24, page 576.)

■ Try This For Fun

For many, m any hundreds of years m athem aticians and pseudom athem aticians have tried to solve three construction problems th a t have come to be known as th e “T hree Fam ous Problems of A ntiquity .” T hey are

(1) T h e squaring of a circle.(2) T he duplication of a cube.(3) T h e trisection of an angle.

Specifically, these are the problem s:

(1) Is it possible to construct a square whose area will be the same as th e a rea of a given circle?(2) Is it possible to construct a cube whose volume is twice the volume of a given cube?(3) Is it possible to construct an angle whose measure is one-third the m easure of a given angle?


O ur first reaction is shocked surprise: of course it is possible to do these things! W hy should m athematicians have been puzzled over these constructions for so long a time? Thus, in answer to problem (3) all we need do is measure the angle with a protractor, then but divide that num ber by 3 and,lo, we have an angle one-third the measure of the original angle! But hold, we have been too hasty and failed to listen to the statem ent of the complete problem. Not only must these figures be constructed but our freedom of movement has been severely restricted: these constructions m ust be done with the aid of only two instruments, the straightedge (no markings on the straightedge, please!) and the compass.

M athem aticians have proved by means beyond our present depth of the subject that these constructions are impossible with the restrictions placed upon them. Has this deterred the would-be angle trisectors from busily plying their trade? By all means, no! For year after year newspapers in various parts of the nation report the am azing success achieved by local Jo e Spivis, the child prodigy,- who has trisected the angle, succeeding where m athem aticians for centuries have failed!

L et us take a look a t one of these constructions.

/A B C is the given angle. A t B construct D B JL BC; a t A construct A E X BC;

a t A construct A H II BC. Hold the ruler at B and move it in a way such that F H — 2 AB, then draw BH.

(1) C an you prove that m /H B C = \m /A B C ?(2) W here in the construction did we cheat and not use the instruments to w hich we were restricted?

Areas of Polygons and Circles

M EA SU R EM EN T, E X C E PT FO R VERY FEW SIT - uations, is the process of com paring the object whose size we would like to know to an object_whose size we do know. T h e object whose size is known is called standard unit, j i t is determ ined, usually, by some governmental decree orppo5stbfyr5y"a m athem atical definition. In any event, its size is fixed an d the m easurement of o ther similar objects consists in determ ining how m any standard units are contained in the object being measured.

Thus, the measure of a line segment is a num ber that shows a comparison betw een this line segm ent and th e standard un it called, perhaps, th e inch, the yard, tl)f ceritimgter, o r any one of m any others. The statem ent

That the length of a line segm ent is 15 feet implies that the measure of this segm ent is 15 times the m easure of th e standard unit called the foot. T h e im portan t feature, however, is the fact th a t both the object and the standard un it are creatures of the same classifications; they are both line segments.

Now, le t us tu rn to th e subject m atter of this chapter. O ur first objective is to determ ine a method for assigning a num ber to the region enclosed by such curves or polygons as those below. Since the word “ regao^w H l appear several tim es in this chapter, w e should reach someamderstanding as to its m eaning. Although we can and will define ,a triangular region, defining the regions such as in Figures 17-1 and 17-2 w tid d involve ~GT in more diffi-

AREAS OF POLYGONS AND CIRCLES 581

culty than is w arranted by the importance of this word. W e will accept an intuitive understanding of the general term “ region” based on the definition of a triangular region.D e f in it io n 97: A t r i a n g u l a r r e g io n is th e u n io n o f t h e s e ts o f p o in t s c o n s i s t

in g o f t h e t r i a n g le a n d i t s in t e r io r .

In the illustration given earlier concerning the length of a line segment the standard un it belonged to the same classification, a line segment, as the object being measured. But to determine a unit of m easure for the creatures above seems as though it would be an impossible task, for they obviously belong to different categories. It is not the objects themselves, how ever, for which we are seeking a standard of measure, but ra th e r the region bounded by these creatures! Hence, any object that itself bounds part of the plane m ight act as our standard of measure. Several of these are pic

tured in Figure 17-4.

Figure 17-4.

Thus, the second of these m ight be called the triangu lar un it; the fourth, the circular unit; the fifth, the hexagonal unit. And to say th a t the size of the region bounded by the curve in Figure 17-1 is 9 triangular units, would, as before, imply

(1) that the size of the region enclosed by the curve is 9 times as great as the region enclosed by the sides of the triangle or(2) th a t the triangle can be made to fit exactly 9 times into the regionenclosed by the curve. A>C > ’f ' i , J C

Sim ilar comparisons could be made were we to use any of the rem aining four units. As you well know from other courses in m athem atics, the square unit was selected to be the standard for m easuring the size of the region bounded by closed curves or polygons.

D e f in it io n 98; T he area of the region enclosed by a curve or polygon is , the num ber of square units contained within this region.

582 AREAS OF POLYGONS AN D CIRCLESP o s t u l a t e 42: If the intersection of two polygons is a line, then the. area

of the region bounded by these polygons is the sum of the areas of the two polygons.

I t is im portant that wc realize that just as the measure of a line segment was a num ber, and the m easure of an angle was a num ber, so too is ihs area of a region simply a number. It is the num ber assigned to that region in terms of the num ber of square units th a t are contained in the region. As you recall, the measure of a line segment was the num ber assigned to that line segment by the coordinate of one of its endpoints when th e other endpoint was the zero value on the num ber line.

T o simplify the way of expressing ourselves, we shall henceforth speak of finding the “ area of a triangle” or the “ area of a rectangle” ra ther than the “ area of a triangular region” or the “ area of a rectangular region” as we should.

As has been the case over and over again in the development of each segment of this course, it has been necessary to assume certain properties about some of the simpler figures. Having made these assumptions, we were then in a position to prove more complex properties about more complex figures. So, too, is our present problem. O f the m any polygons whose area we m ay have to investigate, the simplest of these is the rectangle.

O bservation indicates that the num ber of square units in the rectangle can be found by counting th e num ber in the first row and then m ultiplying this num ber by the num bst-ef-wass. Thus, we would say th a t the area of this rectangle is 24^(square inchesyor that the area of this rectangle is

V ■X.

8 inches Figure 17-5.

24 times the area of 1 square inch. This same num ber, however, could have been obtained by finding the product of the measure of its base, 2?C, with th a t of its altitude, ZX?. Hence, we seem to be on fairly safe ground if we assume that

P o s t u l a t e 43; T h e area of a rectangle is equal to the product of the measures of its base and altitude.

I t is im portant to realize th a t it is not 8 inches th a t is being m ultiplied by 3 inches, but simply 8 by 3. This product, 24, being the area, implies either

(1) th a t there are 24 square inches in the region enclosed by the rectangle or

AREAS OF POLYGONS AND CIRCLES 583

(2) th a t the region enclosed by the rectangle is 24 times as large as theregion enclosed by 1 square inch.

EXERCISES

1. Find the area of a rectangle whose base is 10 yards 1 foot and whose

altitude is 6 yards.2. T he floor of a rectangular living room was scraped and finished a t

15 cents per square foot. The dimensions of the room are 18 feet by 12 feet 4 inches. W hat was the total cost of this work?

3. T h e dimensions of an asphalt tile are 9 inches by 9 inches. T he tile selected to be laid on the floor of a rectangular/deft costs 9 cents per tile. Assuming no waste, w hat was the cost of the tiles for this room whose

dimensions were 17'3’' by 12'6"?4. If each square below is considered as a square unit, determ ine the

5. If each side of a square is a linear units in length, w rite a form ula expressing the area of a square in terms of a.

6. (a) T he dimensions of a rectangle are 32 inches by 2 inches. W hat isthe length of the side of a square whose area is equal to th a t of this rectangle?

(b ) If the dimensions of the rectangle had been 8b by 2b, w hat w ould the length of the side of the square have been?

7. (a) T he area of a rectangle is 216 square feet. W hat is the base of therectangle if its altitude is 18 feet?

(b ) If the area of the rectangle was 1 2 square units, w hat would the altitude be if the base was 3a linear units?

8. Find the area of the rectangle whose diagonal is 39 inches and whose base is 36 inches.

9. Find the area of the square whose diagonal is 6 inches.4 10. A rectangle is inscribed within a circle having a diam eter of 26 inches.

W hat is the area of the rectangle if its altitude is 10 inches?11. (a) The bases of two rectangles are each 15 inches. If the a ltitude of

the first is 7 inches while that of the second is 10 inches, w hat is the ratio of the areas of these rectangles?

584 AREAS OF POLYGONS AND CIRCLES

(b ) T h e bases of two rectangles are each a linear units. If the altitude of the first is b l in ' i r units while that of the second is c linear units, w hat is the ratio of the areas of these rectangles?

(c) M ake up a scatement showing what you have proved in pa rt (b) of this problem .

12. Using the m ethod presented in Problem 11(b), prove that if the altitudes of two rectangles are congruent, then the ratio of their areas is equal to the ratio of the measures of their bases.

13. T h e area of a square inscribed in a circle is 64 square inches. W hat is the radius of the circle?

14. T h e m easure of the side of one square is three times the measure of the side of a second square. W hat is the ratio of their areas?

15. If the measure of the diagonal of a square is a, w hat is the area of the square in terms of a?

■ Area of the Parallelogram, the Triangle, and the TrapezoidT here are only three special polygons, other than the

rectangle, whose areas are considered im portant enough to investigate. These polygons are the parallelogram , the triangle, and the trapezoid. T h e area of any other polygon is found by drawing lines so_as_to divide it into a com bination of these four polygons.

You m ay have noticed that When we expressed the area of a rectangle, the statem ent was in terms of the “ base” and “ altitude” rather" than in terms of the “ sides.” Normally we think of BU (see Figure 17-6) as being

F ig u r e 17-6. F ig u re 17-7.

the base, while either AB or DC is the altitude. Had the rectangle been ro tated 90° in a counterclockwise direction, as in Figure 17-7, the base w ould then appear to be A B and the altitude BC. Thus, the roles of the measures of the segments would have been reversed. It is apparent, then,

AREA OF PARALLELOGRAM, TRIANGLE, TRAPEZOID 585

th a t any side of a rectangle can be considered as its base, while one of its ad jacent sides is the altitude.

Sim ilar considerations are m ade when we exam ine the areas of the parallelogram, the triangle, and the trapezoid. If in the parallelogram (Figure 17-8) A B is referred to as the base, then EF, the common perpendicu lar segment to AB and DC, is its corresponding altitude. If on the other h and BC is the base, then Cttf m ust be considered as the altitude of the par-

— • • . i • i _ t xi. _allelogram. In the same way,

Do t r y L

I


Figure 17-10,

triangle can be considered as the base of the triangle. O nce the base has been specified, then the a ltitude of the triangle is simply the a ltitude to th a t side. Thus, if PR is called the base, the altitude of the triangle would be

Q T. If RS is the altitude of the triangle, w hat will the base be?In a trapezoid, the bases, as before, are still the parallel sides, while

the a ltitude is the common perpendicular segment to the bases. In Figure

17-10 the bases are K J and GH, while LM is the altitude.Ju s t a word about the symbols that will be used. T he statem ent th a t

the area of polygon ABCD is equal to the area of polygon X Y Z W T will be abbreviated to read as, ABCD = X Y Z W T . I t is im portant to rem em ber th a t this implies only that the areas of these two polygons are the same.

One further postulate still remains to be stated before it is possible to continue.P ostulate 44: If two triangles are congruent, then their areas are equal. T H E O R E M 136: T h e a rea o f a p a ralle log ram is equal to th e p ro d u c t

of the m easures of its base a n d co rresp o n d in g a lt i tu d e ' (A - bk). — ...............................


Given: O A B C D with altitude A E Concl.: Area CJABCD = A D -A E

Figure 17-11.

A n a l y s i s : T h ere is bu t one postulate a t our disposal upon which to base the proof of this theorem. This concerns itself with the area of a rectangle. Hence, it is necessary to show that some relationship exists between the area of a parallelogram and that of the rectangle. This we will do by proving th a t ABCD = AEFD.

( p r o o f STATEMENTS REASONS

1. ABCD is a parallelogram . 1. Giveni—►

2. Let DF be the line through D 2. W hy possible?>

parallel to AE.

3. Extend BC until it intersects 3. Why possible?<—►

DF.

4. A E is an altitude. 4. Given5. AEFD is a rectangle. 5. Reverse of def. of a rectangle6. A E == D F (/) 6. Opposite sides of a parallelo

gram are congruent.1. A B ~BC (h) 7. Same as 68. A A B E — A D C F 8. H.L.9. .-. A A B E =. A DCF 9. If 2 triangles are congruent,

then their areas are equal.10. AECD = AECD 10. Reflexive property of equality11. A A B E + AECD = 11. Addition postulate

AECD + A DCF12. But 12. Postulate 42; see page 582.

A A B E + AECD = O A B C D13. AECD + A DCF = 13. Same as 12

rectangle AEFD14. O A B C D = rectangle AEFD 14. Substitution postulate15. However, 15. Why?

rectangle AEFD = A D -A E16. ABCD = A D -A E 16. Substitution postulate

T H E O R E M 137: T h e a rea o f a tr ia n g le is e q u a l to o n e -h a lf th e p ro d u c t of the m easures of its base a n d a ltitu d e (A = \b h ).

A D

Given: A A B C w ith a ltitude A E to BC Concl.: A rea A A B C ~ j(B C -A E )

B E C

F ig u r e 17-12.



1. L et A D be the line through 4i—►

parallel to BC.

2. Let CD be the line through C

parallel to AB.3. ABC D is a parallelogram.4. A E is an altitude to BC.5. CJABCD = BC X AE

6. O A B C D = A A B C + A CD A7. A A B C + A CDA = B C -A E8. But A A B C £* A C D A9. A ABC = A CDA

10. A A B C + A A B C = B C -A E or, ABC = B C -A E

11. Thus, A A B C = j(B C -A E )

T H E O R E M 138: T he a rea o f a trap ezo id is equal to o n e -h a lf th e p ro d uct of the m easures of its a ltitu d e an d th e sum of th e m easures o f its bases; A = \h (b i + bi).

Given: T rapezoid ABCD 'w ith altitude A E to the bases AD and BC

C oncl.: Area of trapezoid ABCD = \A E {B C + AD)

F ig u r e 17-13.


1. AE is an altitude to bases of trapezoid ABCD.

4—►2. Let AC be the line through

points A and C.<->

3. Extend AD.

4. Let CF be the line through>

C perpendicular to AF.

5. AE j| CF

6. AECF is a parallelogram .7. A E = CF8. Area A C D A = \C F -A D9. Area AC D A = \A E -A D

10. Area A ABC — \A E -B C11. A A B C + A C D A =,

IA E -B C + { A E -A D12. ABCD = A A B C + A CDA13. ABCD = | A E -B C + i A E -A D or

ABCD = \A E (B C - f AD)

588

Illustration:AREAS OF POLYGONS AND CIRCLES

P R O O F | ' STATEMENTS•f—>

1. L et A D be the line through

A perpendicular to BC.

2. A M is the m edian to

3. M is the m idpoint of BC.4. B M = M C5. A A M B = IB M -A D6. A A M B = iM C -A D7. A A M C iM C -A D

A A M B = A A M C

Given: A ABC with m edian A M Concl.: Area A A M B = area A A M C

REASONS

1. W hy possible?

2. Given

3. Def. of a median4. Def. of a midpoint5. Theorem on area of a triangle6. Substitution postulate7. Same as 5

Transitive property of equality------- v ui e q u a l i ty

T H E O R E M 139: I f two triang les have co n g ru en t bases a n d c o n g ru en t a ltitu d es, th en th e ir areas w ill b e equal.

T h e proof of this theorem is very m uch the same as th a t of the illustration above. I t will be left for you to do.

EXERCISES

1. F ind the area of a triangle whose base is 42 inches and whose a ltitude is 17 inches.

2. (a ) T h e legs of a right triangle are 5 and 12 respectively. F ind thearea of this right triangle.

(b ) Find the altitude to the hypotenuse of this right triangle.

3. I f th e hypotenuse and one leg of a right triangle are 52 and 48 respectively, find the area of the triangle.

4. F ind the a rea of a rhom bus whose diagonals are 16 and 20 respectively.5. T he a rea of a triangle is 195 square inches. I f the altitude is 15 inches,

w h a t is the length of the base?

6. T h e a ltitude to th e hypotenuse of a right triangle divides the hypotenuse


into two segments of 8 inches and 2 inches in length. Find the area of the triangle.

7. If a leg and the base of an isosceles triangle are 10 and 12 respectively, find the area of the triangle.

8. The hypotenuse of an isosceles right triangle is 8. Find the area of the triangle.

9. Find the area of an equilateral triangle under each of the following conditions:(a) if a side is 6 inches. (b) if an altitude is 6 inches.

^ 10. T he length of the line segment from a point to the center of a circle is 25. If the diam eter of this circle is 14, wha<v;s the area of the triangle whose sides are this segment, the tangent segm ent from this point, and the radius to the point of contact of the tangent?

11. T he area of a rhombus is 442 square feet. If one of the diagonals is 34 feet, what is the length of the other diagonal?

/ 12. Two sides of a triangle are 25 and 26, while the a ltitude to the th ird side is.24. Find the area of the triangle.

13. A triangle is inscribed in a circle such that one of its sides is a diam eter of the circle. If the radius of the circle is 30.5 inches and one of the sides is 60 inches, find the area of the triangle.

✓ 14. Two adjacent sides of a parallelogram are 8 and 14 respectively. If the angle between them is 45°, what is the area of the parallelogram?

> 15. T he bases of a trapezoid are 8 and 11 respectively, while the a ltitude is 6. Find the area of the trapezoid.

16. T he area of a trapezoid is 42 square feet, while its upper and lower bases are 6 feet and 7 feet respectively. W hat is the length of the altitude of the trapezoid?

17. T he median of a trapezoid is 25, while the altitude is 8. F ind the a rea of the trapezoid. (H int: See Problem 12(c), page 405.)

18. T he area of a trapezoid is 480 square inches. If the a ltitude is 15 inches, w hat is the length of the median of the trapezoid?

19. T he area of a trapezoid is 80 square feet. If the lower base is 11 feet and the altitude is 8 feet, what is the length of the upper base?

, 20. T he upper and lower bases of an isosceles trapezoid are 10 and 16 respectively. If one of the lower base angles is 45°, w hat is the area of the trapezoid?

' 21. T he upper and lower bases of an isosceles trapezoid are 20 and 36 respectively. If one of the nonparallel sides is. 10, w hat is the area of the trapezoid?

22. T he altitude of an isosceles tiapezcid is equal to 4, while one of thecongruent sides is 5. If the area of the trapezoid is 14, find the lengths of the upper and lower bases.


1. Given: ABCD is a parallelogram.

C o n c l: A ABC = A DBC

3 . Given: A BCD is a parallelogram.

Concl.: A A B E = A AED

B

Giver. •. A BCD is a trapezoid

with AD II BC. Concl.: A ABC = A D BC

A -------------------- - 0

Given: E is any point on m edian AD.

Concl.: A E B D - A E D C A ,

5.

6.

Using the same Given Data as in Problem 4, prove that A A B E = A AEC. (H int: Use the information proved in Problem 4.)

Given: M edian AD of

A A B C was extended to point E.

Concl.: A A B E = A A C E

G iven: B E is the m edian to AC. 7 .

SZ) is the m edian to AS. Concl.: A D B C = A EBC

(H int: See Theorem■ 7 CZ

AREA O F PARALLELOGRAM , TRIANGLE, TRAPEZOID

G iven: A D X BC8 . Given: ABCD is a parallelogram.

AF X BC «-> *-*

AE X CD Concl.: B C - A F -

CD-AE

B E 1 AC Concl.: B C -A D =

A C -B E

591

9.

10. Given: PQRS is a O -M is the m idpoint

o f fS .

R M meets QP at *4.

Concl.: A AQR = O P Q R S (H int: Prove A A P M S A R SM and use addition postulate.)

12. Given: P midpt. of BD Concl.: ABCP = ADCP

G iven: 0 1 inscribed within A A B C

Concl.: A ABC *=%r{AB + AC + BC)

11.

Given: AC X BD 13.C oncl.: ABCD = $ A C -B D

592 AREAS OF POLYGONS A N D CIRCLES

14. G iv e n : ABCD is a p ara l

lelogram .M is the m idpoint of AD.•V is the m idpoint of HU.

C o n c l.: B N D M = \A B C D

A . . . M 0

16. G iv e n : ABCD is a trapezoid

w ith AD li BC.BC = IA D

C o nc l.: A D B C — 3 A B D A

18. G iv e n : AD bisects ZBAC .

C o nc l.: A A B D -.A A C D = A B ’.AC

G iven: B E is the m edian to AC. 15. CD is the m edian to AB.

Concl.:(1) A B D F = A C E F

(See Problem 7.)(2) A A B E = A ACD

A.

Given: ABCD is a trapezoid

w ith AD II BC.M is the m idpoint of S C

Concl.: A M D A - f A M C B = A M A B (H int: Prove A M D A + A M C B = \ABCD.)

17.

Given: D m idpt. of AB E midpt. of AU

Concl.: A F B C .= ADFE

19.

' S O . i G i v e n : A ABC w ith m edians AM , B N , and UP intersecting at G.

A^EAi OF PARALLELOGRAM, TRIANGLL, TRAPEZOID

GF 1 BC

A H 1. BCConcl.:

(1) A H = 3GF (H int: See Problems 19 and 20, page 392.)

(2) A G B C = \A A B C

Given: ABCD is a p a ra llelogram.

593

21.

CE intersects DA a t F.

C oncl.:(1) A DEC = \O A B C D(2) A A D E = A F E B

J

1. If two. parallelograms have congruent bases, the ratio of their areas is equal to the ratio of the measures of their altitudes, f

2. I f two parallelograms have congruent altitudes, the ratio of their areas is equal to the ratio of the measures of their bases.f

3. (a) Make up two statem ents similar to those in Problems 1 and 2 th a trefer to triangles ra ther than parallelograms.

(b ) Prove the statements that you wrote as your answer to pa rt (a).

4. The area of a rhombus is equal to one-half the product of the measures of its diagonals.

5. If a line passes through the midpoint of a diagonal of a parallelogram , •it will divide the parallelogram into two equal quadrilaterals.

6. The diagonals of a parallelogram will divide the parallelogram into four equal triangles.

7. * If two polygons are congruent, then their areas are equal, f8. * If the three medians of a triangle are drawn to their point of concur

rency, then ihe triangle will be divided into three Squal triangles. (H int: See Problem 20 in Group B.)

t T h is statem ent often appears as a theorem .

594

✓

AREAS OF, POLYGONS A N D CIRCLES

Areas of Similar Triangles

T he areas of similar triangles bear a special relation to each other th a t is im portant enough in mathematics to be worthy of our attention. A comparison can be m ade of the areas of two sim ilar triangles— or two similar polygons, in fact—without the necessity of having to find these areas. T he next theorem will give us this tool.

T H E O R E M 140: T h e ra tio of th e areas of two sim ilar trian g les is e q u a l to th e ra tio o f th e squares o f th e m easures of an y two co rrespond ing sides.

Given: A A ,B ,C , ~ A A 2B2C2

Concl. AAiB jC i&AzB%C%

1 • Let h be the perpendicular from A\ to 3,.

2. Let be the perpendicular from Az to a2.

3- A A iB tC i = \a\hi4.

5.

6.

7.

REASONS

A A iB tCi = \a ih iA_AiB,C, \a xhi _ a ht

\ aih2 as ' /,2 B u t A ^ S , C , ~

. £i Ci Q% Ci

8- Z 5 29. Z A XD XB-, ^ /A iD iB t

AAiBiDi ~ AAtBiDi

A A2B2C2

1. W hy possible?

2. W hy possible?

3. Theorem on the area of a triangle4. Same as 35. Division postulate

6. Given

7. Def. of similar polygons

8. Sam e as 79. Why?

10. A.A. theorem on sim ilarity

11. Same as 7

12. Why? (See steps U and 7.)


It is interesting to note th a t had we replaced j by ra th e r than vice

versa, step 13 would have been

AAjBiCi _ hi ' hi AAzBsCz hi h2

A A \B \C \ _ or A A iB tC i ~ V

Hence, it appears that we have also proved that

T H E O R E M 141: T h e ra tio o f th e areas o f two sim ila r trian g les is equal to the ra tio o f th e squares o f th e m easures of any two correspond ing a ltitudes.

These two theorems pave the way for the proofs of several theorem s in space geometry. As usual, however, terms will have to be defined before we cari proceed.

AREAS O e SIMILAR TRIANGLES 595

D e f in it io n 99: A pyramidal surface is a surface th a t is generated by a line that moves so as to always pass through a given point and always in tersect a given polygon where the given point is not in the plane of the polygon.

In the figure above, the fixed point, P, is called the vertex,, while the given polygon is ABCDE. Any one of the m any positions th-at'the moving line may assume is called an element of the surface. Notice th a t the surface extends above point P as well as below. Each of these two sections is called a nappe of the surface.

To define the term that we are seeking will require the need for another and final undefined term.

S o lidM O W ! R ather than attem pt to describe, the term sol if!, we will assume that all of us have a common understanding of it. ■


p

Figure 17-17.

D e f i n i t i o n 100: A p y r a m i d is a s o l id b o u n d e d b y o n e n a p p e o f a p y r a m i d a l

s u r f a c e a n d a p l a n e t h a t in t e r s e c t s e v e ry e l e m e n t o f th i s s u r f a c e b u t d o e s n o t c o n t a in th e v e r t e x o f t h e s u r f a c e .

D e f i n i t i o n 101: T he base of a pyram id is tHe polygon formed as th e in tersection of the plane with the pyram idal surface.

D e f i n i t i o n 102: T he altitude of a pyram id is the perpendicular segment th a t exists from the vertex to the p lane of the base of the pyram id.

D e f i n i t i o n 103: A triangularjpyram id is a pyram id whose base is a triangle.

I t would seem as if we had gone to extraordinarily great lengths m erely to build the background for one or two simple theorems. U nfortunately, as is the case in any area of the sciences, we can not speak intelligently unless w e learn the language!

T H E O R E M 142: If a p lan e is p a ra lle l to th e base o f a tr ia n g u la r p y ra m id an d in tersects its py ram id al surface, b u t no t th e vertex , th e n th e in te rsection w ill b e a tr ia n g le s im ila r

G iven: Plane D EF II plane ABC Concl.: A A B C ~ A D E F

to the base.

B

Figure 17-18.

AREAS OF-SIMILAR TRIANGLES 597

PRO O F ! STATEMENTS

1. Plane DEF II plane ABC<r-¥ <->

2. EF II BC

ZP E F = ZPBC Z.PFE ^ ZPCB

A P E F ~ A PBC Hence, EF: BC = PE'.PB

REASONS

1. Given

2. I f a p lan e intersects tw o p ara lle l

planes, the lines o f intersection are

p aralle l. (T h eo re m 43, p age 281)3. Corresponding angles of parallel lines4. Same as 35. A.A. theorem on sim ilarity6. D ef. o f s im ilar polygon s

In the same m anner, b y p ro v in g A P D E ~ A P A B it w ill fo llow that

7. D E '.AB = PE'.PB 7. Sam e as 68. EF'.BC -■ D E '.AB 8. Transitive .property of equality

Again, in ihe same m anner it can be shown thatEF'.BC = D E '.AB = DF'.AC

9. H ence, A A B C ~ A D E F | 9. S.S.S. theorem on sim ila rity

TH E O R E M 143: If a p lan e is p a ra lle l to the base of a t r ia n g u la r p y ra m id , th e n th e ra tio of the a rea of th e tr ia n g le o f in te r section to th e a re a o f th e base is e q u a l to th e ra tio o f th e square of th e d istance of th e p lan e from th e v e rtex to th e square of th e m easure of the a lt i tu d e .

Given: Plane D EF II p lane A BC

PR X plane ABC

A D E F (PQ)1C Concl A A B C (P R y

PRO O F STATEMENTS REASONS |

1. Plane DEF 11 plane ABC 1. Given

2. PR J- plane ABC 2. Given

3. PR 1 plane DEF 3. If a line is perpendicular to one oftwo parallel planes, it is perpendicu-

598 AREAS:OF POLYGONS A N D CIRCLESp

4. DQ || A R

5. ZP Q D S* Z PRA and ZP D Q S ZA 4/J

6. A W ) Q ~ A /M A . PQ PD1' T r = 7 a

8. D E II AB9. ZJPDE ~ Z P A B

and APED S A P D £ ~ A P A B D E PD A 3 = A4 D E _ PQ A B ~ PRBut A D E F ~ A A B C ■ A D E F (D E )1

" A /4 f iC ~ 04B)1

10.

11.

12.

13.

14.

F ig u r e 17-19,

| u la r to the other also. (Theorem 46,j page 284)

4. If a plane intersects two parallelplanes, the lines of intersection areparallel.

5. Why?

6. A A . theorem on similarity

7. Def. of sim ilar polygons

8. Same as 49. Why?

10. Same as 6

11. Why?

12. T ransitive property of equality

13. Theorem 142

14. T he ratios of the areas of two sim ilar triangles is equal to the ra tio of the squares of the measures of any two corresponding sides.

15. Substitution postulate, 5 Hence ^ Wis . Hence, A A B Q {pR)i

Illustration 1;

Two corresponding sides of two similar triangles are ' 6 inches and 8 inches respectively. If the area of the second triangle is 45 square inches, w hat is the area of the first triangle?

AREAS OF SIMILAR TRIANGLES 599

S o l u t i o n : — = (Theorem 140)Ai a%

= <>? = 6 X 6 = 2 X 2 = 4 45 92 9 X 9 3 X 3 9

Ai = 20

Illustration 2:A triangular pyramid with an altitude of 20 inches has a base of 144

square inches. How far from the vertex m ust a plane be passed such that the area of the triangle of intersection will be 81 square inches?

S o l u t i o n : • 4* = 3 (T heorem 143)Ai U2

i ! = A144 202

* 202 X 81* - 144

d\ — 15 inches

EXERCISES

LA]1. W hat is the ratio of the areas of two sim ilar triangles if two correspond

ing sides are respectively(a) 5 inches and 4 inches (b ) 2 inches and 6 inches(c) 10 feet and 20 feet (d ) 6 feet and 8 feet

2. W hat is the ratio of two corresponding sides of two similar triangles if their area's are respectively(a) 1 and 4 (b) 4 and 36

3. A plane is passed parallel to the base of a triangular pyram id. If thedistance from the vertex of the pyram id to the plane and the altitude of the pyramid are given by the figures below, w hat is the ratio of the area of the triangle of intersection to the area of the base?(a) 14 feet and 35 feet (b) 72 inches and 96 inches

4. A plane is passed parallel to the base of a triangular pyram id. If thearea of the triangle of intersection and the area of the base are given


by the figures below, how does the distance from the vertex to the p lane com pare w ith the a ltitude of the pyramid?(a) 108 and 147 (b) 14 and 50

5. Two triangles are similar. If a side of one is four times a corresponding side of the other, w hat is the ratio of their areas?

6. Two triangles are similar. If the area of one is four times the area of the other, w hat is the ratio of any two corresponding sides?

7. If two corresponding altitudes of two similar triangles are in the ratio of 3 :5 , w hat is the ratio of their areas?

8. If the areas of two similar triangles are in the ratio of 5 :9 , w hat is the ratio of any two corresponding altitudes?

9. Tw o triangles are similar. If the area of one is 16 times the area of the other, w hat is the ratio of any two corresponding altitudes?

10. T he angles of a triangle rem ain unchanged, but the sides of the triangle a re doubled. W hat happens to the area of the triangle?

11. T he area of a triangle is 63 square inches, while one of its sides is3 inches. In a similar triangle the side corresponding to the 3-inch side is 5 inches. W hat is the area of the second triangle?

12. T h e areas of two similar triangles are 48 and 60 square inches respectively. If a side of the first is 8 inches, what is the length of the side corresponding to this in the second triangle?

13. A plane is passed parallel to the base of a triangular pyram id such th a t the area of the bace is 9 times the area of the triangle of intersection. How far from the vertex does the plane intersect the altitude?

14. A plane is passed 4 feet from the vertex of a triangular pyram id whose altitude is 6 feet. If the area cf the base of the pyram id is 108 square feet, w hat is the arp'a of the triangular intersection?

15. A plane is passed parallel to the base of a triangular pyram id to make a cross section whose area is 56 square inches. If the base and altitude of the pyram id are 72 square inches and 12 inches respectively, w hat is the distance from the base to the plane?

16. * An altitude of one equilateral triangle is congruent to a side of another. W hat is the ratio of the areas of these two triangles?

H1. If a line joins the midpoints of two sides of a triangle, it cuts off a tri

angle whose area is one-fourth of the area of the original triangle.2. If two triangles are similar, then the ratio of their areas is equal to the

AREAS OF SIMILAR TRIANGLES 601ratio of the squares of the measures of any pair of corresponding angle bisectors.

3. If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of the measures of any pair of corresponding medians.

4. Construct a triangle whose area is equal to one-sixteenth the area of a given triangle by drawing a line parallel to the base of the given triangle.

5. Given: /LA is an angle ofA -4 £ F an d A A B C .

A F -A EConcl.

A AEF = _______A ABC AC -AB

(H int: Draw altitudes from E and B in each of the triangles.)

6. Given: Points D, B, and C are collinear

„ , A A B C B A -B CC o n c l . ' --------------A D B E B E -B D

(H int: Draw the altitudes from C and D in each of the triangles.)

7. If the area of a triangle is one-half the product of the measures of two sides of the triangle, then the triangle is a right triangle. (H in t: Use

@the indirect m ethod of proof.)If two triangular, pyramids have equal bases and congruent altitudes, then sectiohsf m ade by planes parallel to the bases and equidistant from the vertices are equal.

9. Given: Plane DEF II plane ABC

A D E F {VE)1 A A B C (V B Y

Concl.

t A “ section" of a p lane and a solid is the intersection o f the two.

602

10. Given:

Cone!.:

AREAS OF. POLYGONS A N D CIRCLES

Plane D E F II plane A B C

VP is any line through V intersecting D E F and A BC in P and Q respectively.

A D E F _ (VPY A A BC (VQ)"

■ A r e a s of R « g u U f Polygons

O u r study of the areas of polygons has been lim ited to areas of either 3-gons or special 4-gons. W hat can be said of the areas of other polygons? In general, very little. If enough information is known, the polygon can always be divided into triangles and the area of the polygon can be found by com puting the sum of the areas of all the triangles. T here is, however, a special class of polygons whose areas can be found m ore readily.

Consider the circle below th a t has been divided into n congruent arcs.

T o be as general as possible, we have stated the num ber as n ra ther than the 10 that you observe. Since the arcs are congruent, their corresponding chords are congruent and, hence, th e polygon ABCD . . . is equilateral. By drawing in the radii the triangles can readily be proved to be congruent by the 1W .1S'. congruency theorem, From this it would follow that Z l = Z 3 = Z5 = Z7 , . . and, too, th a t Z 2 S Z4 = Z6 . . . . By the addition postulate Z A B C — ZB C D = Z C D E . . . ; hence, the polygon is, also, equiangular. Polygons such as these are called regular polygons.

AREAS OF REGULAR POLYGONS 603

D efin itio n \0 4 : A regu lar polygon is a polygon th a t is b o th eq u ila te ra l a n d

equiangular.In the illustration above we started with a circle and showed th a t there

existed a regular polygon that was inscribed within a circle. T he converse of this is also true; that is, a circle can be circumscribed about a regular polygon. Since we have no need for this statem ent, we do not intend to prove it.

Let us return to the regular polygon on page 602. Since all the triangles were shown to be congruent, the altitudes from point 0 , such as OP and 0 3 , will be congruent. Each of these altitudes is called an apothem of this polygon; the point from which they are draw n is the center of the regular polygon. A radius, such as OA or OB , is said to be the radius of a regular polygon. And the perimeter, of course, is the sum of the measures of the sides of the polygon.T H E O R E M 144: T h e a rea of a reg u la r po lygon is equal to o n erh a lf th e

p ro d u c t o f the m easure o f th e ap o th em a n d th e p e rim eter.

. C 8

Given: ABC D . . . is a regular polygon.a is the measure of the apothem .

Concl.: Area of ABCD . . . = \ap

Figure 17-21.

PRO O F I STATEMENTS REASONS

1. Let OA be the line through points 0 and A.

2. Same for 0 1 i, OC, . . .3. A O AB = \a -A B

A O B C = ■ BC etc.4. A OAB + A O B C + . . . =

\ a ' A B %a ' BC -j- . . .5. Area of ABCD . . . =

A O A B + A O B C + • ■ ■6. Area of ABCD . . . —

%a(AB + BC + . . .)1. p = A B + BC + . . .8. Area of ABCD . . . = \<ip

1. Why possible?

2. Same as 13. Area of a triangle


5. Why?


7. Definition of perim eter8. Same as 6

604 AREAS OF POLYGONS AND-CIRCLES'Illustration:

W hat is the area of z regular hexagon, one of whose sides is 8?

M e t h o d : T h e perim eter can imm ediately be found as 48. Finding the apothem is a bit more involved. By drawing the radii of the hexagon, we find th a t each central angle is 60°. Since UA = OB, each of the angles

Figure 17-22.

OAB and OB A will also have to be 60°. Hence, A O A B is equiangular and, therefore, equilateral. Thus, OA = 8. T h e apothem OP bisects the base; therefore, AP = 4. Using the theorem of Pythagoras, a is found to be 4V 3. Hence,

area = \ap = ^•4V/3 '4 8 = 9 6 ^ 3

EXERCISES

1. T he perim eter and apothem of a regular polygon are 48 and 6 respectively. W hat is the area of the polygon?

2. Find the area of a regular hexagon if the length of one of its sides is(a ) 10 (b ) 16(c) 5 (d ) la

3. If the radius of a regular hexagon is 12, what is the area of the hexagon?

4. If the apothem of a regular hexagon is 7V,3, w hat is the area of the hexagon?

| L eave answ ers in rad ica l form.

5. T he radius of a square is 3. W hat is the area of the square?6. Find the area of an equilateral triangle if its radius is

(a) 8 (b) 12(c) 7 (d) 2a

7. Find the area of an equilateral triangle if its apothem is(a) 6 (b) 24(c) 5 (d ) a

8. A circle with a radius of 14 is inscribed within an equilateral triangle.Find the area of the triangle.

9. A circle with a radius of 9 is inscribed within a regular hexagon.W hat is the area of the hexagon?

10. A sqViare is inscribed in a circle of radius R. W hat is the area of the square?

11. (a) A circle is inscribed within one square and circum scribed aboutanother. If the radius of the circle is 6, what is the ratio of the areaof the larger square to that of the smaller square?

(b ) If the radius of the circle is la, what is the ratio of the areas of thetwo squares?

12. A circle is inscribed within one regular hexagon and circumscribedabout another. If the radius of the circle is 4, w hat is the ratio of the

- - ^ r e a of the larger hexagon to that of the smaller hexagon?

13. In the pyramid at the right, ABCD , is a regular polygon; in this case,

a square. T he altitude VP passes through the point of intersection of the diagonals. A pyram id such as this is called a regular pyramid. VQ, the altitude in face VCD, is called the slant height. If BC = 8 and VQ = 12, what is the lateral surface a rea o f the pyramid? (T h e lateral surface area will be the total surface

AREAS OF REGULAR POLYGONS 605

area excluding the_area, of.the ba^£.)14. Refer to the pyramid above for each of the following problem s:

(a) BC = 6, VQ = 10. Find the total surface area.(b ) BC = 12, VP — 8. Find the lateral surfacc area.(c) BC ~ 8, m /.PVQ = 45. Find the lateral surface area,(d) BC = a, VQ = s. Find the total surfacr a re i.

15. (a) T he base of a regular pyram id is a regular hexagon, each of whose


sides is 8. If the slant height of the pyramid is 10, w hat is the lateral surface area of the pyramid?

(b) W hat is the total surface area of this pyramid?

16.* The base of a regular pyram id is a regular hexagon, one of whose sides is 6. If the altitude of the pyram id is 5, what is the laterai surface area of the pyramid?

17.* The base of a regular pyramid is an equilateral triangle, one of whose sides is 12. If the altitude of the pyram id is 2, w hat is the total surface area of the pyramid?

A1. A radius of a regular polygon bisects the angle to which it is draw n.2. T he perpendicular bisector of a side of a regular hexagon passes through

the center of its circumscribed circle.3. If two regular polygons have the same num ber of sides, then they are

similar.4. If two regular polygons have the same num ber of sides, then the ratio

of the measures of their radii is equal to the ratio of the measures of their apothems.

5. A square and a regular hexagon are inscribed within the same circle. Prove: The ratio of the measure of a side of the square to the m easure of a side of the hexagon is v /2:1.

6. Prove: The lateral surface area of a regular pyram id is equal to one-half the product of the measure of the slant height and the perim eter of the base.

\ j ■ Circumference of a CircleWe could not very well leave the topic concerning the

regular polygon without showing how some of its properties can be extended to formulate properties of the circle. Notice in Figure 17-23 what appears to be happening when the num ber of sides of a regular polygon is increased from 4 to 8 to 16 and finally to 32. As the num ber of sides increases, the

CIRCUMFERENCE OF A CIRCLE 607

polygon itself resembles more and more the circle in which it is inscribed. Yet no m atter how large the num ber of sides may be, the polygon will always differ from the circle, for the sides of a polygon are line segments while any portion of a circle between two points of the circle must be an arc of that circle.

M athem aticians frequently encounter situations such as this where for “ all practical purposes” one quantity can be m ade to so closely resemble another that a replacement of one for the other is permitted. Thus, in the situation here, as the number of sides of the regular polygon is increased, the perimeters of the polygons approach closer and closer to the circumference of the circle. If this is so, and it seems reasonable to suspect that it is, then the mathematician would say, “ Why bother to pursue this endless sequence of numbers representing the perim eters! These num bers can be made to differ so little from the circumference of the circle itself that for all practical purposes we may as well use the circumference of the circle to represent the perimeter of the regular polygon when the num ber of sides of that polygon is very, very large.” . . . And of course we have assumed that you recall from your work in elem entary m athematics that,

D e f i n i t i o n 105: The circumference of a circle is the measure of the circle (in linear units).

In view of our analysis it seems that we would not be going too far astray by postulating the following,

P o s t u l a t e 45: W hen the number of sides of a regular polygon inscribed in a circle is very large, then the circumference of the circle can be used as a replacement for the perim eter of the polygon.

O ur discussion above has been on an extremely elem entary level. .Some day you may examine this topic m uch more thoroughly in the subject called calculus. The topic is the theory of limits which, needless to say, is quite beyond the scope of our work a t this time. Postulate 45, however, enables us to prove the theorem below. T his theorem , in turn, leads us to a method for finding the circumference of a circle.

T H E O R E M 145: In any two circles th e ra tio o f th e c ircu m feren ce to th e rad ius of th e first is e q u a l to th e ra tio of th e c ircum ference to the rad iu s of thg second.

Given: OO and O P where Co and Cp represent the circumference of e&clv respectively.

• Concl.: Co'-OA = Cp '.PQ



1. L et regular polygons of n sides be inscribed in each of the circles.

2. OA = OB

3. PQ = PR4. OA'.PQ = OB'.PR

5. m AB = m BD = m D E = . . .

= 360n

360 f!

6. m ZA O B =

1. A regular polygon of n sides can be inscribed in a circle.

2. T he radii of a circle are congruent.

3. Same as 24. Division Postulate

5. I f in a circle two chords are congruent, their corr. arcs are congruent.

6. T h e measure of a central angle of a circle is the measure of the m inor arc it intercepts. (Rev. of Def. 82.)

7. m QR = m RS = m S T — . . . 7. Same as 5

= I™n

8. m ZQ PR = —n

9. ZA O B S ZQ PR10. A / i O B ~ A QPR11. AB-.OA = QR-.PQ12. n(AB):OA = n(QR):PQ13. But n(AB ) ■= pi

and n(QR) = pi14. pi'.OA — pz’.PQ

Now if we allow the num ber of sides in the two polygons to rem ain equal bu t become very large, then

15. Co’-OA = Cp:PQ | 15. Postulate 45

We have ju st shown that the ratio of the circumference of a circle to the measure of its rad ius is always the same no m atter what the size of the circle m ay be. Thus, we can set this ratio equal to a constant that we will call 2 t (read as two pi). Therefore,

8. Same as 6

9. Transitive Property

10. iS’,/4.6'. Theorem on Sim ilarity11. Def. of similar triangles12. M ultiplication Postulate13. Def. of perim eter of a regular

polygon14. Substitution Postulate

Hence, by applying the multiplication postulate we find that,

C = 2irr

This is the formula frequently ysed for finding the circumference of a circle.


But what is the value of the symbol x? I t was approxim ated by Archimedes, an ancient Greek m athem atician, to fall somewhere betw een 3f$- and 3-fft. More recently, m athem aticians have approxim ated ir far more accurately. This symbol can never be expressed as an exact num ber—either in the form of a term inating or a non-term inating decim al. But approxim ations do exist that will give as great a degree of accuracy as one might desire. Some of these approxim ations are,

3, 3.14, 3.1416, 3.1415927, 3.14159265359

There is also, of course, the im proper fraction that you may have used in elem entary school:

Illustration 1:W hat is the circumference of a circle whose diam eter is 20 inches?

M e t h o d : T he formula for the circumference of a circle, calls for the measure of the radius. By dividing the measure of the diam eter by 2 we find that the measure of the radius is 10 (inches). Hence,

C = 2irr = 2tt(10)= 20ir

Unless otherwise called for, the answer is usually left in term s of ir.

By applying the comm utative property of m ultiplication, it is possible to rewrite the formula for the circumference of a circle as,

C = ir2r

T hen, by realizing that the measure of the diam eter of a circle is twice the measure of the radius of the circle, it is possible to replace 2r by d. This will lead to the following variation of the formula that, is often used.

C = ird

We would like to take a moment now to tie together the two measures that we have created for a circle. O n the one hand we created the a rc degree (see page 453) wherein we state arbitrarily that the m easure of every circle is 360 arc degrees. O n the other hand we have just discovered th a t the circumference of any circle can be found by using the form ula C = 2irr. The arc degree unit is what is called a non-denominate unit. In reality, the arc degree is not a unit in the sense of such things as the pound, the yard, or the gallon—for it is not fixed in size. T o say that an arc contains 60 arc degrees is merely to d raw a com parison between the length of that arc and the length of the entire circle. In this case it would imply that the arc is -yW of the length of the circle. T hus its length is shown to be some fraction of the length of the circle rather th an some definite quantity. However, by saying that an arc is 6 inches we are applying an absolute unit as its measure. Now we

610 AREAS OF: P O LY G O N S -A N D CIRCLES

have indicated that the arc— if stretched tight—will be 6 times as long as the linear unit called the inch. As we know, the inch is a very definite.length.

Consider now the arc AB whose measure in arc degrees is n. We w ant to determine the number of linear units in this arc. T o do this we realize

that an arc of n arc degrees has a length ■— of the length uf its circle.

Since the length of a circle is but another name for the circumference of the

circle, arc AB is ^ of 2-irr, where r is the measure of the radius of the circle

of which AB is an arc. Using symbols, this can be expressed as,

m A B = n

therefore, the length of /IB = of its circle

hence, the length of AB = ^ • 2irr

TH E O R E M 146: T he len g th of an arc of a c irc le is g iven b y th e fo rmula,

' - 3 S 5 ' 2"w here n is the n u m b er of a rc degrees in the arc an d r is the n u m b er of lin e a r units in th e rad iu s of th e circle .

Illustration 2:If the radius of a circle is 5 inches, find the length of an arc of this circle

whose central angle has a measure of 40.M e t h o d : Since the measure of the central angle is 40, the measure of the arc is 40. Hence, by Theorem 146,

‘ = 350 • 2 "

40 0 = 360 ' 2* 5

= — ir (inches)V

EXERCISES(Unless otherwise stated, leave all answers in terms of it .)

1. Find the circumference of each of the following circles.(a) r = 20 (b ) r = 5£(c) a = 16 (d) d = 6 \

2. Find the length of each of the following arcs.

(a) m AB = 60; radius is 10 inches(b ) m AB = 90; radius is 12 inches


(c) m AB = 150; radius is 25 feet

(d ) m AB = 120; diam eter is 40 feet

(e ) m AB = 80; diam eter is 35 feet

3. O ne of the sides of a square is 8 inches long. T he square is inscribed in a

circle.(a) W hat is the radius of the circle?(b) W hat is the circumference of the circle?(c) W hat is the length of an arc of the circle cut off by one of the sides

of the square?

4. An equilateral triangle is inscribed in a circle. If the apothem of the triangle is 5 inches, what is the length of an arc of the circle th a t is cut off by one of the sides of the triangle?

5. An isosceles right triangle is inscribed in a circle.(a) If the length of the hypotenuse of the triangle is 10 feet, w hat is

the length of the arc cut off by one of the legs?(b) If the length of one of the legs of the triangle is 10 feet, w hat is the

length of the arc cut off by this leg?

■ 6. A wheel travels the distance of its circumference in one revolution.(a) How far will a bicycle having a 28 inch diam eter wheel travel if

the wheel makes 1,000 revolutions? (Use as the approxim ation for ir.)

(b) How m any revolutions will a wheel of this bicycle make during a trip of 1 mile? (Use as the approxim ation for ir.)

7. An arc of a circle has a measure of 30 arc degrees and is 5ir inches in length.(a ) W hat is the radius of the circle?(b) W hat is the circumference of the circle?

8. T he back wheel of a motor bike is placed on a stand and spun about its axle. T he diam eter of the wheel is 22 Jnches. How far does a point on the wheel travel when a spoke on the wheel rotates through an angle of 72 degrees?

9. A rope is stretched around the two pulley wheels as shown in the diagram.


10. T he “ 30th parallel” on the earth ’s surface is a circle determined as shown in the diagram.

(a) If the diam eter of the earth is approxim ately 8,000 miles, what is the circumference of the 30th parallel? (Use 3.14 as the approxim ation for ir.)

(b ) D uring one day the earth makes one revolution abn-.it its axis. How m any miles per hour is a tree traveling if it is located on the 30th parallel?

(c) How m uch faster would the tree in (b) be traveling if it were located along the equator?

11. A circle is inscribed in a square one of whose sides is 12 inches. Find the length of the arc whose endpoints are two successive points of tangency.(a) If the polygon had been an equilateral triangle, what would the

length of the arc have been?(b) If the polygon had been a regular hexagon, what would the length

of the arc have been?

12. Im agine the earth to be a perfect sphere and consider a thin sheet of metal draw n tightly about the surface at the equator where the circumference is approximately 24,000 miles. T he length of the metal is increased by 50 feet. It is then held away from the surface by the same distance throughout the 24,000 miles. Is it possible for a man who is 6 feet tall to stand beneath the metal sheet? Justify your answer.

I Area of a CircleExamine Figure 17-20 on page 602 and imagine the num

ber of sides of the regular polygon to increase indefinitely. W hat conclusion did we draw concerning the perimeter of the polygon when this occurs? W hat do you believe will occur to the radius of the polygon under this condition? And finally, what will happen to the apothem CKj?

Actually, nothing will happen to the radius of the polygon for the radius of the polygon is also the radius of the circle, and as the polygon approaches the circle in appearance, the radius of the circle does not alter. However, the apothem of the polygon becomes closer and closer to being the radius

AREA OF A CIRCLE 613

of the circle as the num ber of sides increases. We need ju st such a postulate before it is possible to prove a theorem relating to the area of a circle.

P o s t u l a t e 4 6 : W hen the num ber of sides of a regular polygon inscribed in a circle is very large, then the measure of the radius of the circle can be used as a replacem ent for the measure of the apothem of the polygon.

T H E O R E M 147: T h e a rea of a circle is given by th e fo rm ula ,

A = irr!

PRO O F

Through our analysis on page 602 we know that it is possible to inscribe a regular polygon within a circle. T he area of this polygon is eriven by the formula,

A = \ap

where a is the measure of the apothem and p is the perim eter of the polygon. Now, by making the num ber of sides of the polygon sufficiently large,

by Postulate 45, C can replace pand by Postulate 46, r can replace aHence, A - VCbut since, C = 2irrthen, A = \r-2irror A = ir r2

And this is what we set out to prove.

Illustration:In the figure at the right, a circle is inscribed in a

square one of whose sides is 8 inches. Find the area of the shaded region.

M e t h o d :Area of square = 8 X 8 = 64 sq. in.Area of circle = irr2 = ir '4 J = 16ir sq. in.Area of shaded region = (64 — 16ir) sq. in.

Before completing our discussion of area as related to a circle, it would seem only natural that we examine the area of a region such as the “ pie slice” pictured in Figure 17-24. Earlier o'.:r discussion led from the circum ference of a circle to the length of an arc of a circle— the arc being but a section of the circumference. Now our discussion is leading us from the area of a circle to the “ pie slice” —where the “ pie slice” is but a section of the area of the circle.


Obviously, the actual name of this region is not a “ pie slice” but rather, a sector oj a circle.

D e f i n i t i o n 106: A s e c to r o f a c i r c le 0 is t h e u n io n o f t h e s e ts o f p o in t s

c o n s is t in g o f 0 2 , AB a n d th e s e t o f p o in t s i n th e i n t e r i o r o f Z A O B w h e r e t h e d is ta n c e f r o m e a c h o f th e s e p o in t s to 0 is le ss t h a n OA.

In finding the formula for the area of a sector of a circle we m ust apply a slight variation of Postulate 45. T h at is, not oniy is it possible to replace the perim eter of a regular inscribed polygon with the circumference of the circle when the num ber of sides become sufficiently large bu t we are also able to replace any fraction of th a t perim eter by an equal fraction of the circum ference. Thus, intuitively we m ight gather from Figure 17-25 th a t as the

Figure 17-25;

num ber of sides of the regular polygon increases the sum of the measures of the sides between points P and Q would come very close to the measure in linear units of PQ. But the sum of the areas of the region bounded by the red line segments is,

A = \at

where a is the measure of the apothem and I is the sum of the measures of the sides from P to Q. Hence, as the num ber of sides of the polygon increases, a can be replaced by r and i can be replaced by

360 ' 2,rr

since this expression represents the num ber of linear units in the length of the arc of a circle (see Theorem 146). Hence,

a rea of a circle615

n360

T H E O R E M 148: T h e area of a sector of a c irc le is g iven b y th e fo rm ula ,

A - h " *w h ere n is the n u m b er o f a rc degrees in th e a rc a n d r is th e m easure of t t e ra d iu s of th e circle .

Illustration:A square inscribed in a circle has a side of 6 inches. Find the a rea of

th e region bounded by a side of the square and its corresponding arc.

Analysis: O u r problem resolves to one in w hich we are seeking the area of the shaded region in the diagram .T o do this we will find the area of A O A B and the area of the sector bounded by the radii T)A and UE and the arc AB. T hen the difference between these areas will be the area of the region we have been asked to determine.

M e t h o d •. Since the sides of the square are congruent, their corresponding arcs will be congruent and hence, the m easure of each arc is 90. In view ofthis, m A.AOB = 90 and by using the T heorem of Pythagoras we can find'

that OA = 4 -V 2

Hence, Area of sector = • iff2

90 / 6 V“ 360 ‘ T ' { V 2 j

1 364 ' r ‘ 2

9jt . 1= square inches

Also, Area of A OAB = ^ OA-OB

= 1 . JL . 62 . V 2 V 2

= 9 square inches

Therefore, the area of the shaded region is,

9v — 18 . .= ------ ^------s q u a r e in c h e s

T he shaded region in the illustration above is called a segment cf a circle.


EXERCISES

A

(Leave all answers in terms of ?r unless otherwise stated.)1. Find the area of each of the following circles.

(a) r = 7 (b ) r = 4^ (c) d = 24 (d ) d = 15

2. Find the area of each of the following sectors of a circle.

Measure oj Arc Radius of Circle(a) 60 8 inches(b) 40 12 inches(c) 90 4 j inches(d ) 120 3} inches

3. Using 3.14 as the approxim ate value of ir, fold the approxim ate area of the segment of the y rc le in the illustration on page 615.

4. An arc of a circle has a measure of 90. Find the area of the segment of the circle bounded by this arc and its corresponding chord if the radius of the circle is 16 inches.

5. (a) An isosceles right triangle is inscribed in a circle that has a diameterof 10 inches. Find the area of the segment of the circle bounded by one of the legs of the triangle and its corresponding arc.

(b ) If the right triangle in (a) had been a “ 30-60 degree” right triangle, what would be the area of the segment bounded by the shorter leg and its corresponding arc? (See problem 10, page 370.)

6. (a) A regular hexagon is inscribed in a circle that has a radius of 12feet. Find the area of the segment of the circle bounded by a side of the hexagon and its corresponding arc.

(b ) If the polygon in (a) had been an equilateral triangle, what would the area of the segment have been? ------ .

7. T he shaded region bound by two concentric circles is called an annulus.(a) Find the area of an annulus if the radius of the larger circtfc-4s-lil-

inches while the radius of the smaller circle is 7 inches.(b ) Show that the area of an annulus can be expressed by the formula,

AREA OF A CIRCLE 617

A = ir (R — r)(R + r) where R is the measure of the radius of the larger circle and r is the measure of the radius of the sm alk ' circle.

8. (a) A sector of a circle has an arc of 40 degrees. If the area of the sectoris 5ir, what is the area of the circle?

(b ) A circle has an area of 36jt square feet. W hat is the area of a sector of this circlc if the arc of the sector is 20 degrees?

9. Using the dimensions shown, find the area of the shaded region of each of the figures below.

f

J ,

10. (a) Show that the ratio of the areas of two circles is equal to the ratioof the squares of their corresponding radii.

(b ) Show that the ratio of the areas of two circles is equal to the ratio of the squares of their corresponding circumferences.

11. Show that the area of a circle can be expressed by the form ula A = \trrP where a is the measure of the diam eter of the circle.

0Prove each of the following statements.

1. T he area of the circle circumscribed about a square is twice the area of the circle inscribed within the square.

2-. The area of a circle circumscribed about an equilateral triangle is four times the area of the circle inscribed within the triangle.

3. In a right triangle if semicircles are constructed on each cf the sides as a diameter, then the area of the semicircle on the hypotenuse is equal to the sum of the areas of the semicircles on the two legs.


4. Prove that the area of the annulus a t the right is equal to 7r(/lC)2.

5. In the diagram at the right, ~AC is a diam eter of the larger circle while A B and BU are diameters of the smaller circles. Prove that the area of the shaded region is equal to the area of the unshaded region. (See problem 11 of group A.)

6.* Prove that the area of A ABC is equal to the area of the shaded region in the figure a t the right, where A B and BC are d iam eters of the smaller semicircles and AC is a diameter of the largest semicircle.

I Test and Review

A1. (a) Find the area of a right triangle, one of whose legs is 12, while the

hypotenuse is 13.(b) Find the altitude to the hypotenuse of this triangle.

2. T he altitude to the hypotenuse of an isosceles right triangle is 8. Findthe area of the tri-angle.

3. (a) The ratio of the measures of two corresponding sides of two similartriangles is 3:5. Find the ratio of their areas.

(b ) The ratio of the areas of two similar triangles is 27:64. Find theratio of the measures of a pair of corresponding sides.

4. A rectangle and a parallelogram have equal areas. T he base and a ltitude of the rectangle are 16 and 12 respectively. If the m easure of the base of the parallelogram is to the measure of the base of the rectangle as 5:8 , then w hat is the a ltitude of the parallelogram?

5. (a) Find the area of a parallelogram if its sides are 8 and 12 respectively,while the measure of the angle formed by these sides is 45.

(b ) If the measure of the angle were 30, what would the area be?(c) If the measure of the angle were 60, what would the area be?

TEST AN D REVIEW 619

6. T he coordinates of the vertices of a triangle are (0, 0), (8, 0), and (5, 7).

Find the area of the triangle.

7. T he side of a regular 8-gon is b. If the measure of the apothem is a, express the area of the polygon in terms of a and b.

8. T he tangent segments to a circle from an external point form an angle of 120°. If the radius of the circle is 6, what is the area of the triangle whose sides are the two tangent segments and the line segment joining the points of tangency?

9. T he measure of the lower base of an isosceles trapezoid exceeds the measure of the upper base by 8, while the measure of each of the nonparallel sides is 5. If the area is 36, w hat is the measure of each base?

10. The measures of two sides of a triangle are 10 and 12, while measure of the angle formed by these sides is 30. Find area of the triangle.

11. (a) Find the area of a regular hexagon if the length of one of its sides is 6.(b) Find the area of a regular hexagon if the length of its apothem is 10.(c) Find the measure of the side of a regular hexagon if its area is

108V3.

12. (a) T he upper and lower bases of an isosceles trapezoid are 18 and 24.If the lower base angles are 45°, what is area of the trapezoid?

(b ) I f the lower base angle were 30°, what would the area be?(c) If the lower base angle were 60°, what would the area be?

13. A plane is passed parallel to the base of a pyramid in which this base is an equilateral triangle. The measure of the altitude of the pyram id is 4,- while the distance from the vertex to the plane is 3. If the area of the base is 64, what is the measure of a side of the intersection?

14. Find circumference and area of the circle whose d iam eter is 18 feet.

15. A .“ 30-60 degree” right triangle is inscribed in a circle. If the sideopposite the 30 degree angle is 5 cm, find length of its corresponding arc.

16. (a) Show that the ratio of the circumferences of two circles is equal tothe ratio of the measures of their corresponding diameters.

(b) If the measure of the radius of one circle is 3 times the measure of the radius of a second circle, how do their circumferences compare?

(c) How do the areas of the two circles in (b ) compare?

17. A square, one of whose sides is 8 feet, is inscribed in a circle. F ind area of segment bounded by a side of the square and its corresponding arc.

18. Two tangent segments are drawn from an external point to a circle having a diam eter of 10 inches. T he angle formed by the rays of these two segments has a measure of 60. Find the area of the region bounded by the tangent segments and the m inor arc of the circle.


BProve each of the following:

1. Given: ABCD is a O .Cone).: A A BP = A ADP

3 . G iven: AD is the m edian to ~B€.

E is the m idpoint of AD. Concl.: A A B E = A D C E

5 . G iven: A BCD is a O . Concl.: A PAB + A P D C =

hABCD

Given: E is the midpoint of. AD.

Concl.: A A B C = A D B C

2 .

Given: Q uadrilateral ABCD with diagonal BD

B F ^ S E Concl.: A BCE = AFCD

Given: ABCD is a £3.M is the m idpoint of BD.

Concl.: A B M E - CFMD


JLProve each of the following statem ents:

1. If the altitudes of a triangle are congruent, then the triangle is equilateral.2. The median to a side of a triangle separates the triangle into two equal

triangles.3. T he product of the measures of the legs of a right triangle is equal to the

product of the measures of the hypotenuse and the a ltitude to the hypotenuse.

4. The sum of the measures of the perpendiculars from any point within an equilateral triangle to the sides is equal to the measure of an altitude of the triangle.

5. If two regular polygons have the same num ber of sides, then the ratio of their areas is equal to the ratio of the squares of the measures of any two corresponding sides.

6. T he area of a sector of a circle is equal to one-half the m easure of the radius times the length, of the arc of the sector.

B Try This For FunAt the time we studied the theorem of Pythagoras, it was

pointed out that the early proofs of this theorem were very likely those th a t relied on the areas of polygons. The figure below was supposedly the one used by Leonardo DaVinci, the famous Italian painter, to prove this theorem. Using his diagram , can you prove that, “ The square on the hypotenuse of a right triangle is equal to the sum of the squares on the legs” ?

Given: Right A A C B withZA C B the right angle.

PQ II BC

PQ II AC Concl.: A B R P = A STC +

C B Y W(These are the squares on each of the sides.)

Suggestion: Prove A B Y S ~ APQC and T S Y W ^ RQCB.

IT SEEM S ONLY N A TU R A L T H A T H A V IN G learned how to determ ine th e measure of a plane figure, we would tu rn to space geometry and try to devise some means of com puting the measure of a space figure. O ur first objective will be to establish an appropriate un it for determining this measure. W ith this as a tool, we can then turn our a ttention to certain special solids and develop theorems th a t will enable us to com pute their measure in term s of this unit.

O f th e m any space figures th a t exist, perhaps the most im portant of these are solids such as those in Figure 18-1.

Figure 18-1.

622

VOLUMES 623

Each of these solids is called a prism. Before a pyram id was defined, it was necessary that the pyramidal surface first be defined. So, too, before the prism can be defined, the prismatic surface will have to be defined.

D e f in i t i o n 107 : A prismatic surface is a surface th a t is generated by a line th a t moves so as to always be parallel to a fixed line and always intersect a fixed polygon. T he fixed line does not He in the plane of the fixed polygon. (Note the similarity between this definition and th a t of a pyram idal surface.)

F ig u r e 18-2.

In Figure 18-2 the fixed line is the line /, while ABCD is the fixed polygon. T h e lines a, b, c, d, e, f, and g are various positions of the moving line.

D e f in i t i o n 108: A prism is a solid bounded by a prism atic surface and two parallel planes.

F ig u re 18-3.

T he lines AiAi, B A . CiA are called th e latsral edges of the prism. Since they represent the moving line in various positions, they must be parallel to each other, for lines in space th?t are parallel to the same line are parallel to each other.

624 VOLUMES

T H E O R E M 149: T h e la te ra l edges of a prism are p a ralle l. \

T he intersection of a plane and a prismatic surface is called a iectjon. ^ In the prism above, the sections <4i5iC; and A-.BiC} are the bases of the prism.'’

If the section is m ade by a plane parallel to the bases, then- this section is a i f cross section of the prism. If the section is m ade by a plane perpendicular to a1/ lateral edge, then it is a right section of the prism.

T he polygons A3A iB ,B i , B iB xCxC%, and AtAiC:Cj are called the faces of the prism. T he common perpendicular to the bases of a prism is its

■ v altitude. Should a_ lateral edge be perpendicular to the b'Sses, then the prismi / would be a right prism.

Prisms are also classified by the polygons that form their bases, Thus, aS triangular prism is one in which the bases are triangles. The prism o^ page 611 }

is a triangular prism. W ere it an equilateral triangular prism, the bases-w-euld be equilateral triangles.

EXERCISES

A

A num ber of the statements to be proved in the following group of problems should be considered as^theorems, Nas they will be -used in the p ro o fs jh a t are to follow. Each of these statem ents will be marked with a n (as’te risk \ r

1. T he faces of a prism are parallelograms.*2. T he lateral edges of a prism are congruent. *3. T he bases of a triangular prism are congruent triangles.*4. A cross section of a triangular prism is a triangle congruent to either

base of the prism. *5. A cross section of a prism is a polygon c o n g T u e n t to either base of the

prism .* (H in t: See the definition of congruent polygons and then use the inform ation in Problem 4.)

6. T he lateral area of a prism is equal to the product of the perim eter of a right section and the measure of a lateral edge. * (The lateral area of a prism is the sum of the areas of the faces.)

7. I f a plane is passed through two nonadjacent lateral edges of a prism, the intersection of the plane with the prism will be a parallelogram.

8. T h e faces of a right prism are rectangles.9. I f a plane is passed parallel to an edge of a triangular prism, the inter

section of the plane with the upper and lower bases of the prism will be congruent line segments.

VOLUMES 625

10. Given: Prism A B C -D E F is a right equilateral triangular prism. Plane QRS II plane D E F Plane P R S contains RS.

Concl.: A P R S is isosceles.

11. Given: ABC D -EFG H is a prism. Concl.: If D F and H B are drawn,

they will bisect each other.

12. Use the same diagram as in Problem 11. Given: R ight prism ABCD -EFG H

EFGH is a rectangle.Concl.: (H B)2 = (H E)2 + (EF)* + (BF)1

BEach of the problems in this group will depend upon the

theorem you proved in Problem 6 of the exercises in G roup A.

1. T he base of a right prism is,an eight sided equilateral polygon, one of whose sides is 5. Find the lateral area of the prism if the lateral edge is 12.

2. A side of the base of a right equilateral triangular prism is 10. If the lateral edge of the prism is 15, what is the Lateral area of the prism?

3. T he base of a right triangular prism is a right triangle whose legs are 18 and 24 respectively. If the lateral edge of the prism is 38, w hat is the lateral area of the prism?

4. T he iateral area of a prism is 448. If the lateral edge is 7, w hat is the perim eter of a right section?

5. The lateral area of a prism is 684, while the length of a lateral edge is 12. If a right section is an equilateral triangle, w hat is the length of one of its sides?

6. T he base of a right triangular prism is a right triangle whose legs are

626 VOLUMES24 and 32 respectively. If a lateral edge is 14, w hat is the total area of the prism? (The total area of a prism is the sum of the lateral area and the areas of the bases,)

7. A side of the base of a right square prism is 16. The length of a lateral edge is 24. W hat is the total area of the prism?

8. A side of the base of a right equilateral triangular prism is 4. If a la teral edge is 6, w hat is the total area of the prism?

9. A right section of a prism is an equilateral triangle, one of whose sides is 12 inches. T he prism ’s altitude, whose length is 8 inches, makes an angle of 45° with a lateral edge of the prism. W hat is the lateral area of the prism?

10. * T he right section of a prism is a rhom bus whose diagonals are 6 inches and 8 inches respectively. T he lateral area of the prism is 140 sq u are ’ inches. If a lateral edge of the prism makes an angle of 45° w ith the altitude of the prism, w hat is the length of the altitude?

■ Volume of a PrismPerhaps the most commonly observed prism is the simple

“ box.” 'Since its base is a rectangle and its lateral edges are perpendicular tQ. the base, it could be called a right rectangular prism. I t is, however, known as a rectangular parallelopiped.

D e f i n i t i o n 109: A parallelepiped is a prism whose base is a parallelogram . D e f i n i t i o n 110: A rectangular parallelepiped is a right prism whose base is

a rectangle.

In w hat two ways does a rectangular parallelepiped differ from the general parallelepiped?

D e f in it io n 1 1 1 : A c u b e is a r e c t a n g u l a r p a r a l l e l e p ip e d w h o s e e d g e s a r e c o n

g r u e n t .

Figure 18-4 shows the general rectangular parallelepiped, while Figure 18-5 is a drawing of a cube. In terms of the letters of these diagram s w hat are sorae of the properties of these solids?

VOLUME OF A PRISM 627

Each of these definitions brings us somewhat closer to the objective of this chapter: to determine a means of expressing the measure of a solid.

T here were a num ber of units available to us at the tim e we were seeking a way tu express the measure for the region enclosed by a plane figure. The unit selected was one that was, possibly, the “sim plest” polygon that enclosed a region of the plane. It'was the square unit. So, too, in selecting the unit through which the measure of a solid can be com puted, we reach for a solid that appears to us to have the “ simplest” form. This solid is the cube. Thus, a cube, each of whose edges is one unit in length, is called a cubic unit.

D e f i n i t i o n 112: The volume of a solid is the num ber of cubic units contained by the solid.

Let us examine what this will imply in terms of the rectangular pa rallelepiped. By actual count we find that there are 24 cubic units in the

/ /11

/ \ ) f" 7 7---- 7------7------7-------- 7 ' 7 / ~ 7c L J

4 units

/ / / / / / / / — -ft- — -j£ — — -jL. — ^ —

/ / / / / / / /' - ./•■ / - t . / / . J . . /

^ / 3 units

8 units

F ig u r e 18-6.

bottom layer of this box. Since this box is 4 units high, there will be a total of 4 layers, or 96 cubic units, within the box. T he number of cubic, units in the first layer could have been determined by finding the area of the base of the rectangular parallelepiped, for there exists a one-to-one correspondence between the cubic units in the first layer and the squares in the region of the base. To find the volume of this solid would then be a m atter of simply m ultiplying the area of the base by the measure of the altitude.

P o s t u l a t e 47: T he volume of a rectangular parallelepiped is equal to the product of the area of the base and the measure of the altitude.

It is im portant to realize that “ volume,” like “ area ,” “ measure of a line segm ent,” and “ measure of an angle,” is simply a number. It is the num ber assigned to a solid dependent upon the num ber of cubic units th a t the surface of the solid bounds.

628 VOLUMES

To determ ine tlie-'vclumcs of othersSiids will necessitate assuming the statem ent knowiyas Cavalicri's Principle, y

P o s t u l a t e 4 8 : v th tre exist twojettcfs and a fixed plane such that every plane paralleTTxr-tti^r-ftxed plane intersects the solids in sections that have equal areas, then the solids have equal volumes.

This principle can be illustrated by examining a deck of playing cards. W hether they are stacked vertically or ruffled slightly and placed on the table such as in Figure 18-7b, the volume of the deck will not be altered.

No m atter how we may twist or distort the position of the cards, just so we do not separate the cards, the volume of the deck will always be the total volume of the individual cards.

This same conclusion can be justified in terms of Cavalieri’s Principle. W ere a plane passed parallel to the table on which the cards were lying, it would intersect each of the decks in a single card. As the decks are identical, the cards of intersection will have equal areas. Since every plane p a rallel to the table and cutting both decks must always do the same, then C avalieri’s Principle grants us the right to say that the volumes of the decks will be equal.

In this illustration the polygons or cards happened to be congruent. This is not of importance. W hat is vital, though, is that the sections m ade by each plane parallel to the fixed plane always be equal in area. Below, one of the sections of equal area is a quadrilateral, while the other is a triangle.

V P

F ig u re 18-8.

P o s t u l a t e 49: For any given polygon there exists a rectangle of equal area.

T H E O R E M 150: T h e volum e of a prism is eq u al to th e pro<fu m easure of its a ltitu d e a n d the a rea o f its base.

Given: Prism with base C and altitude £ Concl.: V — Ch

VOLUME OF A PRISM


1. Prism with base C and altitude Jt2. Let rectangle D have the same

area as the base of the prism and be in the same plane as the base of the prism; thus,jC = D.

3. At each vertex of the rectangle let the lines in the diagram be the perpendiculars to the plane that exist at those points.

4. At a distance h from D let the plane drawn be the one that is parallel to the plane of the base.

5. Let any plane be passed parallel to the plane of the base intersecting the prisms in A and B.

6 . B = C , / i S C

7. B = D, A = C

8. A = B9. Hence, volume of prism P =

volume of rectangular parallelepiped R .

10. But volume of R = Dh11. Volume of prism P = Ch

1. Given2. Postulate

3. There exists only one line perpendicular to a given plane a t a given point on the plane.

4. Why possible?

5. Same as 4

6. A cross section of a prism is a polygon congruent to the base. (See Problem 5, pagfe 612.)

7. T he areas of congruent polygons are equal. (Prob. 7, p. 593.)

8. Why?9. C avalieri’s Principle

10. Postulate 4511. Substitution postulate

630

EXERCISES

VOLUMES

------1

A

1. T he altitude of a prism is 12, while its base is a right triangle whose legs are 24 and 10 respectively. W hat is the volume of the prism?

2. T he base of a prism is a square 15 on a side. W hat is the volum e of the prism if the altitude is 25?

3. A side of the base of a right equilateral triangular prism is 8. W hat is the volume of the p r \m if a lateral edge is 14?

4. T he base of a right triangular prism is an isosceles triangle whose leg and base are respectively 15 and 24. If a lateral edge of the prism is 16, w hat is the volume of the prism?

5. T he base of a right prism is a parallelogram in which two adjacent sides are 8 and 12 respectively. The angle between these sides is 45°. W hat is the volume of the prism if the altitude is 17?

6. A prism whose altitude is 18 has a base with an area of 72. T he volumeof a rectangular parallelepiped is equal to the volume of the prism. If the dimensions of the base of the rectangular parallelepiped are 8 by 6, what is the length of its lateral edge?

7. (a) If the volume of a cube is 64, what is the length of one of its edges?(b ) W hat is the length of a diagonal in a face of the cube?(c) W hat is the length of the line segment that joins a pa ir of opposite

vertices of the cube? (This is called a diagonal of the cube.)

8. A diagonal of the base of a cube is 5V 2, W hat is the volume of the cube?9. How m any cubic feet of w ater could

be stored in a tank such as the one shown in the diagram a t the right?

10. T he front yard of the Evans home was rectangular, having dimensions of 100 feet by 45 feet. In planning his front lawn M r. Evans decided to cover the area with top soil to a depth of 9 inches. If the cost of top soil is $3.50 per cubic yard, w hat was the total cost of the soil?

11. A watering trough for horses has an end that is in the shape of an isosceles trapezoid whose lower base is 2 feet and whose upper base is 3 feet. The distance between the bases is 18 inches. If the length of

VOLUME OF A PYRAMID 631

the trough is 8 feet, what is the weight of the w ater when the trough is full? (1 cu. ft. of water = 62.4 pounds)

B

1. If two prisms have equal bases, then the ratio of their volumes is equal to the ratio of the measures of their altitudes.

2. If two prisms have congruent altitudes, then the ra tio of their volumes is equal to the ratio of their bases.

■ Volume of a Pyramid

T o develop the theorem concerning the volume of a pyram id, we again fall back upon Cavalieri’s Principle. I t is not possible, though, to prove this theorem without paving the way with two prior theorems and a postulate.

P o s t u l a t e 5 0 : If the intersection of two solids is a region of a plane, then the volume of the solid formed from these two solids is equal to the sum of the volumes of the two solids.

T H E O R E M 1 5 1 : If two trian g u la r p y ram id s have e q u a l bases an d cong ru e n t altitudes, th en th e ir volum es a re equal.

G iven: B 1 and Bi lie in the same plane a. Bi = Bi hi = hi

Concl.: Volume of V = volume of P

V P

632 VOLUMES

F ig u re 18-10.


1. Pass any plane parallel to a th a t intersects the two pyramids.

2. B\ = B%, hi = h23. Ai = A2

4. Hence, volume of V = volum e of P

1. W hy possible?

2. Given3. If two triangular pyramids have

equal bases and congruent altitudes, then sections m ade by planes pa rallel to the bases and equidistant from the vertices are equal. (See Problem 8, page 601.)

4. Cavalieri’s Principle

T H E O R E M 152: T h e volum e of a t r ia n g u la r pyram id is equal to one- th ird th e p ro d u c t of th e m easure of th e a ltitu d e a n d th e a rea of th e base.

G iven: Pyram id P -S T W Concl.: Volume of P -S T W = \B h

- 7 W

Figure 18-11.

A n a l y s is : Ak prism is constructed around the given pyramid. O ur objective then becomes one of showing that the given pyramid is one of three equal pyram ids in this prism. Perhaps the most difficult part of this proof is visualizing the three pyramids in the prism. They will be P -S T W , W -PQ R, and P -T Q W . T he pyram id W -PQR will have to be observed from two different points of view: once with vertex W and base P Q R ; the second time with vertex P and base QRW .



1. In plane let PR II S W

and W R II SP.

2. In plane S P T let P Q W S T

and TQ II SP.3. Through P let plane PQR

be parallel to plane S T W .4. P Q R -S T W is a prism.5. A S T W == A P Q R

6. A S T W = A PQR

7. Altitude of P -S T W = altitude of W -PQR

8. .'. P -S T W = W -PQ R

9. A T Q W ^ ARW Q

10. A T Q W - A R W Q11. Altitude of P -T Q W = al

titude of P -R W Q

12. /. P -T Q W = P -R W Q13. But W -PQR = P -R W Q14. .-. P -S T W = P -T Q W15. P -S T W + P -T Q W

W -PQ R = P Q R -S T W16. P -S T W + P -S T W +

P -S T W = P Q R -S T W17. But P Q R -S T W = Bh

18. Hence, 3 P -S T W = Bh19. .'. P -S T W - \B k

1. Why possible?

2. W hy possible?

3. Why possible?

4. Why?5. The bases of a triangu lar prism are

congruent triangles. (See Problem 3, page 612.)

6. If two triangles are congruent, 'then their areas are equal.

7. Reflexive property of equality (Both altitudes are the a ltitude of the prism.)

8. If two triangular pyram ids have equal bases and congruent altitudes, then their volumes are equal.

9. iS'.j'.j’. theorem on congruence (Prove the triangles congruent.)

10. Same as 611. Reflexive property of equality (The

altitudes of both pyram ids is the perpendicular segm ent from P to plane Q T W R .)

12. Same as 813. Reflexive property of equality14. Transitive property of equality15. Postulate 48, page 619.


17. The volume of a prism is equal to the product of the measure of its altitude and the area of its base.'

18. Same as 1619. Division postulate

634 VOLUMEST H E O R E M 153: T h e vo lum e of a py ram id is equal to o n e -th ird th e

p ro duct o f th e m easure of its a ltitu d e a n d th e a rea o f its base.


1. Let V T R be the plane th a t exists through V, T, and R.

2. Let VTQ be the plane th a t exists through V, T, and Q.

3. V -T tiS = \A T R S - h

4. V -T Q R = \A T Q R -h V -TP Q = \ A T P Q h

5. V -P Q R S T = V -T R S + V -T Q R + V-TPQ

6. V -P Q R S T = l& T R S 'h +J A TQ R-h + JA TPQ-hor, V -P Q R ST = JA(A T R S +A T Q R + A TPQ)

7. But P Q R S T = A T R S + A T Q R + A T P Q

8. Hence, V -P Q R ST = ih -P Q R S T or, V -P Q R ST = ihB

1. W hy possible?

2. W hy possible?

3. T he volume of a triangular py ram id is equal to one-third the product of the measure of its altitude and the area of the base.

4. Same as 3

5. If the intersection of two solids is a region of a plane, then the volume of the solid formed from these two solids is equal to the sum of the volumes of the two solids,


7, Postulate 42, see page 582.

8. Same as 6


EXERCISES

1. Find the volume of a pyramid whose base is a square with 12-inch sides and whose altitude is 15 inches.

2. T he base of a pyramid is a rectangle whose diagonal and one side are 35 inches and 28 inches respectively. If the a ltitude is 16 inches, w hat is the volume of the pyramid?

3. T he solid a t the right is a cube whose edge is 9 inches. W hat is the volume of the pyram id A -EF H ?

5.

7.

H E4. Find the volume of an equilateral triangular pyram id with 12-inch

sides and whose altitude is 8 inches.A pyramid whose base is a square has an a ltitude of 15 feet. If the volume of the pyram id is 180 cubic feet, w hat is the length of one of the sides of the square?

A pyram id whose base is an equilateral triangle has a volume of 80v j cubic inches. If the altitude of the pyram id is 10 inches, w hat is the length of one of the sides of the base?A pyram id has a base, of 48 square inches and an altitude of 8 inches. A plane is passed parallel to the base and 6 inches from the vertex. W hat is the volume of the pyramid that was cut off the top of the original pyramid?

8. A plane is passed parallel to the base and 8 feet from the vertex of a pyram id whose base is 45 square feet and whose altitude is 12 feet. W hat is the volume of that part of the pyram id th a t rem ained after the top had been removed?

9. A pyram id has a square base, each of whose sides is 12 inches. A plane is passed parallel to the base and 2 inches from it. If the altitude of the pyram id is 8 inches, what is the volume of the pyram id that was removed?

636 VOLUMES

10. In the pyram id a t the right ABCD is

a square. VE, the a ltitude of the pyramid, passes through the point of intersection of the diagonals. m ZE V C = 45. I? CD = 6, what is the volume of the pyramid?

lil1. If two pyram ids have equal bases, then the ratio of their volumes is

equal to the ratio of the measures of their altitudes, t2. If two pyram ids have congruent altitudes, then the ratio of their volumes

is equal to the ratio of their bases, f3. 1/ the bases of two pyramids are similar triangles, then the ratio of the

volumes of the pyramids is equal to the ratio of the products of the measures of their altitudes and the squares of the measures of a pair of corresponding sides of the bases.

4 . I f a plane is passed parallel to the base of a triangular pyramid, then the ratio of the volume of the top pyramid to the volume of the original pyram id is equal to the ra tio of the cubes of the measures of their altitudes.

5. A plane bisects the altitude of a triangular pyram id and is parallel to the base. T he ratio of the volume of the pyram id above the plane to the

■original pyram id is 1 : 8 . (H int: Use Problem 4 .)

| Surface Area and.Volume of a Cylinder and a ConeW ith bu t slight variations of the definitions that we already

have, it is possible to extend the theorems developed for prisms and pyramidto theorems th a t apply to circular cylinders and circular cones. First, however,w hat are these latter solids?

D e f in it io n 1 1 3 : A circular cylindrical surface is a surface generated by a line that moves so as to always be parallel to a fixed line and always intersect a fixed circle. T he fixed line does not lie in the plane of the fixed circle. (How does this definition differ from th a t of the prism atic surface?)

V

t T hese s ta tem ents often ap p ear as theorems.

AREA AND VOLUME OF CYLINDER AND CONE 637

D e f i n it i o n 1 1 4 : A circular cylinder is a solid bounded by a circular cylindrical surface and two parallel planes.

Term s such as section of a circular cylinder, bases, cross section, right section, altitude, and right circular cylinder have meanings similar to those used with the prism.

Right Circular Cylinder

F ig u r e 18-14.

Applying methods similar to those employed in the proof of Theorem 147, it is possible for us to prove the theorems below. As honors work, you m ight w ant to determ ine what postulates will have to be assumed and how each of these theorems can be proved.T H E O R E M 154: T h e volum e of a c ircu lar c y lin d e r is e q u a l to the

p ro d u c t of th e m easure of its a ltitu d e a n d th e a rea of its base.

Since the base of a circular cylinder is a circle, then the area of the base is represented by A — wr'K Hence, it follows that,

T H E O R E M 155: T h e volum e of a c ircu la r cy lin d e r can b e expressed b y th e form ula:

V = vr*h

w here r is th e m easure of th e ra d iu s o f th e base a n d h is th e m easure of the a ltitu d e of th e cy lin d er.

638 VOLUMEST h e circular cone th a t we mentioned earlier has a definition m uch

like that of the pyramid. And it, too, is derived from a surface sim ilar to the pyram idal surface. Thus,

D e f in it i o n 1 1 5 : A circular conic surface is a surface generated by a line that moves so as to always pass through a given point and always intersect a given circle where the given point is not in the p lane of the given circle. (Figure 18-15.) ■

Again, such terms as vertex, nappe, and element mean m uch the same when related to the circular conic surface as they do in the pyram idal surface (p. 595).

D e f i n i t i o n 116: A circular cone is a solid bounded by one nappe of a circular conic surface and the plane of the circle used in generating the surface. (Figure 18-16.)

F ig u r e 18-15. F ig u r e 18-16.

Theorem s 142 and 153 apply equally well for the circular cone as they do for the pyram id.

T H E O R E M 156: If a p lan e is passed p a ra lle l to the base of a c ircu la r cone and in tersects its surface, b u t n o t th e v e r te x , th e n th e in tersection w ill b e a circle.

T H E O R E M 157: T h e vo lum e of a c ircu la r cone is e q u a l to o n e -th ird th e p ro d u c t of th e m easure of its a ltitu d e a n d th e a rea o f its base.

Since the base of a circular cone is a circle, then the area of the base can be found by the formula A = nr1. Hence, it follows that,T H E O R E M 158: T h e vo lum e of a c ircu la r cone can be expressed by

th e form ula,V = f r M

w h ere r is th e m easure o f th e rad ius o f th e base a n d h is the m easure o f the a ltitu d e .


T h e regular pyramid discussed in problem 13 on page 605 finds its counterpart in the right circular cone.

D e f i n i t i o n 117: A right circular cone is a circular cone in which one endpoint of the a ltitude is the center of the base, (Figure 18-17.)

p

T he slant height of the regular pyram id which is an altitude of one of the faces now becomes the slant height oj the right circular cone.

D e f in i t io n 118: T he slant height of a right circular cone is the set of line segments whose endpoints are the vertex of the cone and a point of the base.Notice that the slant height has been defined in terms of a right circular

cone only. W hy was this definition not broadened to include all circular cones?

W ith this as a background, Theorem 159 below appears to be a natural extension of problem 6 on page 606.

T H E O R E M 159: T h e la te ra l surface a rea o f a r ig h t c ircu la r cone is eq u al to one-half th e p ro d u c t of th e m easure o f th e s lan t h e ig h t a n d th e c ircum ference of th e base.

Needless to say, the circumference of the base is 2w. Hence, one-half of this becomes r r and therefore, Theorem 159 becomes,

T H E O R E M 160: T h e la te ra l surface a rea of a r ig h t c ircu la r cone can b e expressed b y the form ula,

A = irrl

w here r is the m easure of the rad iu s of the base a n d 1 is th e m easure of th e slan t h e ig h t.

Should, we now add the area of the base to that of the lateral surface, we come up with,

640 VOLUMEST H E O R E M 161: T h e to ta l surfacc a rea of a r ig h t c ircu la r cone can

b e expressed by the form ula,

T.A. = irr! + ■* rl or T.A. — tsriy -}-1)

w h ere r ts the m easure of th e rad iu s of the base and I is th e m easure o f the slan t h e ig h t.

O ur work would not be complete if we failed to consider the area of the surface of a right circular cylinder. 'As an outgrow th of problem 6 on page 624 we can say that,

T H E O R E M 162: T h e la te ra l surface a rea of a r ig h t c ircu la r cy lin d eris eq u a l to the p roduct of th e c ircu m feren ce of th e basea n d th e m easure of th e a ltitu d e .

T H E O R E M 163: T h e la te ra l surface a rea o f a r ig h t c ircu la r c y lin d e r can b e expressed by th e fo rm ula ,

A = 2irrh

T H E O R E M 164: T h e to ta l surface a rea of a r ig h t c ircu la r cy lin d ercan b e expressed by th e fo rm ula ,

T.A. = 2irr2 + 2ir rh or T.A. = 2irr(r + h)

Illustration 1:In a right circular cone, the altitude forms an angle of 30 degrees

w ith the slant height. If the slant height of the cone is 12 inches, find (a) the total surface area (b ) the volume

M e t h o d : Using the diagram , we can determ ine the measures of the altitude and the radius of the base. Hence,

(a ) T .A . = t r(r + 0= *6(6 + 12)= 108tt square inches

(b ) V = \xr-h

= 6J ’6 v /3= 72ttV/3 cubic inches

Illustration 2:A rectangle whose dimensions are 6 feet by 8 feet is rotated about its

longer side. Determ ine

(a) the volume (b ) the total surface area of the solid generated.


M e t h o d : In the diagram at the right we perceive that the solid generated is a right circular cylinder whose radius and altitude have measures of 6 and 8 respectively. Hence,

(a) V = irrVi— tt62-8= 288ir cubic inches

(b ) T.A. = 2,rr(r + h)= 2ir • 6 (6 + 8)= 168tt square inches

EXERCISES

(Leave all answers in terms of ir unless otherwise stated.)

1. Find the volume of each o f the following circular cylinders.(a) r = 7", A = 5" (b ) r. = 12', h = 4§'(c) d = 18', A = 6 '4" (d ) d = 24', h = 5 '3" ;;

2. Find the' volume of each of the following circular cones.(a) r = 6 cm, h = 8 cm (b ) r — 15 yards, h - 5 yards(c) d = 8 feet, h = 4 feet 6 inches (d ) d = 3 yards 2 feet, h = 6 yards

3. Find the lateral surface area of each of the following right c ircular cylinders.(a) r = 10', I, = 4 '3" (b) d = 14', h = 5 '6"

4. Find the total surface area of each of the following right circular cyl

inders.(a) r = 5", / = 8" (b ) r = 6", I = 4*"(c) r = 6 cm, // = 8 cm (d ) d = 10 feet, h = 12 feet

5. A right circular cylinder has a lateral surface area of 168tt square inches. T he altitude of the cylinder is 14 inches. W hat is the area of one of the bases?

6. A right circular cone has a lateral surface area of 32ir square inches. The^siantjieight) of the cone is 8 inches. W hat is the total surface area of the cone?

7. T he volume of a right circular cone is 96tr cubic inches. If the altitude of the cone is 8 inches, what is the lateral su^'ace area?

8. The. altitude, of a right circular cone makes an angle of 45 degrees w ith the slant height. If the radius of che base is 6 feet, what is the volume of the cone?

642 VOLUMES

9. (a ) If the altitude of a right circular cone is doubled while the radiusof the base remains the same, what happens to the volume of thecone?

(b ) Tf the radius of the base of a right circular cone, is doubled while the altitude remains the same, what happens to the volume of the cone?

(c) If both the altitude and the radius of the base of a right circular cone are doubled, w hat happens to the volume of the cone?

10. A plane is passed parallel to the base of acone and 6 inches from the base. Thealtitude and radius of the base of the original cone are shown in the diagram.(a) W hat is the radius of the upper cone?( b ) W hat is the volume of the upper

cone?

11. A pile of sand is in the shape of a right circular cone where the radius of the base is 10 feet and the altitude is 6 feet. A board is passed parallel to the ground removing all the sand within 2 feet of the vertex.-W hat is the volume of the sand that remains?

12. T h e bases of a circular cone and a circular cylinder are exactly the same and the vertex of the cone lies in the same plane as the upper base of the cylinder. Com pare the volumes of the two solids.

13. T h e area of the base of a circular cone is equal to the lateral area of the cone. How does the radius of the base compare to the slant height?

14. An isosceles right triangle is rotated about its hypotenuse. If the measure of the hypotenuse is 14, w hat is the volume of the solid th a t is generated?

15. A hot water tank is designed in the shape of a right circular cylinder large enough to hold 75 gallons of water. If the d iam eter of the tank is to be 2 feet, approxim ately how high will the tank have to be? T here are 7§ gallons in 1 cubic foot. (Use as an approxim ation for jr.)

16. (a ) Show that the ratio of the area of the base of a right circular coneto the lateral area is equal to the ratio of the radius to the slant height.

(b ) Show that if two cylinders have, equal altitudes, then the ratio of their volumes is equal to the ratio of the squares of their corresponding radii.

■ Volume and Surface Area of a SphereIn developing the theorem for determining the volume of a

sphere we fall back on Cavalieri’s Principle. Hence it is necessary for us to

VOLUME AND SURFACE AREA OF A SPHERE 643

find a solid where every, cross section of that solid has an area th a t is equal to the area of the corresponding cross section of the sphere. T o do this we work with half the sphere rather than the total one. Therefore, when the proof is completed we will, have to double the volume found to determ ine the volume of the sphere.

T o create the solid needed, we construct a right circular cylinder where the radius of the base and the altitude of the cylinder are both congruent to the radius of the sphere. We then construct a right circular cone whose base is the upper base of the cylinder while its altitude is the altitude of the

F ig u r e 18-18,

cylinder. I t is the solid a t the left above to which we are referring.Specifically, our objective is to prove that the volume bounded between

the cone and the cylinder is equal to the volume of the hemisphere (halfsphere). T o do this we will show that the areas of the shaded regions in the two solids are equal. Since they represent sections formed by any plane parallel to the plane containing the bases of the two solids, then by applying Cavalieri’s Principle we can conclude that the solids have equal volumes.

For the shaded region of the hemisphere:Since this is a circle its area will be,

A \ = t y 2

but, / = r2 — x2therefore, Ai = ir(r2 — x-)

For the shaded region of the solid at the left:

Since A A B C ~ A A D E , x'.AB = z:r

Eut, AB is the measure of the altitude of the cylinder and hence it is the sameas the m easure of the radius.Therefore, x;r = z:r or x = z

T he area of the shaded region at the left is,

A2 = r r1 — t:z‘ or A.i — ir(r2 — z2)Since x = zT he area is, A2 = 7r (r* — x4)

644 VOLUMES

Thus, the areas of the shaded regions are equal and hence the volumes of the solids are equal. But the volume of the solid at the left is the difference between the volume of the cylinder and th a t of the cone. Hence,

Vh — irr2 r — %wr2-r or Vh — rr3 — jirr3 = jirr1

Therefore, the volume of the hemisphere is,

Vk = W

And, in turn, the volume of the sphere is,

T H E O R E M 165: T h e volum e of a sp h e re can be expressed by th e fo rm ula ,

V =

w h ere r is th e m easure o f th e rad iu s o f th e sphere.

T he proof of the last statem ent we would like to develop is somewhat beyond the scope of this course. However, it is possible to justify it intuitively by making use of Theorem 164.

Consider the possibility of cutting u p a sphere as shown in Figure 18-19 below. Each of the solids resembles a pyram id except for the fact that the

Figure 18-19,

base is a region of the surface of a sphere ra ther than a polygon. However, should each of these regions be taken small enough, then for all practical purposes the areas of the bases of these solids could be considered to be the same as the areas of the bases of pyramids. By accepting this, we can find the volume of the sphere by adding the volumes of all the “ pyramids.” Hence,

Volume of sphere = $A,/ii + iA ,/ii + lA 3h3 + . . . + \A nh„

But the altitude of each of the “ pyramids” is the radius of the sphere. Therefore,

V = + 3^jr + . , . + \ A nr

VOLUME AND SURFACE AREA OF A SPHERE 645

And by factoring,V = 3r(/li + A% + A-j + . . . + A„)

However, (A\ + A, + A, + . . . + A n) represents the surface area of th ; sphere which we call A. Hence,

V = \rAT he volume, though, can be replaced by ^ivr1. Thus,

^jrr3 = \rA

And finally we can conclude that,A = 4nfl

In view of the above, the following postulate appears justifiable. P o s t u l a t e 51: T he surface area of a sphere can be expressed by the fo r

mula,A = 4 AT2

where r is the measure of the radius of the sphere.

Illustration:

L iquid storage tanks are frequently constructed in the shape of a sphere. How m any gallons can a spherical tank hold if it has a d iam eter of 21 feet3 (Use as the approximation for ir.)

M e t h o d : ^ = $ ,rrS

- f V ' W - ) *= 4,851 cubic feet

Number of gallons = 4,851 X 7; = 3 6 ,3 8 2 |

EXERCISES

(Leave all answers in terms of ir unless otherwise stated.)

1. Find the volume and surface area of each of the following spheres(a) r = 4" Y . (b ) r = 2J"(c) d = 4 '6" = y 6'.U;f £ (d ) d - 6 yds. 2 ft.

2. T he surface area of a sphere is lOOir square inches. Find the volume of the sphere.

3. T he volume of a sphere is 2887T square feet. Find the surface area of the sphere.

4. A circle of radius 6 inches is rotated about a diameter.(a) Find the surface area of the sphere that was generated.(b ) How does the surface area of the sphere com pare with the area o:

the generating circle?(c) Is the conclusion found in (b ) true of any sphere and its generating

circle? Justify your answer.

£ r < c h ~ 2- , S ' ^ 1 hj , /a /bftA r Zd,

Trr~ie

f ■

646 VOLUMES5. A spherical balloon is inflated to the point where the m easure of its

radius is three times its original measure.

(a) How does the new surface area comparc with the original surface area?

(b ) How does fhe new volume compare with the original volume?

6. (a ) Show that the ratio ol" the surface areas of two spheres is equal tothe ratio of the squares.of their corresponding radii.

(b ) Show that the volumes of two spheres compare as the cubes of their corresponding radii.

7. Tw o glass marbles having diameters of 4 inches and 6 inches respectively are m elted down and then the molten glass is reshaped into a single marble. Determ ine the diam eter of the new marble. (Leave answer in radical form.)

8. A hollow plastic ball has an inner diam eter of 12 feet and an outer d iam eter of 18 feet. I f a cubic foot of the plastic weighs 6 ounces, w hat is the total weight of the ball? (Use 3.14 as the approxim ate value of ir.)

9. A sphere is inscribed in a right circular cylinder.(a ) Show that the volume of the right

circular cylinder is 1^ times the volume of the sphere.

(b ) Show that the lateral surface area of the cylinder is equal to the surface area of the sphere.

T he upper and lower bases of a right circular cylinder are circles of a sphere. The radius of the sphere is 15 inches while the radius of the base of the cylinder is 12 inches. How much of the volume of the sphere lies outside the cylinder?In the diagram at the right, how does the surface area of the sphere compare with the total surface area of the right circular cylinder?

12.* How does the volume of the sphere in the adjacent diagram compare with the volume of the right circular cor.e?

10

11.

■ Test and Review1. T h e base of a right prism is a triangle whose, sides.are 4, 5, and 7. If

the lateral edge of the prism is 12, what is the lateral area of the prism?2. (a) T he base of a right prism is a rhom bus whose diagonals are 10 and

24 respectively. )f the lateral edge is 8, w hat is the lateral area?(b ) W hat is the totai area of this prism?

3. T he base of a right prism is a quadrilateral whose sides are 3, 4, 7, and 8. If the lateral area of the prism is 132, what is the measure of a lateral edge?

4. A right section of a prism is a right triangle with legs of 9 and 12. The altitude of the prism makes an angle of 45° with the lateral edge. If the measure of the altitude is 10, what is the lateral area?

5. T he volume of a rectangular parallelepiped is 105. If the dimensions of the base are 3 and 7, what is the total area of the parallelepiped?

6. T h e base of a right prism is an equilateral triangle that is inscribed in a

circle of radius 6. If the lateral area of the prism is 9 0 \/3 , w hat is the m easure of a lateral edge?

\ / 7. T he dimensions of the base of a rectangular parallelepiped are 6 and 5 respectively. If the total area is 412, find the volume of the solid.

8. T he length, width, and height of a rectangular parallelepiped are in the ratio of 2 :3 :4 . If the volume of the solid is 648, what is the total area?

9. If the measure of the diagonal of a cube is ^ 7 5 , find the volume of the cube.

10. Tw o cubic feet of liquid plastic is poured into a rectangular m old whose base is 20 feet by 10 feet and allowed to cool. How thick will the sheet of plastic be?

11. T h e base of a right prism is an isosceles right triangle whose hypotenuse is 4. If a lateral edge of the prism is 7, w hat is the volume of the prism?

12. T he base of the prism is a square with diagonal 4 v /2. A lateral edge whose measure is 8 makes an angle of 60° with the altitude. W hat is the volume of the prism?

13. One of the great Egyptian pyramids has a square base, one of whose sides is approximately 233 meters, while its height is approxim ately145 meters. If the average weight of th<> m aterial from which it was constructed is 2.8 tons per cubic meter, w hat is the approxim ate weight of the pyramid? (Assume that it is a solid.)

14. The base of ?. pyramid is an equilateral triangle with a side whose m easure is 8. The altitude of the pyram id is 10. A plane is passed parallel

TEST AND REVIEW 647

A

648 VOLUMES

to the base and 5 units from the vertex, removing a pyramid off the top. W hat is the volume of the rem aining solid?

! 15. Find the volume and lateral surface area of a right circular cylinder ! that has a diam eter of 8 cm and an altitude of 7 cm. i 16, Find the volume and total surface area of a right circular cone where

the d iam eter of the base is 24 inches and the altitude is 16 inches.17. Find the volume and surface area of a sphere that has a diam eter of

10 feet:18. If an object sinks when placed in a basin filled with water, then the

am ount of w ater th a t spills out of the container is equal to the volume of the object. An iron ball having a d iam eter of 3 inches is lowered into a hollow right circular cone that was filled with water. If the radius of the base or'che cone is 8 inches while the altitude is 6 inches, how m any cubic inches of water remain in the"cone?

19.* Tw o cubic feet of m etal is drawn into a wire having a diam eter of ■j inch. How m any inches of wire will there be?

20.* T he base of a right circular cone is the base of a hemisphere while its vertex is a point of the hemisphere. Show that the volume of the hemisphere is twice the volume of the cone.

0Prove each of the following statem ents:

1. Two right sections of a triangular prism are congruent triangles.2. If a plane is passed through two diagonally opposite edges of a paral

lelepiped, these edges and the intersections in the two bases will form a parallelogram .

3. If a p lane is passed through two diagonally opposite edges of a parallelepiped, it will divide the solid into two equal triangular prisms.

4. If a line segment contains the point of intersection of the diagonals of a parallelepiped and terminates in the bases, then that point is the m idpoint of the line segment.

■ Try This For FunM athem aticians seem to go out of their way in order to

throw obstacles in their path. To illustrate, earlier we had learned th a t a lth o u g h . they have m any instruments a t their disposal, they insist that geometric constructions be made with the use of but the straightedge and compass. T o limit the scope of their m ovement even further, they will


frequently lay aside one of these two instruments and a ttem p t to seek out the solution to their problem with the aid of the o ther alone.

Thus, in the two intersecting circles below the objective was to find the center of each circle by using only the straightedge. T o do this, a point P

S—fwas selected on the circle a t the left. PA was drawn and extended until it

intersected the second circle a t Q. This was followed by draw ing QB and extending it to the point of intersection with the circle a t R. A second point

P ' was then selected and the process was repeated. This, in turn , was followed

by finding the points of intersection of PP' with R R ' and P R w ith P 'R ';

these being M and / / . Now, were we to draw M jV, this line would pass through the center of the larger circle.

(a) Can you prove this?(b ) W hat further construction is necessary to find the center of the

larger circle?

Index

N um erals show n in italic ind ica te page num bers of definitions.

A b s o lu te v a lu e , 56, 384 A l t i tu d e

o f a prism , 612 o f a rec tang le , 582, 584 o f a trapezo id , 5*85 of a triang le , 747, 549, 584, 585

A n a ly t ic p ro o fs , 423 A n g le , 18

acu te , 30bisector of, 35, 33, 147 cen tra l, 457construc tion o f an angle congrucn t to a

g iven ang le , 540 degree, 453 d ihed ra l, 290 existence postu la te , 154 ex terio r angle o f a polygon, 220, 221. inscribed , 483, 4155 in te rio r of, 142 m easure of, 27-29 m easure o f a d ih e d ra l angle, 297 obtuse, 30 p la n e angle, 291 r ig h t, 30, 90 sides, 18 s tra igh t, 30, 95 vertex , 18

A n g le s ad jacen t, 173a lte rn a te ex terior, 237, 240, 250 a lte rn a te ir .te rio r, 236} 239, 248 com plem en tary , 31, 96, 97 congruen t, 38, 103, 201 corresponding angles o f co ng ruen t poly

gons, 115corresponding angles of parallel lin^s,

237, 239, 249 difference of, 61 o f a polygon, 300-322 rem o te in te rio r angies of a triangle, 220 sura, 60

supp lem en ta ry , 31, 95. 96 vertical, 101

A n te c e d e n t , 85 A pothem , 603 A rc

cong ruen t, 454 degree, 453 in tercep ted , 452 length , 610 m ajor, 451 m easure of, 452 m inor, 451 sem icircle, 451

A re ala te ra l a rea o f a prism , 624of a circle, 612-613of a para lle logram , 585o f a polygon, 580-606o f a rectang le , 582o f a region, 581o f a sector o f a circle, 615o f a trapezo id , 587of a triangle , 586o f sim ilar triangles, 594, 595

B etw eenness points, 13, 55 rays; 59, 173

B is e c to r o f an angle, 34, 38, 526, 542 o f an angle o f a triang le , 549 o f a line segm ent, 25, 38, 524

C a t e o o r i c a l s ta te m e n ts , 191, 193 C a v a l ie r i ’s P r in c ip le , 628, 631, 642 C e v a ’s T h e o re m , 374 C h o rd s , 450

equ id is tan t from center, 461-462 intersecting, 498rad ius pe rp en d icu la r to cho rd , 462 unequal, 570-571

652 INDEXC ir c le , 156, 440, 440-445 , 450^513, 530,

612area* 612, 613 area o f a sector, 615 center, 156, 441 circum ference, 606, 607, 6055 circum scribed , 547, 548 congruent, 454 equation , 441 exterior, 482 im aginary , 443 inscribed, 547, 548 in terior, 482 leng th o f an arc of, 610 point,' 443 radius, 156, 441 sector, 614, 615 tangen t, 467

C i r c u l a r co n e , 638 la te ra l surface area , 639 r ig h t c ircu la r cone, 639 s lan t he igh t, 639 to ta l surface area , 640 volum e, 638

C i r c u l a r co n ic s u r f a c e , 638 C i r c u l a r c y l in d e r , <537

la tera l surface area , 640 to ta l surface area , 640 volum e, 637

C i r c u l a r c y u n d r i c a l s u r f a c e , 636 C o l l in e a r p o in ts , 201

C o m m ensu rab le l i n e segm ents, 334 C o m p le m e n ta ry a n o le s , 31

com plem ent, 31, 96, 97 Com pound lo c i , 436, 533-537 C o n c u r r e n c y

o f altitudes o f a triang le , 559 o f ang le b isec tors o f a tri.an.gte, 517 o f m edians o f a triang le , 392 o f pe rp en d icu la r bisectors o f th e sides of

a triang le , 536, 559 C o n d i t io n a l s ta te m e n ts , 85, 191-192 C o n g ru e n c e , 62

o f angles, 38 of arcs, 454 of circles, 454 of line segm ents, 38 of polygons (see triang les), 118 postulates, 73

C o n s e q u e n t , 85 C o n s tru c t io n s , 539-554

com pass, 539straigh tedge, 539, 626-627

C o n tr a p o s i t iv e , 517-522, 562-563

C o n v e rse , 248, 269, 517-522, 562-563 C o o r d in a te g e o m e try , 375-449

abscissa, 379 ana ly tic geom etry, 375 collinear points, 396 coord inate, 377, 378 co o rd inate system, 23, 28, 55, 379- m idpo in t of a line segm ent, 388 o rd ered pairs, 379 o rd in a te , 379 orig in , 376p o in t th a t divides a line segm ent in to a

given ratio , 386 q u a d ra n t , 377 slope, 395

C o p la n a r po in ts , 201

C o rre s p o n d e n c e , 23, 28, 1 1 2 -U 6 , 380 chords a n d m inor arcs, 452 correspond ing angles, 115 correspond ing sides, 115 equ ivalen t, 116

C ube, 626

D aV in c i, L e o n a rd o , 621 D e f in i t io n s

conno ta tive , 6, 7 construc ting a definition, 4 -6 deno ta tive , 7 need for, 2 o f geom etry , 540 properties , 6 reverse of, 40 synonym ous, 6

D e g re e an g u la r , 31, 453 arc, 453

D e s c a r te s , 375 C a rte s ia ir coord inate system, 379

D e s c r ip t io n , 9 D e te rm in e , 200

a line, 200a p lane , 201, 204, 205, 280

D i h e d r a l a n o le , 290 cong ruen t, 291 edge, 290 faces, 290m easure, 291 ,p la n e an g le of, 291

D is ta n c e , 179 betw een a po in t and a line, 378 betw een tw o points, 180, 382-385 equally d istan t, 180 eq u id is tan t, 180-186 geodesic, 179

E q u a t io n s , 22 ^

dependen t, consistent, inconsistent, 422 equal products, 353 equal rntios, 353 of a ciiclc, 441 of a line, 441 p roportion , 329

E q u a l i ty , 22 reflexive property, 72 sym m etric p roperty , 72 transitive property, 72

E u c lid , 246, 247, 314, 316 fifth postu la te, 247

E x is te n c e , 255-257, 540 E x is te n c e an d uniqueness o f o r d e r , 316,

564

G eodesic , 179 G e o m e tr ic f ic u ri- , 12 G r a p h , 411

circle, 442-445 o f an equa tion , 414 of inequalities, 425-433 line, 414-418

H a lf - p l a n e , 289, 377 , 431-432

If -c la u s e , 191Inco m m en su rab le l in e secm ents, 334 In e q u a l i ty , 219, 426-428, 564-579

angles o f a triangle, 566 chords, 570ex terio r angle o f a triangle , 221, 303 existence and uniqueness o f o rder, 316,

564graphs, 425-431 o p en sentence, 427 sides of a triangle, 565 transitiv ity o f order, 316, 564

In sc r ib ed a n o le s , 483 in a sem icircle, 485 in the sam e arc, 491 m easure of, 483

I n te r s e c t io n o f sets , 11, 419-422, 437 In v e rs e , 517-522, 563 Isoscei.es t r i a n c l e , 142

base, 142 base a n g le s /l 42

. legs, 142 theorem s, 145-146 vertex angle, 142

L a w o f C o n tr a d ic t io n , 227 L a w o f t h e E x c lu d e d M id d le , 227

INDEXv - M......... .'.T1 S®

L ig h t w aves, 52 : ' • ' i i l i lL ines, 10

bisector of a line sep m -n t * 2 || line of centers, 457 ' ■'■**line segment, 7 5 .....oblique, 45 -*■'* fHopposite sides o f a line, 28.9 parallel, 234 **parallel to the sam e line , 2 5 0 , perpendicular to a ' | S ^ | p § $ perpendicular to th e sam e iine 94® same side of a line, 289 secant, 4SGskew, 230 ;i1vitangent, 466 transversal, 235trisectors of a line segm ent, 27

L ine segm ents, 15 altitude, 147angle bisector o f a trian g le , 747, 549 bisector of, 25 chord, 450com m ensurable, 334 congruent, 38, 103 construction of, 540 corresponding sides of con g ru en t tr i

angles, 115 diagonal o f a po lygon, 259 diam eter, 451 difference, 58hypotenuse, 161, 360,. 361, 366 lateral edge, 623 • \m easure of, 23-24 , 338 ‘ fm edian, 147, I 'll m idpoint, 25, 38 secant segm ent, 471 sum , 57tangent segm ent, 471

L obachevsky , 247, 313, 314 L ocus, 434

com pound, 436, 533-537 in coordinate geom etry , 434 -4 3 7 in synthetic geom etry , 514-563

L o g ic a l in c o n sis ten cy , 2'27

M easure

of a central angle, 452of ? line segm ent, 23 -24 , 50, 338c f an angle, 27-29of an angle w hose vertex lies ou tside a

circle, 487 of an arc, 452of an inscribed angle, 483, 485

654 INDEXof a tangen t-chord angle, 27-29 u n it of, 333

M e d i a n

o f a trapezo id , 392, 405 o f a triangle , 747 , 277, 392, 549

M i d p o i n t o p a l< n e s e g m e n t , 25, 3 8 . 184 m idpo in t form ula, 388

N on-E uclidean ceometry, 245-247, 313— 319

Lobachevsky, 247, 313, 314 R ie m an n , 247, 314, 315

O peration al postulates, 57, 62-70 O rdered pa ir , 379

P a r a g r a p h p ro o f , 228 P a r a l l e l e p ip e d , 6 2 6

rec tangu la r, 6 2 6 cube , 6 2 6

P a r a l le l i s mconstruc tion o f a line parallel to a given

line, 546 in a p lane, 218-278 in space, 279-299line parallel to a side of a triang le , 324,

325, 334, 336, 404 para lle l lines, 2 3 4 , 279, 281, 397, 398 p ara lle l planes, 2 7 9 , 283, 596, 597

P a r a l l e l o g r a m , 2 5 8 , 258-273 diagonals, 259, 262 rec tang le , 2 5 9 rh o m b u s 26 0 , 263 square , 260, 262

P a r a l l e l P o s tu la t e , 247 P a s c h ’s Axiom, 144, 255 P e r p e n d ic u la r i ty

betw een a line an d a p lane, 198-217, 284 betw een tw o lines, 172-197, 400, 401 betw een tw o planes, 2 9 3 construc tion o f perpend icu la r lines, 542-

544perp en d icu la r lines, 3 4 , 240 p lanes p erpend icu la r to the sam e line,

282P i, 608-609 P la n e , 1 9 9 , 201

faces o f a d ihed ra l angle, 290 half-p lane , 2 8 9 , 377 p e rp en d icu la r planes, 2 9 3 ta n g en t to a sphere, 4 7 9

P l a y f a i r , 246 P o in t , 9

co llinear, 2 0 7 , 396

coplanar, 2 07

endpoints of a line segm ent, 15 foot of a line, 207 m id p o in t, 2 5 of tangency, 467 on the same side of a line, 2 8 9 origin, 376 plotting, 376 trisection points, 27

P o l y c o n s , 772 area, 580-606 circum scribed, 5 4 7 congruent, 778 convex, 3 0 9 decagon, 312 duodecagon, 312 hexagon, 312 inscribed, 5 4 7 n-gons, 311, 312 octagon, 312 pentagon, 312 quadrilateral, 2 5 8 regular, 6 0 3 , 602-604 sim ilar, 3 4 2

sum of the measures o f the angles, 309 sum of the m easures of the ex terior a n

gles, 311 P o n s A s in o r u m , 171 P o s t u l a t e s , 53

addition , 62 assumptions, 54 axioms, 54 division, 68 equality, 72 m ultiplication, 68 operational, 57, 62-70 Pasch’s Axiom, 144 reasoning, 85-87 subtrac tion , 65

P r is m , 6 2 3

cross section, 624 faces, 624 lateral edges, 623 parallelepiped, 6 2 6 righ t prism , 624 righ t section, 624 triangu lar, 624 volum e, 626-629

P r i s m a t i c s u r f a c e , 623 P ro o f

appeal to au thority , 2 by elim ination, 226 by force, 1in d ire c t, 144, 224-229

INDEX655

p a ra g ra p h , 228 tw o-co lum n, 91

P r o p o r t io n , 329 extrem es, 330 fou rth p ropo rtiona l, 330 genera l p rop o rtio n , 330 m ean p ro p o rtio n , 330 m eans, 330 p roportionals , 330 p ro p o rtio n by ad d itio n , 331

P r o t r a c t o r , 28 P y ram id , 201, 596

a ltitu d e , 596 base, 596 reg u la r, 605 s lan t he igh t, 605 tr ia n g u la r, 596 volum e, 631-634

P y r a m id a l s u r f a c e , 595 elem ent, 595 n ap p e , 595 p y ram id , 596 vertex , 595

P y th a g o r a s , 365-366converse of th eo rem of, 370 P y th ag o rean Society, 365 th e o rem of, 366, 507, 621

Q u a d r a n t , 377 Q u a d r i l a t e r a l , 258

R a d i u s o f a c i r c l e defin ition of, 156 th e o rem , 156

R a d iu s o f a r e g u l a r p o ly g o n , 603

R a t io , 329 equality, 353, 354

R a y , 76 b isec to r o f an ang le, 35, 38 en d p o in t of, 16 opposite rays, 17

R e a so n j n o correct, 86, 90 in co rrec t, 87

R e c ta n g l e , 259 R e p la c e m e n t p r o p e r t y , 173 R hom bus, 260, 263 R iem ann , 247: 314, 315 R jo h t a n g le s , 30

th eo rem , 9U RtOHT t r i a n c l e , 742

a ltitu d e to hypotenuse , 360, 361, 507 arm s, 1 6 1

construction of, 552 H .L, theorem, 161 hypotenuse, 161 legs, 161

R o p e s t r e t c h e r s , 365

S a c c h e r i , 314-319 quadrilateral, 315

S e c a n t segm en ts , 476, 498-502 S e c t io n , 601, 624 S e l f - e v id e n t t r u t h s , 53 S e t , 10, 409, 514

elements or members of, 10 intersection, 11, 419-422, 437 of ordered pairs, 410, 411 solution set, 41 i subsets, 10 union, 11

S h o r t e s t p a t h ’ between two points, 179

, from a point to a line, 378 S k ew lin e s , 280

‘ S lo p e , 393-403 ' .. of a line, 395, 416

c f a line segment, 395 of lines parallel to the a- or^-axis, 402 of parallel lines, 397 of perpendicular lines, 400 one over zero slope, 403

S o l i d , 595 prism, 623

• \ volume, 627 S p a c e , 199S p h ere , 477, 477-480, 642-649

center, 477intersection o f a plane and a sphere, 478,

479 radius, 477 surface area, 645 ta n g en t planes, 479, 480 volume, 644

S q u a r e , 260, 262 S t r a i g h t a n g le s , 30

theorem , 95 S u b s t i tu t io n p r o p e r t y , - 175 S u p p le m e n ta ry a n g le s , 37

supplem ent, 31 theorems, 95, 96

S u r f a c e , 198 plan*, 799 prismatic, 623 pyramidal, 595 sphere, 477

geometry harry lewis

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