elements of geometry - forgotten books

ELEMENTS OF GEOMETRY,

INCLUDING

PLANE, SOLID, AND SPHERICAL GEOMETRY.

BY

GEO. W. HULL, M. A PH.D.,

Pnornssoa or MA'ms

-n u los i n m s F lRS ‘

l' PE NNSYLVANIA ST AT !

NORI AL Scuoou, M tLLs nsm s, PA. ; Au'rnoa or

Hum /

s Amm an-

n ee Ann Amma n.

BUT LER, SHELDON COMPANY

PHILADELPHIA, NEW YORK, CHI CAGO, BOS TON

P R E FA C E .

IN the preparation of this work the author has keptconstantly in view not only the object to be attained by

the study of Geometry , which is the power of deductive

reasoning, but also the needs of the student in the acquisi

tion of this power.

A careful study of the work in the class- room indicates

that,if symbols are used too extensively , the language and

frequently the logic of the successive steps of a demonstra

tion are impaired ; and, on the other hand,if no symbols

are used,the subject becomes tedious and the student is

discouraged . T he endeavor herehas been to avoid both

of these errors, by the present form and arrangement of

the demonstrations.

A student beginning the study of Geometry requires a

great deal of aid before he can give a clear and logical

demonstration ; therefore, at first, the reason for each step

of the demonstration is given in small type immed iatelybelow the statement. But too much aid of this kind has

been found injurious, and leads to careless habits of

study . T hese references have therefore been limited to

the first nine theorems of Book I . In all other cases

the numbers have been given to indicate the references.

3

C ON T E N T S .

PLANE GEOMET RY.

INT RODUCT ION

BOOK I .

Le,ANGLES , AND POLYGONS

Perpendiculars and Angles

Parallel Lines

T riangles

Quadrilaterals

Polygons

Plane Loci

Practical Examples

Original T heorems

RAT IO AND PROPORT ION

Definitions

Prepositions

Measurement ofQuanti ties

T heory of Limits

Practical Examples

Original T heorems

Definitions

Propositions

Original T heorems

Problems of Construction

Problems for Construction

Loci of the Circle

6 CONTENTS .

BOOK IV.

AREA AND RELAT roNs or POLYGONSDefinitions

Area of Polygons

Squares on Lines

Proportional Lines

S imilar Polygons

Relation of Lines in a Circle

Original T heorems

Problems of Construction

Harmonic Proportion

BOOK V.

REGULAR POLYGONE AND MEAsUREN ENT or T HE CIRCLE

Definitions .

Propositions

Problems for Construction

Original T heorems

Pmctical Examples .

SUPPLEMENT TO PLAN E GEOMETRY.

MAXIMA AND MINIMA

SOLID GEOMETRY

BOOK VI .

PLANES AND POLYHEDRAL ANGLES

Definitions

Lines and Planes

Dihedral Angles

Polyhedral Angles

Original T heorems

BOOK VII

POLYHEDRONSDefinitions

T he Prism

CONTENTS .

S imilar Polyhedrons

Regular Polyhedrons

Practical ExamplesOriginal T heorems

BOOK VIII .

T E E CYLINDER, CONE, AND SPHERET he Cylinder

T he Cone

T he Sphere

Practical Examples

Original T heorems

SPHERICAL GEOMETRY.

BOOK IX.

SPREEICAL SURFACES

Pmpositions

Equal and Symmetrical T riangles

Measurement of Spherical Polygons

Original T heorems

SUPPLEMENT TO SOLID

T HE WEDGE AND PR ISMOID

T heWedge

T he Pr ismoid

Practical Examples

M ISCELLANEOUS OR IGINAL T HEORENSFoam or MENSURAT ION

ANSWERS T o NUMERICAL EXERCIS ES

GrammyBIOGRAPH ICAL NOT ES

IND“ o 0 0 0 O 0 O O 0

SUGGESTIONS TO TEACHERS.

T HE attention of the teacher is called to the following sug

gestions

1 . T he pupil should be required to fix the fundamental

definitions of geometry firmly in mind , and to illustrate them

so fully that there can be no doubt of his clearly comprehend

ing them .

2. All definitions, axioms,theorems

, corollaries, etc., should

be memorized and repeated until they can be given without

conscious effort . Demonstrations should not be memorized .

T he pupil should be taught to think clearly and logically , and

to express the thought in concise and elegant language .

3 . T he construction of the figure is an important part of the

demonstration ,and no assistance should be given the pupil.

T he'

figures should all be constructed with a ruler and compasses.

4. In going over each book for the first time the Exercises

can be omitted ,and taken in connection with the review .

5. S tudents should be taught the Sequence of Propositions. T hus,

the area of a triangle depends upon the area of a parallelogram

the area of a parallelogram depends upon the area of a rectan

gle, etc. T his can be used in any book with excellent results.

6 . S tudents are sometimes at a loss to know in what form

written exercises in geometry should be given . We therefore

suggest that in all written exercises care be taken to arrange

the work to secure clearn ess of thought and brevity of expres

sion . Symbols and abbreviations can be used with great profit .

It is well to begin each statement on a separate line, giving the

8

S UGGES T IONS T O TEACHERS . 9

reason for each step briefly in parentheses immediately below

the line .

We recommend two forms, the first for B eginners, Reviews, or

E saminations,and the second for Advanced Pupils.

Advanced pupils should be required to write on the black

board only those proportions and equationswhich are necessaryfor a clear and rapid demonstration . References, constructions,

etc. , should be given orally .

1. For B eg inners , Review s, or Examinations .

PROPOS IT ION XVI I . BOOK I .

The sum of the ang les of a tr iang le is equal to two

r ight ang les .

G iven—T he A AB C'.

T o Prov e— l A Ad Z0 two right angles.

Dem—Draw BE IIAO, and produce AB to D.

T hen AA Za ,

(Being ext .- int . ang .)

AC A(3.

(Being alt .- int . ang . )

But As.

(T he sum of all the angles about a point on the same

of a straight line equals two rt . 4

Substituting 4 A for Aa,and AC for Ac

, we have

Q. E . D.

IO S UGGES T IONS T O TEACHERS .

PROPOS IT ION VI I . Boox IV .

The a rea of a trap ez oid is equ a l to one half thesum of i ts bases mu ltip lied by i ts a lti tude.

Giv en—AC'DE a trapezoid ,whose altitude is a, and upper and

lower bases b and B respectively .

T o Prov e—T he area ACDE “ B b)a.

Dem—Draw the diagonal AD.

T hen the area of A AC’D x a. (1)

(T he area of a A is equal to one- half the product of its base

and altitude .)

T he area of t he A AED abx a .

(For the same reason .)Adding (1) and the area AOBE “ B b)a.

2. For Advanced Pupils .

PROPOS IT ION X . Boox IV .

I n any tr iang le, the squ a re of a side opp osi te an

acu te ang le is equ a l to the sum of the squa res of the

other two s id es , m inu s tw ice the p rod u ct of one ofthese s id es and the p roj ect ion of the other sid e

up on it .

FIG. 1. FIG. 2.

S UGGES T IONS T O TEACHERS . 11

G iv en— B an acute angle of the triangle AB C,

pendicular to AB .

T o Prov e 2737 + BU? 2AB x DB .

Dem— In Fig . 1 , AD AB DB .

In Fig . 2, AD BD AB .

In the right AABO,

m =m +m —2A3 x DB ,

W =m +W —2AB X DB . Q . E . D.

PROPOS IT ION XXIX . Boox IV .

Two sim ila r polygons a re to each other

squares of any two homologou s sides.

Given—P and P' the areas of the similar polygons EB and

LG respectively .

P EB“

Dem—S ince

T hen

Hence Q. E . D.

PLANE AND SOLID GEOMETRY.

INT RODUCT ION .

DEFINIT IONS OF T ERM S .

l . EVERY material object occupies a definite portion of

space. T his definite portion of space, considered apart

from the body which filled it, is a geometrical solid. T he

material body which filled the space is a physical solid.

A physical solid . A geometrical solid .

A geometrical solid is therefore merely the form of a

physical solid . S ince geometry deals only with geomet

t ical forms,the word solid in this work is used for geo

metrical solid .

A S olid is a definite portion of space, and has length,breadth

,and thickness.

2. T he limitations of a solid are surfaces,the limitations

ofa surface are lines,and the limitations ofa line are points.

A S urface is that which has length and breadth, without

A Line is that which has length, but neither breadth nor

INTRODUCT ION. 15

or more angles have the same vertex,they may be desig

nated by placing a letter in each angle, as the angle a,”

or by naming the letters at the extremities of the sides,

placing the letter at the vertex between the other two, as

the angle B CD.

”

T wo angles, such as B CD and DOE,which have the same

vertex and a common side between them,are called Ad

j acent angles.

T wo angles are called vertical, or opposite,when they have

the same vertex and their sides extend in opposite direc

tions, as angle c and angle d .

EqualAngles are angles which have the same divergence,

or when applied to each other their sides w ill extend in

the same direction .

When one straight line meets another straight line,

making the two adjacent angles equal, each of these angles

is called a Right Angle, as EGF and FGH,and the first line

is Perpendicular to the second .

An Acute angle is less than a right angle ; as the angle

An Obtuse angle is greater than a right angle ; as the

angle EGK.

Intersecting lines not perpendicular to each other are

called Oblique I/ines.

When the sum of two angles is equal to a right angle,

each is called the Complement of the other ; thus, HGK is

the complement of KGF.

When the sum of two angles is equal to two right angles,

each is the Supplement of the other ; thus, HGK is the supplement of KGE .

It is evident that

(a) The complements of equal angles are equal.

(b) The supplements of equal angles are equal.

16 PLANE AND S OLID GEOME TRY.

A Plane Surface, or simply a Plane, is a surface such that

if any two of its points be connected by a straight line,

this line will lie wholly in the surface.

A Curved S urface is a surface no portion of which,how

ever small, is plane.

6. F igures .

A Geometri cal Figure is any combination of points, lines,

surfaces, or solids.

A Plane Figure is a plane bounded by lines either straight

or curved .

Rectilinear Figures are figures bounded by straight lines

Curvilinear Figures are figures bounded by curved lines

Mixtilinear Figures are figures bounded by straight and

curved lines.

7. General Definitions .

Geometry is the science of extension,and treats of the

forms and relations of geometrical magnitudes.

A Theorem is a truth requiring demonstration.

A Lemma is an aux iliary theorem.

An Axiom is a self- evident truth .

A Problem is a question to be solved .

A Postulate is a problem whose solution is self- evident.

A Proposition is a general term applied to theorems,

problems, axioms, or postulates.

A Corollary is a truth easily drawn from a proposition .

A Scholium is a remark upon one or more propositions.

T he Hypothesis is that which is assumed in the state

ment of a theorem.

T he Conclusion is that which follows from the hypothesis.

A proposition is the Converse of another when the hy

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18 PLANE AND S OLID GEOME TRY.

T he plural of these terms may be denoted by writing an

8within or without the symbol ; as A , “8, etc.

13.

Def. , definition . Hyp.

,hypothesis.

Ax .

,axiom. Cons.

,construction .

Post, postulate. Adj . , adjacent.

T h., theorem. Ext.

,exterior.

Cor.

, corollary . Int. , interior.

Alt .

,alternate. Comp .

, complementary .

Inc.

,included . Sup . supplementary .

Ex .,

exercise. Dem.,demonstration .

Q. E . D., quod erat demonstrandum which was to be

proved) .

Q. E . F .

, quod erat faciendum which was to be done) .

PLANE GEOMETRY.

B O OK I .

LINE S , ANGLE S , AND POLYGON S .

PERPEND IOULAR S AND ANGLES .

PROPOS IT ION I . T HEOREM.

14. F rom a g iven p oin t in a stra igh t line on ly one

p erp end icu la r can be erected to tha t line .

E

Given—C any point in the straight line AB .

T o Prove—T hat but one perpendicular can be drawn

to AB at C.

Den —Draw DC, making 1 DCB ADCA, and revolve

DC about C toward the position of CB .

1 DCA will constantly increase, and ADCB will con

stantly decrease ; hence there can be one position of DC,

and but one, where the angles are equal.

Let CE be this position ; then CE is the only perpendic

ular that can be drawn to AB at the point C. Q. E. D.

19

20 PLANE GEOME TRY.—BOOK I .

15. COR —All right angks are

equal.

Given— ABC and DEF right

angles.

T o Prove— £ B = AB .

Dem— Apply DEF to ABC, placing E on B,and making

DE fall on AB .

T hen EF will coincide with BC ; for otherwisewe should

have two perpendiculars to AB at 'B,which is impossible.

[From a given point in a straight line only one perp endicular can

be erected to that line .] a14

NOT E — T he method of demonstration employed in this corollary is

called Superp osition .

PROPOSIT ION I I . T HEOREM .

16. If a stra igh t line meets a nother stra ight line ,the sum of the two adj acen t ang les is equa l to two

r ight ang les .

Given—DC meeting AB at C.

T o Prove— éAOB AB CD two right angles.

Dem— Draw CE perpendicular to AB at 0.

T hen AACD rt. angle ACE + ZDCE’

,

[The who le is equa l to the sum of all itsmm]And AB CD rt. angle BCE éDCE .

Adding these two equalities, we have

AAOB ZB CD two right angles. Ax . 2

[If the same operation is performed on equals, the results

equal ] Q. E . D.

PERPENDICULARS AND ANGLES . 21

1 7. COR . 1 .—]f one of the two adjacent angles,AOB or DCB ,

is a right angle, the other is also a right angle.

18. COR . 2.— The sum of all the

angles formed on the same side ofa straight line at a given point is

eq ual to tworight angles.

For

[The whole is equal to the sum of all its parts.] Ax . 4.

AAOB ADCB two right angles.

Q16

19. COR . 3.

— The sum of all the angles formed about a

point in a plane is equal to four

Produce BO to E .

T hen [ a [O [ e two E m

right angles, 18

And Ac Ad two right an

gles. 18

Adding these two equalities, we have

[ a [ b Ac ZAOD= four right angles.

[If the same operation is performed on equals, the results

EXERCIS E S .

1. If one of two supplementary adjacent angles is what is the

other angle ?

2. How many deg rees in each of three angles formed on the same sideof a straight line at the same point, if they are equal

?

8. How many degrees in each of five angles formed about a point in

a plane, if they are equal ?

4. In Q18, if Aa 38° and Ab how many degrees in

22 PLANE GEOME TRK—BOOK I .

PROPOS IT ION I I I . T HEOREM.

20. CONVERS ELY—If the sum of two adj acent ang les

is equa l to two r igh t ang les , the ir ex ter ior sid es form

one and the same stra igh t line .

Given— é a ADCB two right angles.

T o Prov e—AOand CB a straight line.

Dem— If A08 is not a straight line, let ACE be a straight

line .

T hen Aa ADCE two right angles.

Aa ZDCB two right angles.

£ a+ [ DCE = £ a + ADCB .

[Things that are equa l to the same thing are equal to each other ]Ax . 1

Dropping the common angle a,

ADCE = ADCB.

T his is impossible unless CE coincides with CB .

Hence ACE is a straight line. Q. E. D.

PROPOS IT ION IV . T HEOREM.

21 . If two stra igh t lin es i n tersect each other , the

opp osi te or ver tica l a ngles a re equa l.

Given—LT he straight line AB intersecting CD.

T o Prove Za Ac.


Dem— é a Ab two right angles.

[Being sup . adj . 4 a]

For the same reason Ab Ac two right angles

Hence

Dropping the common angle b,

Aa Ac.

In like manner we may prove Ab Aat. Q. E . D.

PROPOS IT ION V. T HEOREM.

22. If a p erp end icu la r is erected a t the midd le

p oin t of a stra ight lin e, then

l st . Any p oin t in th e p erp end icu la r is equ a lly d is

tan t from the ex tremi ties of the lin e .

2d . J ny p oin t w i thou t the p erp end icu la r i s u n

equ a lly d istan t from the ex tremit ies of the line .

I . Given— CD perpendicular toAB at itsmiddle point C.

T o Prove— Any point of CD equally distant from A

and B .

Dem— Let E be any point of CD,and draw AE and BE .

Revolve the figure ACE around CE as an axis,then AC

will take the direction of CB , since AACE = AB CE .

[Being right ang les ] a15

And S ince,by hypothesis, AO= CB

,the point A will fall

on B,

And AE will coincide with BB .

T herefore AE BE .


I I . Given— E X perpendicular to

EF at its middle point K.

T o Prove— Any point without

E X unequally distant from E

and F .

Dem— Let P be any point without E X, and draw PE and

PF,one of which will cut the perpendicular at some point,

as G ; from G draw GF.

T hen PF PC GP .

[A straight line is the shortest distance between two points ] Ax . 5.

But FG E G.

[If a p erpendicular is erected at themidd le point of a straight line,

any point in the perp endicular is equally distantfrom the extremities

of the line.] t 22

Substituting, PF B C GP,

Or Q. E . D.

In apply ing the figure ACE to B CE it was found that

they coincided throughout.

Hence AA AB,

And AAE C‘

= 4 BB C.

T herefore,

23. COR . 1 .

— If a perpendicular is erected at the middle

point of a straight line, and lines be drawn from any point in

the perpendicular to the extremities of the line,1. They willmake equal angles with the line.

2. They willmake equal angles with theperpendicular.

24. COR . 2. The perpendicular at the middle point of a

straight line contains all points equally distant from the ex

tremities of the line.

25. COR . 3.— A line which has two points equally distant

from the. extremities of another line is perp endicular to that lineat its middlepoint.

26 PLANE GEOME TRY.—B OOK I .

PROPOS IT ION VI . T HEOREM.

26 From a g iven p oint w ithou t a stra igh t line onlyone p erp end icu la r can be d rawn to the line .

Given—C the point, and CD the perpendicular to AB .

T o Prove— T hat CD is the only perpendicular that can

be drawn from C to AB .

Dem— If possible, suppose CF also perpendicular to AB .

Prolong CD to E ,making DE equal to CD,

and draw FE .

T he line FD is perpendicular to CB at its middle point

D,hence

[If a perpendicular is erected at the middle point of a straight

line, and lines be drawn from any point in the perp end icular to the

extremities of the line, they will make equal angles with the perpen

423

But,by hypothesis, ACFD is a right angle.

Hence its equal, 4 EFD,is a right angle,

And CFE is a straight line.

[If the sum of two adj acent ang les is equal to two right angles, their

exteri or sides form one and the same straight line.] Q20

T his is true only when CF coincides with CD.

[Only one straight line can be drawn connecting two given points ]Ax . 6

Hence only one perpendicular can be drawn from a

given point to a given line. Q. E . D.

EXERC IS E .

11. A line drawn perpendicular to the bisector of an anglemakes equalangles with its sides .


PROPOS IT ION VI I . T HEOREM .

27. If two lines ar e d rawn from a p oin t to the ex

tr emit ies of a stra igh t line, th ei r sum is g rea ter than

the sum of tw o other lin es s imi lar ly d rawn ,bu t in

clu d ed by them.

Given— CA and CB two lines drawn from C to the ex

tremities of AB,and DA and DB two other lines similarly

drawn,but included by CA and CB .

T o Prove AC CB AD DB .

Dem— Produce AD to E .

T hen AC+ CE AD DE .

[A straight line is the shortest distance between two points ] Ax . 5

For the same reason,

EB DE DB .

Adding these two inequalities, we have

DE + DB .

Dropping DE from both members of the inequality , and

substituting CB for CE EB,we have

AO+ CB > AD DB . Q. E . D.

EXERC IS E S .

12. How many degrees are there in an angle the sum of whose com~

plement and supplement is 160°

13. How many degrees are there in an angle whose supplement is

equal to three times its complement ?

14. If two straight lines in tersect each other making one of the angleswhat are each of the other angles

15 . T he bisectors of two supplementary adjacent angles are perpen

d icular to each other .

16 . T he line which bisects the angle formed by the intersection of two

lines bisects also its vertical angle.

28 PLANE GEOME TRYZ—B OOK I .

PROPOS IT ION VI I I . T HEOREM.

28. If from a p oin t w i thou t a stra ight lin e a p er

p end icu la r is let fa ll to the line , a nd obli qu e lin es

a re d rawn , then

I st . T he p erp end icu la r is shorter than any obli qu e

lin e .

2d . Any two obliqu e lines wh ich ou t of on the line

equa l d istances from the foot of the p erp end icu la ra re equ a l .

3 d . The obliqu e line wh ich cu ts ofl“on the stra igh t

lin e the g rea ter d ista nce from the foot of the p er

p end icu la r is the g rea ter .

V

I . Given— CD p erpendicular to AB,and CF any other

line drawn from C to AB .

T o Pt ove

Dem.

— Produce CD to E ,making DE = CD

,and draw FE .

T hen,since FD is perpendicular to CE at its middle

point D,

[If a perpendicular is erected at the middle point of a straight line,any point in the perpendicular is equally distant from the extremities

of the line.] 622

But CD DE CF + FE .

[A straight line is the shortest distance between two points] Ax . 5

T hen 2CD 2CF,

Or Q. E . D .


I I . Given— MN perpendicular to GH, and MK and ML

oblique lines drawn from M to GH ,cutting Off equal dis

tances from the foot of the perpendicular.

T o Pt ove

Dem.

— S ince MN is perpendicular to XL at its middle

point N,

[If a perp endicular is erected at the middle point of a straight line ,

any point in theperp endicular is equally distant from the extremities

of the line.] Q22. Q. E . D.

I I I . Given— P0 perpendicular to S T , and PR and PQ

Oblique lines drawn from P to S T,cutting off unequal dis

tances from the foot of the perpendicular.

T o v e PQ PR .

Dem— Produce P0 to V,making 0V= P0,and draw RV

and QV.

T hen, since SO is perpendicular to PV at its middle

point 0,PQ= QV, and PR = RV.

[If a perpendicular is erected at the middle point of a straight line,any point in the perp endicular is equally distantfrom the extremities

fi22

But PQ+ QV> PR + R V.

[If two lines are drawnfroma point to the. extremities of a straight

line, their sum is greater than the sum of two other lines simila rlydrawn ,

but included by them.] Q27

Hence 2PQ 2PB,

Or PQ PR. Q. E . D.

COR Only two equal straight lines can be drawn from a

point to a given straight line.

30 PLANE GEOME TRK—B OOK I .

PROPOS IT ION IX. T HEOREM.

29. CONVERSELY— If obliqu e lines be d rawn from a

p oin t to a line, two equa l obli qu e lin es cu t ofi‘

equa l

d istances from the foot of the p erp end icu lar d rawn

from the p oin t to the line .

Given— CD perpendicular to AB , and CE and CF equal

oblique lines drawn from C to AB .

ED= DI".

Den — I f ED were greater than DF, CE would be greater

than CF.

[1] from a point without a straight line a perpendicular is let fallto the line

,the oblique line which cuts of on the line the greater dis

tance from the foot of the perp end icular is the greater.] t 28

And if DE were less than DF,CE would be less than CF .

But both of these conclusions are contrary to the hy

pothesis that CE CF.

T herefore Q. E . D.

80. COR — Of two unequal lines, the greater cuts of the

greater distance from the foot of the perpendicular.

Norm1.—T he method Of demonstration employed in this theorem is

called Reductio ad Absurdum.

”

NOT E 2.—T hus far the proof for each part of the demonstration has

been given in full,

"

immed iately after the statement, and printed in

italic type. In the remainder of this work only the number of the

section will be given , but the pupil should always be required to give

the full statement.

PARALLEL LINES . 31

PA RA LLEL LINE S .

31. Parallel L ines are lineswhich lie in the same plane,

they are produced ; as AB and

CD.

It follows from this definition that straight lines which

are not parallel, lying in the same plane, will meet one

another when sufficiently produced.

32. When two straight lines, AB and CD,are cut by a

third line EF, called a transversal, or secarrt line, then

1 . T he four angles, a, b, g, and

h,without the two lines, are called

Exterior angles.

2. T he four angles, c, d, e, and f,w ithin the two lines, are called In

terior angles.

3. T he angles d and f, or c ande, are called Alternate

- interior angles.

4. T he angles b and h, or a and g, are called Alternate

exterior angles.

5. T he pairs of angles a and e,b and f, d and h

,or c and

g, are called Exterior- interior angles.

When several lines are drawn through the same point,

they are said to have different directions ; as bA,bE

,and d B .

Parallel lines have the same direction when they lie on

the same side of the transversal, or when they lie on the

same side of a straight line joining their correspondingextremities as d B and cD.

Parallel lines have opposite directions when they lie on

opposite sides of the transversal, or when they lie on Op

posite sides Of a straight line joining their corresponding

extremities ; as bA and cD.

32 PLANE GEOME TRY._ B ooK I .

PROPOS IT ION X. T HEOREM.

88. Two stra igh t lines p a ra llel to the same stra ightline are para llel to each other .

A B

C D

E F

Given—AB and CD parallel to EF.

T o Prove —AB parallel to CD.

Dem— If AB and CD are not parallel, they will meet insome point if sufficient]y produced . 31

.

We should then have. two lines drawn through the same

point parallel to the same line EF, which is impossible.

Ax . 7

T herefore AB and CD cannot meet,and are parallel.

Q. E . D.

84. COR .—A line parallel to one of two parallel lines is

parallel to the other also.

PROPOSIT ION XI . T HEOREM.

35. T aro str a ight lines in the same pla ne p erp en

d icu la r to the same stra ight line a re p a ra llel.

Given—AB and CD perpendicular to EF.

T o Prove—AB and CD parallel.

34 PLANE GEOME TE R—B OOK L

PROPOSIT ION XI I . T HEOREM.

37. If two para llels a re ou t by a th i rd line, then

1 . T he alterna te - in ter ior angles a re equa l.

2. T he ex ter ior - in ter ior ang les a re equa l.

3 . The sum of the in ter ior angles on the same sid e

of the transversa l is equa l to two r igh t angles .

Given—AB and CD parallel and cut by EF.

T o v e , let Ac Ab.

Dem— T hrough 0, themiddle point of GH,drawMN per

pendicular to AB .

T hen MN is also perpendicular to CD. 36

Apply the figure HON to the figure MOG, placing OH on

its equal 0G, and the line ON will take the direction Of

0M,since the angles HON and MOG are equal. 21

S ince H falls on G, and E N and GM are both perpen

dicular to MN ,HN and GM must coincide

,or we should

have two perpendiculars from the same point G to the

same straight line MN,which is impossible. 26

Hence Ac Ab.

T o Prove , 2d Aa Ab.

Dem— S ince Aa Ac, 21

And Ab Ac,

1st part, 37

T hen Aa Ab. Ax. 1

PARALLEL LINES . 35

T o Prove, 3d Ab Ad two right angles.

Dem—S ince Ac Ad= two right angles, 16

And Ac Ab, l st part, 37

T hen Ab Ad two right angles. Q. E . D.

PROPOS IT ION XI I I . T HEOREM .

38. If a stra igh t line in tersect two other stra igh t

l i nes , these two lines w i ll be p a ra llel

1 . Wh en the a lterna te - i n ter ior ang les a re equ a l.

2. When the ex ter ior - in ter ior ang les a re equ a l.

3 . Wh en the sum of the in ter ior ang les on the same

sid e of the transversa l is equal to two r igh t ang les ,

Given—AB and CD two straight lines, cut by EF.


I st. - When Ab Ac.

Dena—T hrough H draw MN parallel to AB .

S ince the parallels AB and MN are cut by EF,

T hen 37

But, by hypothesis, Ac Ab.

Hence Ab AGEN. Ax. 1

But this is impossible unless CD coincides with MN.

T herefore CD is parallel to AB .

36 PLANE GEOME TE R—BOOK L

2d.—When Aa Ab.

Dem— S ince Aa A0,

And , by hypothesis, Aa A6,

T hen Ab A c. Ax . I

But when Ab Ac,AB and CD are parallel. lst part, 38

3d.— When Ab Ad two right angles.

Dem— S ince Ac Ad two right angles, 16

And,by hypothesis, Ab Ad two right angles,

T hen Ao+ Ad= Ab+ Ad. Ax . 1

Dropping the common angle d, Ax . 2

Ac Ab.

But when Ac Ab,AB and CD are parallel.

lst part, 38. Q. E . D .

PROPOSIT ION XIV. T HEOREM.

39. If two ang les ha ve thei r s id es r esp ectively p a r

a llel,

1 . They a re equ a l, if their s id es lie in the same or

in oppos i te d i rect ions .

2; They a re supp lemen tary ,if their sides do not lie

i n the same or in opp osi te d irections .

PARALLEL LINES . 37

I . Gi ven— T he angles AB C and DEF with their S ides

respectively parallel and lying in the same direction .

T o Prove Aa Ac.

Dem—Produce FE to G.

T hen Aa Ad, 37

And Ac Ad. 37

Hence Aa Ac. Ax. 1

I I . Given— T he angles HOK and LMN with their sidesrespectively parallel and lying in opposite directions.

T o Pt ove Ala An.

Dem— Produce H0 to W.

T hen Ala Ah,

37

And An Ah. 37

Hence Ala An . Ax . 1

I II . Given— T he angles PQR and S T V with their S ides

respectively parallel and n ot lying in the same or in op

posite directions.

T o Prove—Angles r and v supplementary .

Den —Produce VT to Y.

T hen Ar = As,

And Ap= As.

Hence Ar = Ap .

But Ap and Av are supplementary .

S ubstituting Ar for Ap ,

We have Ar and Ar supplementary . Ax . 1. Q. E . D.

38 PLANE GEOME TRYZ—BOOK L

PROPOS IT ION XV. T HEOREM.

40. If two ang les ha ve their s id es resp ectively p er

p end icu la r ,

1 . They a re equa l , if both a re acu te or both obtu se.

2. They a re supp lementa ry ,if one is acu te and the

other obtuse .

I . Given— AB perpendicular to DE,and CB perpen

dicular to EF.

T o Prove AB = Ao .

Dem— Construct Ac with its sides respectively perpen

dicular to the sides of Au.

T hen the S ides of A0 will be respectively parallel to the

sides Of AB,

35

And Ac AB . 39

By construction, Aa Ab one right angle,

And Ab 4'

Ac one right angle.

Hence Aa + A b= Ab+ Ac.

Dropping the common angle b,

Aa A0.

But AB = A0.

T herefore Aa AB . Ax . 1

NOT E — T his proposition can be read ily proved when both angles are

obtuse, by producing FE an equal d istance below the line EH to F’and

regard ing DEH and GEE'the Obtuse angles.

T he pupil should be required to construct the figure and give the

demonstration in full.

PARALLEL LINES . 39

I I . Given—AB perpendicular to DE ,and CB to EF.

T o Prove—AB and Ab supplementary .

Dem—Construct AGEH with its S ides respectively per

pendicular to the sides of A6.

T hen the sides of AGEH will be respectively parallel to

the sides of AB, 35

And 39

Now,Aa Ac Ab AGEH four right angles. 19

By construction,Aa Ac two right angles.

Subtracting, Ab AGEH two right angles. Ax. 2

But AGEH AB .

Substituting, Ab AB two right angles, and are

therefore supplementary . Q. E . D.

EXERCIS E S .

21 . T he sum of the four angles formed by the intersection Of two lines

is equal to four right angles.

22. I f two straight lines are. cut by a th ird line, making the sum of

the two interior angles on the same side of the transversal less or greaterthan two right angles, the lines are not parallel.

23 T he Shortest line that can be drawn froma point to agiven straight

line 13 perpendicular to the line.

24. T wo angles are complementary ,and the greater exceeds the less

by how many degrees are there in each angle ?

25. T wo angles are supplementary , and the greater is five times theles s ; how many

’

degrees are there in each angle ?

40 PLANE GEOME TRK—B OOK L

T RIANGLE S .

41. A plane T riangle is a portion of a plane bounded

by three straight lines.

42. T riangles are classified according to their sides and

according to their angles.

43. By S ides .— A triangle is called Scalenewhen no two

Of its sides are equal ; Isosceles when two of its sides are

equal ; and Equilateral when all its S ides are equal.

S calene Equ ila teral

44. By Ang les .— A triangle is called Acute when all

its angles are acute ; Obtuse when one of its angles is

Obtuse ; and Right when one of its angles 18 a right angle .

Obtuse R ight Acute

45. Parts of a T riang le .— T he bounding lines, AB ,

B C,

and CA,are called the sides of the triangle, their sum

,

AB B C+ CA,is called the perimeter, and their points of

intersection, A,B and C

,are called the vertices.

42 PLANE GEOME TE R—B OOK I .

PROPOS IT ION XVI I . T HEOREM .

48. The sum of the ang les of a tr iang le is equal to

two r ight ang les .

Given—T he triangle AB C.

T o Prove—AA Ad AC two right angles.

Dem— Draw BE parallel to AC, and produce AB to D.

S ince the parallels AC and BE are cut by AD and B C,

AA = Aa . § 37

AC= Ac. § 37

But Aa Ad Ac= two right angles. 18

S ubstituting AA for Aa,and AC for Ac

,we have

AA Ad A C two right angles. Q. E . D .

49. COR. 1 .

— An exteri or angle of a triangle is equal to the

sum of the two opposite interior angles.

50. COR . 2.

—An exterior angle of a triangle is greater

than either of the opposi te interior angles.

51. COR . 3.

— Any angle of a triangle is equal to two

right angles less the sum of the other two angles.

52. COR . 4.

—A triangle cannot have two right angles, nor

two obtuse angles.

53. COR . 5 .— T he sum of the two acute angles of a right

triangle is equal to one right angle.

54. COR . 6 .

— If two angles of one triangle are respectivelyequal to two angles of another triangle, the third angles are

equal.

TRIANGLES . 43

PROPOS IT ION XVI I I . T HEOREM.

55. Two tr iang les a re equ a l when two sid es and the

i nclud ed a ng le of on e a re resp ect i vely equa l to two

s id es and the includ ed ang le of the other .

Given— T he triangles AB C and DEF,with

AB = DE,AO= DF

,and AA= AD.

i

T o Prove A AB C==A DEF.

Dem— Place the triangle AB C upon the triangle DEF, so

that AB coincides w ith its equal DE ; and S ince AA equals

AD, the line AC will take the direction of DE ; and since

AC equals DF, the point C will fall on F,

And B C will coincide with E F. Ax . 6 .

T herefore AB C and DEF coincide throughout, and are

equal. Q. E . D.

PROPOS IT ION XIX. T HEOREM.

56. Two tr iangles a re equ a l when two ang les a n d

the inclu ded side of one a r e resp ectively equ a l to two

a ngles and the included s id e of the other .

Given— T he two triangles AB C and DEF, with

AB = DE, A A = AD

,and A B = AE .

T o Prove A ABC= A DEF.

44 PLANE GEOME TRK—B OOK I .

Dem— Place the triangle AB C upon the triangle DEF,

so that AB coincides with its equal DE ; and since AA

equals A D,the line AC will take the direction of DF

,and

the point C will fall somewhere on DF or DF produced .

S ince A B equals A E,the line B Cwill take the direction

of E F,and the point C will fall somewhere on EF or EF

produced .

I f the point C falls on both DF and EF,it must fall on

their intersection F.

T herefore AB C and DEF coincide throughout, and are

equal. Q. E . D.

PROPOS IT ION XX. T HEOREM.

57. Two tr iang les a re equ a l when thr ee sid es of one

a re resp ect i vely equa l to th ree sid es of the other .

Given —T he triangles ABC and DEF, with

AB = DE,AC= DE

,and B C= EF.

T o Prove A ABC A DE F.

Dem— Place the triangle AB C so that AB comcides with

its equal DE ,and C falls at G

,on the opposite S ide of DE

from F ; and draw FG.

T RIANGLES . 45

T hen , since DF = DG,and EF = EG

,the line DE has

two points equally distant from the extremities Of FG,

and is perpendicular to FG at its middle point, § 25

And A a = A b. § 23

Hence A DEF = A DEG, § 55

Or A AB C= A DEF. Q. E . D.

PROPOS IT ION XXI . T HEOREM.

58. Two r ight tr ia ngles a re equ a l when the hyp ote

n u se a nd a side of on e a re r esp ecti vely equa l to the

hyp otenu se and a s id e of the other .

Given—T he right triangles ABC and DEF, with

AO= DE,and B C= EF.

T o Prove A ABC= A DEF.

Dent — Place the triangle ABC upon the triangle

so that B C coincides with its equal EF ; and since the right

angle B equals the right angle E the line BA will

take the direction Of ED,and since AC equals DE, the

point A will fall on‘

D. § 29T herefore AB C and DEF coincide throughout, and are

equal. Q. E . D.

59. COR . l . Turo right triangles are eq ual when the hypote

nuse and an acute angle of one are respectively equal to the

hypotenuse and an acute angle of the other.

T he other acute angles will be equal, by § 54, and thetriangles are equal, by

46 PLANE GEOME TRK—BOOK L

60. COR. 2. Two right triangles are equal when a side and

an acute angle of one are respectively equal to a side and an

homologous acute angle of the other.

T his is true by 56 .

PROPOS IT ION XXI I . T HEOREM.

6 1. If two tr iang les ha ve two sid es of the one r e

sp ecti vely equal to two sid es of the other , and the

inclu ded angles unequa l, the th i rd side is g r ea te r

in the tr i ang le having the g rea ter includ ed angle .

Given— The triangles ABC and DEF, with

AC= DE,B C= EF

,and A o> A F.

T o Prove AB DE .

Dena—Place the triangleDEF on the triangleABC, so that

EF falls on its equal B C, and DEF will take the position

of GB C.

In the triangle AOC,

Ao + oc>Aa

In the triangle GOB ,

OB 0G GB . Ax . 5

Adding these two inequalities, and observing that

AO OB ==AB , and 00+ 0G==00, we have

AB + GC> AC+ GB .

Dropping GC and its equal AC, we have

AB > GB,or AB DE . q . E . n.

TRIANGLES . 47

S CHOLIUM.— T he point Gmight fall on AB ; the theorem

would then be true by Ax . 3.

T he point Gmight fall within the triangle. T he demon

stration would then be as follows :

Given— T he triangles AB C and DEF, with

AO= DE,BC= E F

,and Ac>A F.

AB DE .

Dem— Place the triangle DEF on the triangle ABC, so

that EF falls on its equal B C, and DEF will take the

position of GEG.

T hen AB + AC> GB + GC.

Dropping AC and its equal GC, we have

AB > GB ,or AB DE . Q. E . n.

PROPOS IT ION XXI I I . T HEOREM.

62. CONVERSELY— If two tr ia ngles have two sid es

of the on e respectively equ a l to two sides of the other ,bu t the th i rd sid es unequ a l, the included ang le is

g reater in the tr iang le having the g r eater . th i rd

s ide.

Given—T he triangles ABC and DEF, with

AO DE, BC= EF

,and AB>DE .

T o Prove A C> A F.

48 PLANE GEOME TE R—B OOK I .

Dem— If A C were equal to A. F, the triangles AB C

and DEF would be equal, § 55

And AB would equal DE .

If A C were less than A F,

T hen AB would be less than DE . 61

Both of these conclusions are contrary to the hypothesis

that AB is greater than DE .

Hence A C ) A F. Q. E . D.

PROPOSIT ION XXIV. T HEOREM.

63 I n an isosceles tr iang le, the ang les opposite the

equa l sid es a re equa l.

Given—T he triangle ABC, isosceles, with AO= BC.

T o Prove A A A B .

Dent—Draw CD perpendicular to AB .

T hen, in the two right triangles ADC and BBC, CD is

common and AC equals B C. Hyp.

Hence the two triangles are equal in all their parts . 58

T herefore A A A B . 45 (a) . Q. E . D.

50 PLANE GEOMETRK—B OOK L

PROPOS IT ION XXVI . T HEOREM.

68. In any tr iang le theg rea ter angle lies

theg rea ter side.

Given—T he triangle AB C, with AC> BC’.

A B > A A.

Dem—Lay of!"

CD B C,and draw DB .

T he triangle DROwill be isosceles, and

A o A n.

S ince A o is an exteri or angle of the triangle ABD,

A o> A A. § 50

Now, A B >A n (Ax . and A n = A o, and A o>A .

T herefore A B > A A. Q. E . D .

PROPOS IT ION XXVI I . T HEOREM.

69. CONVERSELY— I h any tr iang le the grea ter side

lies opp osi te the g rea ter ang le.

Given—T he triangle AB C, with A B AA.

Ac B C.

TRIANGLES . 51

Dem— Draw DB,making A n equal to A A.

T hen AD DB .

But DB DC' B C.

Substituting AD for its equal DB ,we have

AD DC.or AO> BC. Q. E . n.

EXERC IS E S .

28. How many degrees are there in each angle Of an equiangular

triangle?

29. Any side Of a triangle is les s than half the sum Of the three sides

of the triangle.

PROPOS IT ION XXVI I I . T HEOREM.

70. Any p oin t in the bisector of an a ngle is equ a lly

d ista n t from the s id es of the a ngle ; a nd any p oin t

not in the bisector is u nequa lly d istan t from the sides

of the angle.

I . Given—Any point, as P, in the bisector Of the angle

AB C.

T O Prove—T he perpendicular PE the perpendicular

PF.

Dem— In the right triangles BEP and EFF,BP is com

mon,and A n = A o by hypothesis.

T hen A BEP= A EFF .

Hence

52 PLANE GEOME TRK—BOOK I.

I I . Given—Any point as Q not in the bisector Of the

angle ABC.

T O Prove—T he perpendicular QH less than the perpendicular QE .

Dem—Draw QF.

In the right triangle QE F,

QH QB.

But QF QP PF.

S ubstituting PE for its equal PF, we have

QF QP PE,or QE .

Now, QH QF QE .

Hence QH QE . Q. E . n .

71. S CHOLIUM. The bisector of an angle contains allpoints

equally distant from the sides of the angle.

EXERCIS E S .

80. If two angles Of a triangle are 40°and 60° res pectively , what is

the th ird angle ?

81. One angle of a right triangle is what is the third angle?

82. If two angles of a triangle are each what is the other angle,

and what is the kind Of triangle ?88. T he th ird angle of an isosceles triangle is four times each angle

at the base ; how many degrees in each angle ?

84. If the perpendicular from the vertex to the base Of a triangle

bisects the base, the triangle is isosceles.

QUADRILATERALS . 53

QU ADR ILA T ERALS .

72. A Quadrilateral is a plane figure bounded by four

straight lines.

78. Quadrilaterals are divided into three classes : the

fl apezimn, the Trapezoid, and the Parallelogram.

74. A T rapez ium is a quadrilateral no two of whose

sides are parallel.

A T rapez oid is a quadrilateral which has only two Of

its sides parallel.

A Parallelog'

ram is a quadrilateral whose Opposite

sides are parallel.

Paralle logram

75. Parallelograms are divided into Rectangles and

Rhmnboids.

76. A Rectang le is a parallelogram whose angles are

right angles.

A S quare is an equilateral rectangle.

77. A Rhomboid is a parallelogram whose angles are

Oblique.

A Rhombus is an equilateral rhomboid.

Rectangle Rhomboid Rhombus

78. Parts Of a.QuSd rilateraL— The bounding lines are

the sides of the quadrilateral ; the angles formed by the

sides are the angles of the quadrilateral.

T he Base of a quadrilateral is the side upon which it

seems to rest.

54 PLANE GEOME TE R—BOOK I .

T he bases Of a parallelogram or a trapezoid are the side

upon which it seems to stand and its parallel ; the altitude

is the perpendicular distance between them.

T he Diagonal of a quadrilateral is a straight line joiningany two vertices not consecutive.

PROPOS IT ION XXIX. T HEOREM.

79. The opp osi te sides and ang les of a para llelo

g ram a re equal.

Given—T he parallelogram ABCD.

T O Prove

AB = CD,AD= B C

,A A= A C

,and A

Den — Draw the diagonal DB .

S ince the parallels AB and CD are cut by DB ,

Am= A n.

S ince the parallels AD and B C are cut by DB,

A o A p . 37

T hen,in the triangles ABD and EDO

,DB is common

,

and we have two angles and the included side in each

respectively equal.

T herefore A ABD A EDO. 56

Whence AB = DC, AD= B C,A A=AC

,and A B =AD

,

being homologous parts. Q. E . D.

80. COR . 1.

— A diagonal of a parallelogram divides it intotwo equal triangles.

81. COR . 2.

—Parallels included between parallels are equal.

QUADRILATERALS. 55

EXERCIS E S .

85. If one angle of a parallelogram is howmany degrees are there

in each of the other angles ?

86. If one base of a trapezoid be extended in both directions, the sum

of the exterior angles is equal to the sum of the two opposite interior

angles .

87. I f the angles adjacent to one base of a trapezoid are equal, those

adjacent to the other base are also equal.

PROPOS IT ION XXX. T HEOREM.

82. CONVERSELY— If the opposi te sid es of a qu ad r i

la tera l a re resp ectively equa l, or if the opposite

a ng les a re resp ectively equa l, the figu re is'

a paral~

telegram.

I . Given—T he quadrilateral ABCD,with

AB DC and AD BC.

T o Prove—AB CD a parallelogram.

Den — Draw the diagonal DB .

Then in the triangles ABD and BCD,DB is common,

and,by hypothesis, AB DC and AD BC.

Hence A ABD A DB C,

57

And A m= A n. 45 (a)

T herefore AB is parallel to DC. 38

In like manner it may be shown that AD is parallel

to B C.

Hence the quadrilateral AB CD has its opposite sides

parallel, and is therefore a parallelogram by definition .

§ 74

56 PLANE GE OME TB K—B OOK I .

I I . Given—T he quadrilateral ABCD, with

A A= A C and A B = A D.

T O Prove—AB CD a parallelogram.

Dem— T he diagonal DB divides the quadrilateral into

two triangles the sum Of whose angles is equal to the

sum Of the angles of the quadrilateral.

T he sum Of the angles Of each triangle is two right

angles (l 48) hence the sum Of the angles Of the quad

rilateral is equal to four right angles.

T herefore A A A B AC A D four right angles .

But Hyp .

T hen A A A D two right angles,

And AB is parallel to CD. 38

In like manner it may be shown that AD is parallel to

B C.

T herefore ABCD is a parallelogram by definition. 74

Q.

PROPOSIT ION XXXI . T HEOREM.

83. If two sid es of a qu ad ri la tera l a re equa l and

parallel, the figu re is a p ara llelog ram.

58.PLANE GEONE TRE—B oox I .

PROPosrrION XXXI I I . T HEOREM.

87. CONVERSELY—If the d iagona ls of a qu ad ri

la teral bisect each other,the figu re is a pa ra llelo

g ram.

Given—T he quadrilateral AB CD,whose diagonals AC

and BD bisect each other at E .


Dem— In the triangles ABE and DEC,

AE E C, and BE ED,

And Am A n.

Hence A ABE A DEC,

And AB DC.

In like manner,

AD BC.

T herefore ABCD is a parallelogram. 82.

88. COR . l .

— If the diagonals of a quadrilateral are equal

and bisect each other at right angles, thefigure is a square.

89. COR . 2.

— If the diagonals of a quadrilateral are un

equal and bisect each other at right angles, the fig ure is a

rhombus.

PROPOS IT ION XXXIV. T HEOREM.

90. Two pa ra llelog rams a re equa l wh en two adj a

cen t sid es and the inclu d ed ang le of one a re resp ec

ti vely equa l to two adj acen t sides and the included

angle of the other .

QUADRILATERALS . 59

G iven—T he parallelograms AB CD and EFGH,with

AB = EE,AD = EH

, and A A= A E .

T o Prove D AB CD D EFGH.

Den —Place the parallelogram AB CD on EFGH,SO that

A A coincides with its equal A E,and

,since AB = EF

,

and AD= EH,by hypothesis, the point B will fall on

~

F,

and D on H .

Now,since DC is parallel to AB ,

and HG to EF, the side

DC will take the direction Of HG,and C will fall some

where on HG,or HG produced . In like manner

,C will

fall on FG,or FG produced.

I f C falls on both HG and FG,it must fall on thei r In

tersection , G.

Hence ABCD and EFGH coincide throughout, and are

equal. Q. E . D.

PROPOS IT ION XXXV. T HEOREM.

91. The stra igh t line wh ich j oins the m idd le p oin ts

of the non-

p a ra llel sid es of a trap ez oid is pa r a llel

to the para llel sid es and equ a l to ha lf their sum .

Given—T he,

line GH connecting the middle points of

AD and B C of the trapezoid AB CD.

T O v e—HG parallel to AB and DC,and equal to

l-(AB DC ) .

0 PLANE GEOME TRK—B OOK 1:

Dem—Draw EF parallel toAD through G,the middle

point Of B C, and produce DC to E .

In the triangles FGB and OGE,BG GC

,by construe

tion,

Am A n, 537

And A o A p . 21

Hence 56

And CE EB,and EG FG. 45 (a)

In the parallelogram AFED,AD FE . 79

Hence DH,the half of AD,

is equal and parallel to E G,the half Of EF .

T herefore, DHGE ,having two sides equal and parallel,

is a parallelogram. 83

Hence HG is parallel to DE ,and also to its parallel AB .

Again, 79

And

Adding these two equations, remembering that CE = EB,

we have2HG DC 4» AB ,

And HG t(1) c+ AB ) .

EXERCIS E S .

88. T he diagonals of a rectangle are equal.

88. If the diagonals of a parallelogram are equal, the figure is a rectangle.

40. T he bisectors Of the interior angles of a parallelogram form a rect

angle.

41. If a line joining two parallels be bisected , any line drawn th roughthe point of bisection and included between the parallelswill be bisected

at that point .

POLYGONS . 61

POLYGON S .

92. A Poly gon is a plane figure bounded by straight

lines.

Polygons are classified according to the number of their

sides.

A T riangle , or Trigon, is a polygon of three sides.

A Quadrilateral, Or Tetragon, is a polygon Of four sides ;a Pentagon has five sides ; a Hexagon, six ; a Heptagon,

seven ; an Octagon, eight ; a Nonagon or Enneagon, nine ; a

Decagon, ten ; an Undecagon, eleven ; aDodecagon, twelve, etc.

93. A Convex Poly gon is one in

which no side, when produced , can enter

the Space enclosed by the perimeter ; as

AB CDE .

A Concave Poly gon is one in which

two or more sides,when produced , will

enter the Space enclosed by the perimeter ;

T he interior angle KE G is called a re

entrant angle, and the sides KH and GH

are called re- entrant sides. A concave polygon is sometimes called a re- entrant polygon.

94. Parts of a Poly gon .— T he bound ing lines AB ,

B C,

CD,DE

,EA are the sides of the polygon ; and their sum is

the perimeter .

T he angles AB C, B CD,CDE

,etc.

,are the angles of the

polygon .

ADiagonalOf a polygon is a line joining any two verticesnot consecutive ; as E C, EB ,

etc.

95. An Equilateral Polygon is a polygon whose sides are

all equal. An Equiangular Polygon is a polygon whose

angles are all equal.


96. T wo polygons are mutually equilateral when th eir

corresponding sides are equal. T wo polygons aremutuallyequiangular when their corresponding angles are equal.

In polygons that are mutually equilateral or mutually

equiangular, corresponding sides or corresponding angles

are homologous.

97. T wo polygons are equal i f when applied to each

other they coincide throughout.

PROPOS IT ION XXXVI . T HEOREM.

98. The sum of the in ter ior angles of a p olygon is

equ a l to two r ight ang les taken as many times, less

two, as the polygon has sides .

Given— T he polygon ABCDEF.

T O Prove— T he sum of the angles equal to two right

angles taken as many times as the polygon has sides, less

two.

Dem— If diagonals be drawn from the vertex A,the

polygon will be divided into as many triangles as the

polygon has sides, less two.

T he sum of the angles Of the triangles equals the sum

of the angles of the polygon .

T he sum of the angles of each triangle equals two right

angles. 48

Hence the sum Of the angles Of a polygon is equal to

two right angles taken as many times as the polygon has

sides, less two. Q. E . D.

POLYGONS . 63

I f R represents a right angle, and n the number Of sides,then

99. COR . 1. The s umof the angles of apolygon is (n 2)2R,

or 2nR 4B .

100. COR . 2.— Each angle of an equiangular polygon is

(n 2)2R

n

I f the right angle be regarded as unity , or 1, then

101. COR . 3.— The sum of the interior angles (n

(nand each angle

n

SOHOLIUM.—T he sum of the angles Of the different

polygons, and the value Of each angle when the polygon

is equiangular, are found as follows

NO. of Rt ht Value Of each An le in RI htPolygon .

Angles?

Angles and Di grees.

g

T riangle

Quad rilateral .

Pentagon

Hexagon

EXERCIS E S .

42. Required the number Of degre es in each interior angle of an

equiangular

Heptagon . Nonagon . Undecagon .

Octagon . Decagon . Dodecagon .

48. Required the number of degrees in each exterior angle

equiangular

T riangle. Pentagon .

Hexagon . Octagon .

64 PLANE GEOME TRYZ—B OOK I .

PROPOS IT ION XXXVI I . T HEOREM.

102. If the sid es of any polygon be p rod uced so as

to form one ex ter ior angle a t each ver tex ,the sum of

these ex ter ior ang les is equa l to fou r r igh t ang les .

Given— T he polygon AB CDE ,with its sides produced so

as to form one exterior angle at each vertex .

T o Prove—T he sum Of the exterior angles equal to four

right angles.

Den —T he sum of AA Aa two right angles. 16

T he same is true at each vertex .

Hence, if the polygon has n sides, the sum Of the in

terior and exterior angles equals 2n right angles, or

Int. Ext.

But Int. 2nR 4R.

S ubtracting these two equations, we have

Exterior angles 4R . Q. E . D .

EXERCIS E S .

44. If two angles Of a quadrilateral are supplementary , the other two

angles are supplementary .

45. T he exterior angles at the base of any triangle are together greater

than two right angles.

46 . T he hypotenuse is greater than either of the other sides of a right

47. If the two exterior angles at the base of a triangle are equal, the

triangle is isosceles .

65 PLANE GEOME TRK—B OOK I.

PROPOS IT ION XXXIX. T HEOREM.

105. The bisectors of the angles of a tr iang le meet

in a common p oin t .

Given—T he triangle ABC, and AF,BD

,and CE the bi

sectors Of the angles A,B,and C respectively .

T o Prove—T hat AF,BD

,and CE meet in a common

point.

Dem— AF and BD will intersect at some point, as 0.

S ince 0 is in the bisector AF, it is equally distant from

AB and AC. 70

For the same reason 0 is equally distant from AB and

BC.

T hen 0 is equally distant from the S ides AC and B C

(Ax . and therefore lies in the bisector CE . 71

Hence AF, BD,and CE meet in the common point 0.

Q. E . D;

PROPOS IT ION XL. T HEOREM.

106 . The p erp end icu la rs erected a t the m i d d le

p oin ts of the sides of a tr iang le meet in a common

p oin t .

Given— T he triangle ABC,with EH

,FK

,and DC per

pendicular at the middle points of AB , B C, and AC.

POLYGONS . 67

T O Prove—T hat EH,FK

,and DC meet in a common

point.

Dem—EH and FK will intersect at some point, as 0.

S ince 0 is in the perpendicular EH,it is equally distant

from A and B . 22

For the same reason 0 is equally distant from C and B .

T hen 0 is equally distant from C and A (Ax . and

therefore hes In the perpendicular DG. 24

HenceEH,FK

,and DGmeet in a common point. Q. E . D.

PROPOS IT ION XLI . T HEOREM.

107. The p erp en d icu la rs from the ver t ices of a tr i

angle to the app osi te s id es meet in a common p oin t .

Given—T he triangle ABC, with CK,AG

,and BH per

pendicular to AB ,B C

,and AC respectively .

T o Prove—T hat CK,AG

,and BH meet in a common

point.

Dem—T hrough A,B,and C draw FD

,DE , and EF

parallel to B C, AC; and AB respectively .

T hen AB CF a nd ABEC are parallelograms,.

And FC==AB , and CE AB .

Whence FC CE,and C is the middle point of EF.

Now,since C is the middle point of EF, and CK is per

68 PLANE GEOME TB Y.-BOOK I .

pendicular to AB , it is also perpendicular to the parallel

EF at its middle point. 36

In like manner, AG and EH are perpendiculars erected

at the middle points of FD and DE respectively .

T herefore CK,AG, and BH meet in a common point, 0.

106 . Q. E . D.

PROPOSIT ION XLII . T HEOREM.

108. The med ia l lines of a tr iang le meet in a

common p oin t .

Given—T he triangle ABC, and AG, BD, and CH the

medial lines.

T o Prove—T hat AG, BD,and CH meet in a common

point.

Dem—AG and BD will intersect at some point, as at 0.

Let E and F be the middle points Of AO and B 0 t e

Spectively , and draw DG, DE , EF, and FG.

Now,since DG bisects AC and B C

,it is parallel to AB

and equal to lAB .

S ince EF bisects AO and BO, it is parallel to AB and

equal to lAB . 104, 103

T herefore DC and EF are both equal (Ax. 1) and

parallel hence DEEG is, a parallelogram, 83

And DE and EG bisect each other. 84

By construction, E is the middle point Of AO.

T herefore AH E0 0G, and 0G is QAG.

EXERCISES . 69

Hence BD cuts off i of AG.

In like manner it may be proved that CH cuts ofi'

iOf AG.

Hence AG, BD, and CH meet in the common

109. T he Locus Of a point in a plane is the straight or

curved line which contains all points having a common

property , and no others.

T hus,all points w an angle, and equally distant

from its sides, lie in bisector of the angle. § 71:

Hence the locus of a point within an angle, and equally

distant from its sides, is the bisector of the angle.

M ERC IS E S .

PLANE LOCI .

48. What is the locus of a point equidistant from two parallel lines ?

49. What is the locus Of a point equid istant from two fixed points ?50. Find the locus Of a point equidistant from a fixed point.51. What is the locus of a point equidistant from a given straight

52. Find the locus of a point equidistant from the extremities Of a

given line.

58. What is the locus Of a point equidistant from the sides Of an

isosceles triangle ?

54. What is the locus ofa point equidistant fromapair of intersectingstraight lines ?

PRACT ICAL EXAMPLES .

55. If one Of the two equal angles Of a triangle is what is each of

the other angles ?

56 . If one angle Of a triangle is what is each Of the other angles,

provided they are equal to each other ?

57. If one angle Of a parallelogram' is what is each of the other

anglu

70 PLANE GEOME TRPZ—BOOK L

58. What is the sum of the interior angles Of a pentedecagon ?

58. Howmany sides has the polygon in which the sum of the interior

angles is five times the sum of the exterior ?

EXPLANAT ION .—T he sum Of the interior angles of a polygon equals

(n right angles (i and the sum Of the exterior angles equals

four right angim (QHence (n 5 x 4,

Or 2n

And 2a n 24.

n = 12.

Hence the polygon has twelve sides.

60. How many sides has a polygon in wh ich the sum Of the interior

angles equals the sum of the exterior angles

61. How many sides has a polygon in wh ich the sum Of the exterior

angles is double the sum of the interior angles

62. How many sides has a polygon in which the sum Of the inte rior

angles is three times the sum of the exterior angles ?

68. How many Sides has a polygon in which the sum Of the interior

angles is ten times the sum of the exterior angles ?

64. How many sides has an equiangular polygon five of whose anglesare equal to eight right angles ?

65 . How many sides has an equiangular polygon seven Of whose

angles are equal to twelve right angles

ORIGINAL THEOREMS .

66. If the base of an isosceles triangle is pro

duced in both directions, the obtuse angles formed

will be equal.

67. I f the two equal sides Of an isosceles trian

gle are produced , the Obtuse angles below the base

will be equal.

68. If the angles at the base Of an isosceles tri

angle are bisected by lines meeting within the

tri angle, the triangle thus formed will be isosceles .


76. T he perpendiculars from the middle point of the base of an

isosceles triangle to the sides are equal.

77. If the non -

parallel sides of a trapezoid are equal, its diagonals are

equal.

78. A perpendicular let fall from one end of the base of an isosceles

triangle upon the Opposite sidemakes an angle, with the base, equal to

one half of the vertical angle.

78. T he middle point of the hypotenuse of a right triangle is equallydistant from the vertices of the triangle.

80. If the hypotenuse of a right triangle is double one of its sides,one

of the acute angles is double the other .

81. If one of the acute angles of a right triangle is double the other,the hypotenuse is double the shortest side.

82. T he bisectors of the angles of a rectangle form a square .

88. T he medial line to any side of a triangle is les s than half the sum

of the other two sides.

84. T he sum of the lines drawn from any point with in a triangle to

the vertices is less than the sum, and greate r than the half sum,

of the

three sides.

85. T he sum Of the four sides of any quadrilateral is greater than the

sum of the two diagonals.

86 . Prove that the number of diagonals Of a polygon of n sides is

c)

87. T he altitude Of a triangle divides the vertical angle into two parts

whose d ifference is equal to the difference of the base angles of the

triangle.

88. I f a perpendicular be let fall from the middle point of either side

of an equilateral triangle to the base, it will cut off one fourth of th e

base .

88. T he sum of the angles at the vertices of a six -

pointed star is equal

to four right angles.

80. T he sum of the angles at the vertices of a five- pointed star is equal

to two right angles.

B OOK I I .

RAT IO AND PROPORT ION .

DEFINIT ION S .

110. T he Ratio of two similar quantities is the quotient

obtained by dividing the first quantity by the second .

T hus,the ratio of a to b is and is expressed a b.

T he An tecedent is the first term of the ratio,and the

Consequent is the second term.

111. A P mportion is an equality of ratios.

I f the ratio of a to 6 equals the ratio of c to d, they will

form a proportion ,which may be written

aor _ = e

b d'

T he Couplers of a proportion are the ratios compared .

T he first and fourth terms are called the Extremes ; the

second and third terms are called the Means .

T hus,in the proportion a b e d

,

a and b are thefirst couplet, c and d are the second couplet.

a and d are the extremes, b and c are the means.

Four quantities are reciprocally proportional w hen the

ratio of two values,A and B

,is equal to the reciprocal

of the ratio of two other values,A

’

and B'

.

T hus

RAT IO AND PROPORT ION. 75:

1 12. A Mean Proportional between two quantities is a

quantity which may be made the consequent of the first

couplet and the antecedent of the second .

T hus,in the proportion a b b c

,b is the mean pro

portional between a and c,and c is called the T hird Pro

port ional.

1 13. A F ourth Proportional to three quantities is the

fourth term of a proportion in which the other three quan

tities form the first three terms.

T hus, in the proportion a b c d,d is the fourth pro

portional to a, b, and c.

1 14. A Continu ed Proportion is one in which each con

sequent is the same as the next antecedent .

T hus,

a : b= b : c

1 15. Four quantities are in proportion by Alternation

when antecedent is compared with antecedent,and conse

quent with consequent.

T hus,if a : b= c : d

,by alternation a : c= b : d .

1 16. Four quantities.are in proportion by Inversion

when antecedents are made consequents, and consequents

are made antecedents.

T hus,if a b= o zd, by inversion b z a= d zo.

1 17. Four quantities are in proportion by Composition

when the sum of antecedent and consequent is compared;with either antecedent or consequent.

T hus, if a : b= c : d,by composition a + b zb= c+ d z d.

1 18. Four quantities are in proportion by Div ision when

the difference between antecedent and consequent is com

pared with either f antecedent or consequent.

T hus,

division a — b zb a c— d zd.

RAT IO AND PROPORTION. 77

PROPOS IT ION IV. T HEOREM.

122. A mean p ropor tiona l between two quanti ties is

equa l to the squa re root Of thei r p rod uct .

Let

Whence b= VdE

PROPOSITION V. T HEOREM.

128. If fou r quant i ties a re in p roportion , they w i ll

be in p ropor t ion by Alternation.

Let

ad= bc. § 119

§ 120. Q. E. D.

PROPOSIT ION VI . T HEOREM.

124. If fou r quanti ties a re in p rop ortion , they wi ll

be in prop ortion by Inversion.

Let

ad= bo. § 119

b za - s d zc. § l2o. Q. E . D.

PROPOSIT ION VII . THEOREM.

125. If fou r quan ti ties a re in p rop ort ion , they w ill

be in p roport ion by Composition.

Let a bu c : d.

ad al l be.

Adding db to both members,ad dba be db,

(Kc b) n M0 01)

§ 120. Q. E . D.

78 PLANE GEOME TRY.- B OOK II .

PROPOS IT ION VI II . T HEOREM.

126. If fou r quant it ies a re in p rop or tion , they w i ll

be in p rop or tion by Division.

Let.

a b c d.

ad be.

Subtracting bd from both members,

ad bd= bc bd,

d(a b) b(c d) .

a — b : b= c— d : d. § 120. Q. E . D .


127. If fou r quan t i t ies a re in p rop or tion , they w i ll

be in p rop ortion by Composition and Division.

Let

a + b c+ d

b d

a — b c— d

b d

a + b c+ d

a b c

_

—_

d°

Whence a + b : a —d. Q. E . D.

PROPOSIT ION X. T HEOREM.

128. If fou r quan t i ties a re i n p roportion , l ike

p owers Or like roots Of those quantit ies w i ll be p ro

p ort iona l .

Let a b c d.

T hc

.en

6 d

Raising both members to the nth power,a“

c"

Ax . 2b“ d'“

80 PLANE,GEOMETRK—BOOK II.

PROPOS IT ION XI I I . T HEOREM .

181. If two p roport ions have a coup let in each the

same, the other coup lets w i ll form a p ropor t ion .

Let

And a z b= e f,T hen -c : d= e f. Ax . 1. Q. E . D .

Cos — If two proportions have a couplet in each proportional,

the remaining couplets will be proport ional.

PROPOS IT ION XIV. T HEOREM.

182. The p rod ucts of the cor resp ond ing terms oftwo or more p rop or tions a re p ropor tiona l .

Let a b c d,

And m n p q.

T hen Zd’And

m Kn (I

Multiplying,£

27

: ill

;Whence am bn op dq.


133. In a ser ies Of equa l rati os , the sum Of a ll the

a n teced en ts is to the sum Of a ll the consequ ents as

a ny anteced en t is to its consequ en t .

Let

Let r In the common ratio,

T hena

r,and a br .

b

(

dr, and c= dr .

THE THEOR Y OF LIMI TS . 83

I llustration 2.

— I t will be found in D

267 that ifAB CD is a square, then

AC’

AB

S ince V'

Zcan be expressed only ap

proximately as a decimal,there is no

line,however small

,which is contained

an exact number of times in both AC’ and AB .

I Ience AC and AB are incommensurable.

THE T HEORY OF LIM IT S .

137. S uppose a point to moveAW

from A toward B,under the con

d ition that the

l st second it moves over half of AB to C'

2d half of the remainder to D

3d half of what now remains to E,

and so on indefinitely .

I t is evident that if the point continues to move ao

cording to this law it can never reach the end of the

line.

T he distance fromA to themoving point is an IncreasingVariable Quantity the distance from B to themoving point

is a Decreasing Variable Quantity ; and the distance AB is

a Constant Quantity and the Limit of the increasing variable

quantity .

Hence a Variable Quantity, or simply a Variable,is a

quantity whose value is supposed to change. A Constant

Quantity is a quantity wh ich does not change its value in

the same discussion . T he Limit of a variable quantity is

a constant quantity toward which the variable continues

to approach , but which it can never reach.

8l PLANE GEOME TRK—B OOK I I.

188. As a numerical illustration,let the line AB 137)

be 2 units long. Denote the variable bv x, and the differ

ence between the variable and its limit by v ; then

At the end of the lst second,x 1

,1

At the end of the 2d second,x 1 l , v

At the end of the 3d second,x 1 l l, v

and so on indefinitely

I t is evident that the limit of x is the limit of the

series 1 l l etc.,which is 2. I t is also eviden t

that the limit of v is the last term of the series 1, l, fir,which is zero.

189. COR — The difl'

erence between a variable and its limi t

is a new variable whOse limit is zero.

PROPOS IT ION XVI . T HEOREM.

140. If two va r iables a re a lway s equa l, thei r limi ts

a re equal.

Given— x and 3] two variables,the lim. x AB

,

y = AC,and x= y.

T o v e AB AO.

Dem— T here are three possible suppositions

1. AB AC.

2. AB AO.

30 AB 44Co

THE THEORY OF LIMI TS . 85

Suppose AB AC.

On AC take AD AB .

Now,since the limit of y is AC

’

, y can assume values between AD and AC’

,while x cannot pass beyond AD.

T hat is, x and y can be unequal.

But this is contrary to the hypothesis that x and y are

always equal.

T herefore AB cannot be less than AC.

In like mannerwe may prove that AB cannot be greater

than AC.

Hence AB AO. Q. E . D.

141. COR . l .— The limit of the sum of two variables is the

sum of their limits.

Let lim. x a and lim. 3] b,then ,

by 138,

x= c — v,

x + y = a +b—v - v’.

lim. (a + b—v

But lim. (a + b— v

Hence lim. (x y) a b.

142. COR . 2.

— Tbe limit of the difi’

erence of two variablesis the difi

'

erence of their limits.

Let lim. x=a and lim. 3] b,then

,by 138

,

x a 17,

And

Subtracting, x y a b

lim. (x (a — b— v

But lim. (a— b a — b.

Hence lim. (x—y) a b

EXERC'

ISL'

87

lim. nx lim. (an

lim. (an nv) an.

EXERCIS E S .

PRACT ICAL EXAMPLm.

105. Find amean proportional between 4 and 8. Between a and b.

108. Find a th ird proportional to 4 and 10.

107. Find a fourth proportional to 8, 10, and 24.

108. Write 4 : x 2 y by alternation .

108. Write 8 6 a b by inversion .

110. Write x y y a b b by composition .

111 . Write x + y : x — y= a + b a — b by division .

112. Write a proportion from ab c.

113 . T he sum of two numbers is to their difference as 7 to 3, and their

product is 160 ; find the numbers.

ORIGINAL THEOREMS .

114. If a : b=- c : d , prove that

115 . I f a b= c : d , prove that a’ b’ ac z bd .

118. If a : b - c : d, prove that b—a : b= d — c : d.

117. If a : b=

118. If a z b= b c, prove that —c.

118. cx + dy : c.

120.

121 .

122. Prove the lim. equals i .

123 . Prove the lim. equals 8.

124. Prove the lim. .555 equals 5.

126 . Prove the lim. .333 equals 3.

128. If the lim. z =a, prove

127. If the lim. x= a, prove lim. {75 175 .

128. I f the lim. x a,lim. y = b, and lim. x= c, prove lim. (x + y z )

=a + b + a

129. If the lim. x a, lim. y= b, and lim. z a c, prove lim. xyz abc.

B O O K I I I .

THE CIRCLE .

DEF INIT IONS .

146. A C ircle is a plane figure bounded by a curved

line every point of which is equally distant from a point

Within,called the Centre.

T he Circumference is the boundingline. Any portion of the circumfer

ence is an Arc,as CB . A Quadrant

is one fourth of a circumference.

A Radius is a straight line drawn

from the centre to the circumference ;as CC.

A Diameter is a straight line drawn through the centre,having its extremities in the circumference ; as AB .

A Semi - circumference is one half of a circumference ; a

Semicircle is one half of a circle . A Chord is a straight

line joining the extremities of an arc ;

147. A Tangent is a straight line

which touches the circle at only one

point ; as CD. T he point E is called

the point Of tangency.

T wo circles are tangent to eachother

when they are both tangent to the

same straight line at the same point.

A common tangent to two circles is a straight line which

is tangent to both of them.

88

90 PLANE GEOME TRK—BOOK III .


150. Every d iameter bisects the circle and its ci r

Given—AO the diameter of the circle ABCD.

T o Prove—T hat AC bisects the circle and its

ference.

Dem—Apply the segment ABC to ADC by revolving it

about AC as an axis. T hen will the arc ABC coincide with

the arc ADC,S ince every point of each is equally distant

from the centre 0.

Hence the segments ABC and ADC coincide throughout,

and are equal.

T herefore AC bisects the circle and its circumference.

Q. E . D.

PROPOS IT ION I I . T HEOREM.

151 . The d iameter Of a circle is grea ter tha n any

other chord .

Gi ven—AR the diameter of the circle ACB .

THE CIRCLE . 91

T o Prove—AB greater than any other chord.

Den — Let B C be any other chord,and draw the radius

CO

T hen OB 00 B C.

But OB 00 AB .

T herefore AB BC.


152. .A stra igh t lin e ca nnot in tersect a circumfer

en ce in mor e than two p oin ts .

Given— 0 the centre of the circle and AB a straight line

intersecting the circle.

T o Prove— T hat AB cannot intersect the circumference

in more than two points.

Dem— If AB could intersect the circumference in three

points, as C, E , and F,and we should draw the radii OC

,

OE,and OF

,

T hen OC= OE = OF. Ax . 8

We should then have three equal straight lines drawn

from the same point to a straight line, which is impossible.

28,Cor.

T herefore AB can intersect the circumference in only

two points. Q. E . D.

92 PLANE GEOME TR Ifi—B OOK I II .


158. I n the same circle, or in equ a l circles, equ a l

angles a t the cen tre in tercep t equ a l arcs on the ci r

cumference .

Given—B and O the centres of the circles AOB and

MNP respectively , and AB A0.

T o Prove Arc AC arc MN.

Dem— Apply the sector AB C to MON,so that AB coin

cides with its equal A0.

S ince AB MO,and CB NO

,Ax . 8

A will fall on M,and C on N,

And the arc AC will coincide with the arc MN,S ince all

points in each are equally distant from the centre.

T herefore arc AC= arc MN . 146. Q. E . D .

180. T he longest line that can be drawn from a point within a circle

to the circumference is a portion of the diameter drawn through the

centre and the point .

181. T he shortest line that can be drawn from a point without a circle

to the circumference is the line which when produced passes through

the centre of the circle.

THE CIRCLE . 93


154. CONVERSELY— I n the same ci rcle, or in equa l

ci rcles , equa l a r cs a re in tercep ted by equa l ang les

a t the cen tre.

Gi ven— B and O the centres of the circles AGB and

MNP respectively , and the arc AC the arc MN.

T o Prove AB AO.

Dem— S ince the circles are equal, we may apply the

circle AGB to MNP,so that the centre B will fall on O and

the point A will fall on M ; and S ince the arc AC is equal

to the arc MN,the point C will fall on N,

and BC coincides

with ON. Ax . 6

T herefore the figure AB C will coincide throughout with

AB = AO. § 4. Q. E . D.

EXERC IS E S .

132. T he Shortes t line that can be drawn from a point within a circle

to the circumference is a portion of the diameter drawn through the

po int .

133 . T he longest line that can be dra wn from a po int without a circle

and terminating in the concave arc is the line which passes through the

cen tre of the circle .

134. If an equiangular triangle be constructed on each side of any

triangle , the lines drawn from their outer vertices to the opposite vertices

of the triangle are equal.

94 PLANE GEOME TE R—BOOK I II.


155. In the same circle , or in equa l circles, equ a l

chords subtend equ al a rcs .

Given— AO and MN equal chords in the equal circles

ACD and MNP .

T o Prove arc AC= arc MN.

Dem—Draw the radii AB,B C

,MO

,and NO.

T hen in the triangles AB C and MON,

AR = MO,and B C= ON.

AR = AO.

are AO= arc MN.

PROPOS IT ION VII . T HEOREM.

156. CON-VERS ELY— I n the same. ci rcle , or i n equ a l

circles, equ a l a rcs a r e subtend ed by equ a l chords .

T IIE CIRCLE . 95

Given— AR and MN equal arcs in the equal circles ABD

T o Prove chord AB chord MN.

Dem— Draw the radii AC,B C

,MO

,and NO.

S ince the circles are equal and the arcs are equal,

T hen AC AO. 154

Also AC MO, and B C= NO. Ax . 8

Hence A AB C= A MON . 55

T herefore chord AB chord MN. § 45 (a) . Q. E . D.


157. I n the same ci r cle, or in equ a l ci rcles

, the

g r ea ter a rc is su bten d ed by the g rea ter chord ,each

a r e being less tha n a semi - ci rcumference .

G iv en— T he equal circles ACD and MNP,and arc MN

are AC.

T o Prove chord MN chord AC.

Dem— Draw the radii AB,B C

,MO

,and NO.

I f we apply the circle MNP to ACD,so that the centre 0

falls on B ,and the point M on A

, then since the arc MN is

greater than the arc AC,the point N will fall beyond C

at N’

,

And AAEN’

> AABC’. Ax . 3

96 PLANE GEOME TRK—BOOK II I .

T hen in the two trianglesABN’and ABC

,AB is common

,

B C equals BN'

(Ax . and AABN'

AABC.

T herefore, chord AN

’or its equal MN chord AC.

61. Q. E . D .

PROPOS IT ION IX. T HEOREM .

158. CONVERSELY— I n the same ci rcle, or in equa l

ci rcles , the greater chord subtend s the g rea ter a r c,

each a re being less than a semi - ci rcumference .

Given—T he equal circles ACD and MNP,with the chord

MN chord AC.

T o Prove arc MN are AC.

Dem— If the arc MN were equal to the arc AC,

chord MN chord AC.

And if the arc MN were less than the arc AC,

chord MN chord AC. 157

But both of these suppositions are false,S ince by

hypothesis the chord MN chord AC.

T herefore arc MN are AC. Q. E . D.

159. SOHOI.IUM.

—When each arc is greater than a semi

circumference, the greater arc is subtended by the less

chord , and conve1s e1y the greater chord subtends the

less arc.

8 PLANE GEOME TRK—BOOK III.


162. Th rough any th ree p oin ts not in the same

stra igh t line one can be

and bu t one.

Given— A,B,and C

,any three pOints not in the same

straigh t line.

T o Prove—T hat a circumferencemay be passed through

them.

Dem— Draw AB and B C,and erect EG and FH perpen

dicular respectively to AB and B C at their middle points.

T hen , since A,B,and C are not in the same straight

line, EG and FH will meet at some point, as 0.

Now draw the lines AO, B 0, and CO.

T hen AO B 0 CO. 22

T herefore a circumference described from 0 as a centre,

with a radius equal to AO, will pass through the three

points A,B,and C.

2d . S ince only one perpendicular can be erected at the

middle point Of a line and two straight lines can

intersect in only one point, there can be only one po int

equally distant from A,B,and C

,and therefore only one

circumference can pass through the three points. Q. E . D.

COR . 1.— Two circumferences can intersect in only two poin ts.

COR . 2.

— T here can be but one point equally distant fromthree other points.

THE CIRCLE . 99

PROPOSIT ION XI I . T HEOREM.

168. I n the same ci rcle, or i n equ a l circles , equa l

chord s a re equa lly d istan t from the cen tre .

Given—AB and CD equal chords of the circle ABD.

T o Prove— AB and CD equally distant from the centre.

Dem— Draw OE and OF perpendicular to AB and CD

respectively , and draw B 0 and CO.

T hen E is the middle point of AB and F of CD,160

And EB CF, being halves of equal chords,

And BO CO. Ax . 8

T hen A BOE A COF. 58

Whence CE OF. 45 (a)T herefore AB and CD are equally distant from 0.

Q. E . D.

EXERCIS E S .

135 . Prove that a circumferencemay be circumscribed about a triangle,

and also inscribed with in it.

136 . Prove that the inscribed and circumscribed circles of an equi

lateral triangle are concentri c.

137. T he radius of a circle inscribed in an equilateral triangle is one

th ird of the alti tude of the triangle .

133. A rad ius wh ich bisects a chord is perpendicular to it and bisects

the arc subtended by it .

139. T he radius of a circle circumscribed about an equilateral tri angle

is equal to twice the distance of the Sides of the triangle from the centre

of the circle.

100 PLANE GEOME TRK—BOOK I IL

PROPOS IT ION XI II . T HEOREM.

164. CONVERSELY— I n the same ci rcle , or in equ al

ci rcles , chords equally d istan t from the centre are

equa l.

Given—AR and CD two chords equally distant from the

centre Of the circle ABD.

T o Prove chord AB chord CD.

Dem—Draw OE and OF perpendicular to AB and CD

respectively, and draw B0 and CO.

T hen in the two right triangles BOE and COE,

B0 00,

And,by hypothesis, E0 F0.

Hence A BOE A COF, 58

And BE CF. 45 (a)But BB is one half of AB ,

and CF is one half Of CD.

160

T herefore AB a CD. Ax . 2. Q. E . D.

EXERCIS ES .

140. T he perpendiculars erected at the middle points of the sides of

an inscribed polygon meet in a common point .

141. A radius that bisects the arc subtended by a chord bisects the

chord at right angles

142. If an equilateral polygon be inscribed in a circle, acircumference,concentric with the given circle,may bemade to pass through the middle points of all the S ides.

102 PLANE GEOME TRYZ—B OOK II I .


166. CONVERSELY— I n the same ci rcle, or i n equ al

ci rcles , if two chord s a re u n equally d istant from

the cen tre, the more remote is the less .

Given— The chord AB in the circle ABD

the centre than CD.

T o Prove chord AB chord CD.

Dem—Draw the chord DG equal to AB ,and draw OE

,

OF,and OH perpendicular to AB ,

CD, and DC respectively.

T hen AB, CD, and D0 will be bisected at E,F,and II

respectively, 160

And OH OE . 5163

Draw HF,then in the triangle OHF,

OH OF. Hyp.

And A3 A4. 68

S ince AD110 and ADFO are right angles,

ADHO 4 3>ADFO A4,

Or A1 A2.

T hen DH DF. 69

But DF is one half of DC,and DH is one half of DC .

160

Hence DO’ DC,

But DC AB . Con.

T herefore chord AB chord CD. Q . E . D.

THE CIRCLE . 103

167. COR — The least chord that can

be drawn through a given point is the

chord perp endicular to the diameter

through the point.

For,draw any other chord

,as CD

,through the given

point P ,and draw OE perpendicular to CD.

T hen OP > OE . 28

Hence AB is farther from the centre than CD,which is

any other chord, and is therefore the least chord that can

be drawn through the point P . 166

PROPOSIT ION XVI . T HEOREM .

168. .4 stra igh t li n e p erp end icu la r to a r ad ius a t

i ts ex tremi ty is tang en t to the ci rcle .

Given— AB perpendicular to the radius CO at C.

T o Prove— AB tangent to the circle.

Dem— From the centre 0 draw OD to any point of AB

except C.

T hen OD 00. 28

Hence every point of AB,except C, lies without the

circle .

T herefore AB is tangent to the circle 147. Q. E . D.

104‘

PLANE GEOME TRK—BOOK II I .

PROPOS IT ION XVI I . T HEOREM.

160. CONVERSELY— A ta ngen t to a ci rcle is p erp en

d icu la r to the rad i u s d rawn to the p oin t of con tact .

Given— AB tangent to the circle CE .

T O Prove—AB perpendicular to the radius OC.

Dem— Draw OD to any point of AB except C.

S ince AB is a tangent to the circle,every point of AB

,

except C, lies w ithout the circle . 147

Hence 0C is the shortest line that can be drawn from

the point 0 to AB . Ax . 11

T herefore 00 is perpendicular to AB . Ex . 23. Q. E . D.

1 70. COR — A line drawn perpendicular to a tangent at its

point of contact passes through the centre of the circle.

EXERC IS E .

143. When two tangents meet they are equal, and the line join ingthe point Of intersection and the centre Of the circle bisects the included

angle.

PROPOS IT ION XVI I I . T HEOREM .

171. Two pa ra llel lin es in tercep t equ a l a rcs on a

ci rcumference .

CASE I . When both lines are secants.

106 PLANE GEOME TR Ii—B OOK III .

CAS E III . When both lines are tangents.

Given— AB and CD two parallel tangents.

T o Prove arc HEG arc E FG.

Dem— Draw the secant EF parallel to CD.

T hen arc E 0 arc FG,

Case II .

And are EH arc FH . Case I I .

Adding, arc GEH arc GFH. Ax . 2. Q. E . D.

1 72. COR . The straight line joining the points of contact

of two parallel tangents is a diameter.

PROPOS IT ION XIX. T HEOREM .

1 73. If two ci rcumferences in tersect ea ch other,the

line j oin ing“

the i r cen tres is p erp en d icu la r to thei r

common chord a t i ts m id d le p oin t .

Given—D and C the centres of the two circles whose

circumferences intersect at A and B .

THE CIRCLE .

T o Prov e— T hat DC bisects AB at right angles.

Den — T he point D is equally distant from A and B .

Ax . 8

T he point C is equally distant from A and B . Ax . 8

Hence DC is perpendicular to AB at its middle point.

25. Q. E . D.


1 74. If two ci rcles a re ta ngen t to each other ,the

li n e j oin ing thei r cen tr es p asses through thei r p oin t

of con tact .

Given— A and B the centres Of two circles tangent to

each other at 0.

T o Prov e—T hat the line AB joining their centres passesthrough 0.

Dem— Draw the common tangent CD.

At 0 erect a perpendicular to CD.

T his perpendicular will pass through both centres A

and B 170) and coincide with AB . Ax . 6

T herefore A,O,and B are in the same straight line.

Q. E . D.

EXERC IS E .

144. If two Opposite sides of a quadrilateral inscribed in a circle are

parallel, the other two sides are equal.

PLANE GEOME TRYZ—BOOK II I.


1 75. I n the same ci rcle , or in equ a l ci rcles , tw o

a ng les a t the cen tre ha ve the same ra ti o thei r

in tercep ted a rcs .

CAS E I . When the arcs are commerwurable.

Given—AB C and MON two commensurable angles, and

AC and MN their intercepted arcs.

[ AB C arc ACT o Prove

AMON arc MN

Dem— Let AD be a common unit ofmeasure of the arcs

AC and MN,and suppose it is contained 3 times in AC and

5 times in MN.

T henarc AC 3

are MN 5

I f we draw radu to these points of division, the angle

ABC will be divided into 3 equal parts, and MON into 5

equal parts

AAB C’

3T hen

[MON 5

Hence[AB C arc AC

Ax. 1AMON arc MN

CASE I I .— If the arcs are incommensurable.

PLANE GEOME TRY.- B O0K I II .

PROPOS IT ION XXI I . T HEOREM .

1 76 . The n umer ica l measu re of a n a ng le a t th e

cen tre of a circle equa ls the n umer ica l measu r e ofi ts in tercep ted a rc, w i th cor r esp ond ing u n i ts ofmea su r e .

Given— T he arc AD as the unit of measure of AC,and

the’

AABD the unit of measure of AB C.

T o Prove—T hat the numerical measure of the angle

ABC equals the numerical measure of the arc AC.

Dem—We have proved that

AAB C arc AC

AABD arc AD175

ButAABO

is the numerical measure of AABC, 134AABD

Andare AC

is the numerical measure of arc AC. 134arc AD

T herefore the numerical measure of AAB C equals the

numerical measure of the arc AC. Q. E . D.

1 77. S CHOLIUM 1 .- T his theorem is conventionally stated

thus,An angle at the centre ismeasured by its intercepted are.

In this statement the expression is measured by” is

used for has the same numeri cal measure as.

”

178. S CHOLIUM 2.

— T he unit of measure of arcs is T hpart of a circumference

,called a degree, and denoted by

the symbol T he corresponding un it of angles is { lg-

g

Of four right angles, and is also called a degree.

THE CIRCLE. 1l 1

T he degree (either of are or angle) is divided into 60equal parts called minutes

,and the minute into 60 equal

parts called seconds. T hese are denoted respectively byth e symbols and

When the sum of two arcs is a quadrant, each is called

th e Complement of the other. When the sum of two arcs

is a semicircumference,each is called the Supplement of

th e other. (S ee

PROPOS IT ION XXI I I . T HEOREM .

1 79 An inscr ibed a ng le is mea su red by one ha lf

i ts in tercep ted a rc.

CASE I . When one side is a diameter .

Given—AB a diameter and B C a chord of the circle

T o Prove— T hat AABC is measured by i are

Den —Draw the radius 00 ; then OC= OB,

And AB = AC.

But AAOC= AB + AC’.

Therefore AAOC = 2AB,

Or AB = i AAOC.

But AAOG is measured by are AC.

Hence AB is measured by i are AC.

112 PLANE GEOAIE TRY.—BOOK I IL

CASE I I . When the centre is within the angle.

Given—BB a diameter of the circle ABC.

T o Prove— T hat AABO is measured by l are AC.

Dem—By‘

Case I ., AABD is measured by l are AD,

And ACBD is measured by i are CD.

Hence the sum of the angles ABD and CBD,or AABO,

measured by one half the sum of the arcs AD and DC.

T herefore AAB C is measured by l are AC.

CASE I I I . When the centre is without the angle.

Given— ED a diameter of the circle DCB .

T o Prove— T hat the AAB C is measured by i arc AC.

Dem—By Case I ., ADBC is measured by i are CD,

And ADBA is measured by i are AD.

Hence the di fference Of the angles DB C and DBA,or

114 PLANE GEOME TRYZ—B OOK III

184 COR . 5.

— The opposite angles

of an inscribed quadrilateral are s up

plements of each other.

For the angles A and C are meas

ured by one half of the circumfer

ence,or by an arc of 180

°


185. .dn ang le formed by a tangen t a nd a chor d is

mea su red by one ha lf th e a re i nclu d ed betw een i ts

Giv en—AB a tangent and CD a chord to the circle E CD.

T o Prove— T hat ADCB is measured by l are CD.

Dem— Draw DE parallel to AB .

T hen AD ADCB ,37

And are E C arc CD. 171

But AD is measured by i are E C. 179

T herefore ADCB is measured by i are E C, or its equal

i are DC. Q. E . D.

PROPOS IT ION XXV. T HEOREM.

186. Zh e ver tica l ang les formed by two intersectingchords a re each mea su red by on e ha lf of the sum ofthei r intercep ted a rcs .

THE CIRCLE . 115

Given— AB and CD two chords of the circle ADC.

T o Prov e— T hat AAOD is measured by

i (arc AD arc CB ) .

Dem— Draw CE parallel to AB .

T hen AAOD AC 37

And are AE arc CB . 171

But AC is measured by i are DE . 179

T herefore AAOD is measured by i are DE ,or i (are AD

arc AE ) , or l (are AD arc B C) . Q. E . D.

NOT E —T his theorem may be proved by drawing a line from C to A .

EXERC IS E S .

An angle at the centre of a circle contains howmany degrees

in the arc included between its sides

148. An inscribed angle includes an arc of how many degrees in

the angle

147. T he angles of an inscribed triangle are and 80° respect

ively ; how many degrees in each segment of the circumference ?

148. T he triangle AB C is inscribed in a circle, AA and the arc

AR = find the number of degrees in each Of the remaining arcs

and angles.

149. T he quadrilateral AB CD is inscribed in a circle, AA

A B = 60°, and the arc DC find the number of degrees in the

remain ing angles and arcs .

150. An angle formed by a chord and tangent contains find the

number of degrees in each segment of the circumference.

116 PLANE GEOME TRK—BOOK I II .

PROPOSIT ION XXVI . T HEOREM.

1875 The angle between two secan ts, in tersectingw i thou t the ci rcumference , is measu red by one ha lfthe d if ference of the in ter cep ted arcs .

Given—AB and CB two secants intersecting without the

circle AEC.

T o Prove—That angle B is measured by

l (are AC arc FE ) .

Dent—Draw ED parallel to B C.

T hen AB ADEA, 37

And are DC arc EF. 171

But ADEA is measured by i are AD. 179

T herefore AB is measured by l are AD,or l (are AC

arc DC) , or i (are AC arc FE ) . Q. E . D.

NOT E .—T h18 theorem may be proved by drawing a line from E to C ;

the demonstration is then similar to that of T heorem XXVI I .

PROPOSIT ION XXVI I . T HEOREM.

188. The angle between a secant and a tangent is

mea su r ed by one ha lf the d ifference of the in tercep ted a rcs .

118 PLANE GEOME TRY.—BOOK I II .

PROPOS IT ION XXVI I I . T HEOREM .

189. T he angle between tw o ta ngen ts is mea su red

by one ha lf the d ifi'

erence of the in tercep ted arcs .

Given— AB and CD two tangents of the circle

T o Prove—T hat AB is measured by i (are DEF arc

DE ) .

Dem—Draw DE parallel to AB .

T hen AB AEDG, 37

And are DF arc EF . 171

But AEDC is measured by l arc DE . 185

T herefore AB is measured by i are DE , or i (arcDEFarc EF ) , or i (are DEF arc DE ) . Q. E . D.

NOT E .—T his theorem may be proved by drawing a line from D to F .

EXERCIS ES .

155. T he quadrilateral AB CD is inscribed in a circle. I f the S idesAB

and AD subtend arcs Of 60° and 140° res pectively, and the angle AEB

between the d iagonals is how many degrees are there in each angle

Of the quadrilateral

156 . I f the angles A and B of a circumscribed triangle AB C are 80°

and and the S ides AB , B C, and AO are tangent to the circle at the

points D, E ,and F , find the number of degrees in each angle of the tri

angle DEF formed by connecting D,E,and F .

THE CIRCLE . 119

EXERC IS E S .

ORIGINAL T HEOREMS .

157. T he lines joining the corresponding extremities of two parallel

chords are equal.

156 . If two lines intercept equal arcs on the circumference of a circle,they are parallel.

159. T wo tangents drawn from the same point without a circle are

equal.

160. Any two chords of a circle which cut a diameter at the same

po int and at equal angles are equal to each other.

161 . In any quadrilateral that circumscribes a circle the sums Of the

Opposite sides are equal. (See

162. T he line which bisects the angle formed by two tangents passesth rough the centre of the circle . (Q

163 . If a circle be inscribed in a righ t triangle, the difference between

th e two legs and the hypotenuse is equal to the diameter Of the circle.

164. T he circles described on any two s ides Of a triangle as diameters

intersect on the third side.

Succm xom—Let a perpendicular fall from the opposite vertex upon

th e th ird side .

165 . NO parallelogram bu t a rectangular one can be inscribed in a

circle. (2184 and Q

120 PLANE GE OME TRY.- BOOK IIL

166. If a triangle, ADE , is formed by the inter

section of three tangents to a circumference, twoOf which , AB and AC, are fixed , while the th ird ,DE , is tangent to the circle at a variable point P ,

prove that the perimeter of the triangle ADE is

equal to AB AC, or 2AB . Prove also that the

angle DOE is constant .

167. If the two opposite sides of an inscribed quadrilateral are equal

and parallel the figure is a rectangle.

166. T he circle described on on e Of the equal sides Of an isosceles tri

angle as a diameter bisects the base .

169. If tw o circles are tangent externally . and a straigh t line is drawn

through their po int of contact, terminating in their circumferences, the

radi i drawn to its ex tremities are parallel.

Also, the tangents at its extremities are parallel.

170. In an inscribed trapezoid the non -

parallel sides are equal ; and

also the diagonals .

171. I f a tangent is drawn to a circle at the extremity Of a chord ,the

middle point of the subtended arc is equally d istant from the chord and

the tangent .

172. If a circle is described on the rad ius of another circle as a diam

eter, any chord of the greater circle drawn through their point of con

tact is bisected by the circumference Of the smaller circle.

173. If a circle be circumscribed about a righ t triangle, and another

he inscribed in it, the sum of the two sides containing the right angle is

equal to the sum Of the diameters.

174. If a d iameter and a chord be drawn from the same point in the

circumference of a circle, the angle thus formed will be one- half of the

angle formed by the tangents drawn through the extremities of the

chord .

175. If the base of the triangle AB C be

produced both ways, so that AE AC and

BF = B C, and a circumference be describedthrough the points E ,

0. and F ,the line join

ing its centre 0 with the vertex 0 will bisect

the angle C.

122 PLANE GEOME TRK—BOOK III .

PROPOS IT ION XXX. PROBLEM.

191. Fr om a g iven p oin t w ithou t a stra ight

to d raw a p erpend icula r to that line .

Given—P a point without the straight line AB .

Required—T O draw a perpendicular from P to AB .

Cons—With P as a centre, and with a radius greater

than PF,describe an arc intersecting AB at C and D.

With C and D as centres, and with a radius greater than

CF,describe arcs intersectn at E , and draw PE .

T hen PE is perpendicular to AB .

Proofi—For the line PE has two points each equallydistant from C and D.

Hence PE is perpendicular to CD at its middle point.25. Q. E . F.

PROPOSIT ION XXXI . PROBLEM.

192. At a given poin t in a stra ight line, to erect a

p erp end icu lar to tha t lin e .

PROBLEMS OF CONS TR UCT IONZ

Given— P a point in the straight line AB .

Required— T o erect a perpendicular to AB at P

Cons— On the line AB take CP equal to PD.

With C and D as centres,and with a radius greater than

CP , describe arcs intersecting at E,and draw EP .

T hen PE is perpendicular to AB .

Proof.—For the line PE has two points each equally

distant from C and D.

Hence PE is perpendicular to CD at its middle point.

25. Q. E .

Another Method.

Given— P a point in the straight line AB .

Requ ired— T o erect a perpendicular toAB at P.

Cons —With any point 0, without the line AB ,as a cen

tre, and with a rad ius equal to OP , describe a Circumference

intersecting AB at C and P .

Draw the diameter CE , and draw PE .

T hen PE is perpendicular to AB at P .

Proofi— For AOPE is a right angle, being inscribed in a

semicircle. 180. Q. E . F .

NOT E—m method is preferred when the given point is near the end

of the line.

EXERCIS E .

176 . Bisect a line 6 inches long.

124 PLANE GEOME TRK—B OOK III:

PROPOSIT ION XXXI I . PROBLEM.

193. To bisect a g iven ang le.

Given— ABC an angle.

Required— T o bisect AABC.

(Dims—With B as a centre,and with any convenient

radius,describe an arc intersecting AB at D

,and CB at E .

With D and E as centres,and with any radius greater

than EG,describe arcs intersecting at F, and draw BF.

T hen BF bisects AAB C.

Proofl— For, since B and F are each equally distant from

E and D,B F is perpendicular to the chord of DE at its

middle point. 25

Hence BF bisects AAB C. 160. Q. E. F.

PROPOS IT ION XXXII I . PROBLEM.

194. To bisect a g iven are.

Given—ABC an arc.


PROPOS IT ION XXXV. PROBLEM.

196 . Th rough a g iven p oin t w i thou t a stra ight line,to d raw a para llel to the line.

Given— F a point without the straight line AB .

Requ ired— T o draw through F a line parallel to AB .

Cons — T hrough F draw any line GE .

Construct AGPD AFEB . 195

T hen CD is parallel to AB . 38. Q. E . F.

Proof—For two lines are parallel when the exterior

interior angles are equal. § 38

Another Method.

E

F

Given— E a point without the straight line AB .

Required— T o draw through E a line parallel to AB .

Cons—With the given point B as a centre,and with a

radius greater than EF, describe the indefinite arc OH .

With H as a centre,and with the same radius

, describe

the arc EF.

With H as a centre, and with a radius equal to the chord

of EF,describe an are at G.

PROBLEMS OF CONS TRUCT ION: 127

Draw the lines GE and HE .

T hen CD. is parallel to AB .

Proof. - For 177

And two lines are parallelwhen the alternate angles are

equal. 38. Q. E . F.

197 Given two ang les of a tr ia ng le, to find the

th i rd .

Given—A and B two angles Of a triangle.

Requ ired— T o find the third angle.

Cons— At the point 0 on the indefinite straight line CD

construct ABOD AA,and AEOF = AB

T hen COF is the required angle.

Proof.

— For the third angle of a triangle equals two right

angles minus the sum of the other two. 51. Q. E . F.

PROPOS IT ION XXXVI I . PROBLEM.

198. Gi ven two sid es and the inclu d ed angle of a

tr iang le, to construct the tr ia ng le .

Given—m'

and n the S ides and A the included angle of

a triangle.

PLANE GEOME TRPI—BOOK HI

Required— T o construct the triangle.

Cons —On the line CD take CE equal tom.

Construct AFCE AA.

On CG take CF = n,and draw FE .

T hen FCE is the required triangle.

Proof.— T his is evident from 55. Q. E . F.

PROPOS IT ION XXXVI II . PROBLEM .

199 Given two ang les a nd the included sid e of a

tr iang le, to cons tru ct the tr iang le.

Given—A and B the two angles and m the included

S ide of a triangle.

Required— T O construct the triangle.

Cons—On the line CD take CE equal to m.

Construct AC= AA,and AB = AB .

T hen EG and CF intersect at H,

And CEH is the required triangle.

130 PLANE GEOME TRY.-BOOK III .

Given—m and n the two S ides and O the angle Opposite

n in a triangle.

Requ ired— T o construct the triangle.

Cons—At A,on the line AB ,

construct AA AO,and

take AD m.

With D as a centre,and with a radius equal

”

to n,

describe an arc.

T hen there may be three cases to this problem.

Let p be the perpendicular from D.

First—When n p .

T he are will intersect AB in two points, E and C,

And there will be two triangles, ABD and ACD,either

of which will answer the given conditions.

In this case there are two solutions.

S econd —When n p .

T he arc will be tangent to AB .

In this case there is one solution a right triangle.

T hird.

—VVhen n p .

T he are will not intersect AB .

In this case there is no solution.


CASE IV. When 0 is acute, right, or obtuse, and n m.

Cons —At A,on the line AB

,construct [ A A0, and

take AD m.

With D as a centre,and with a radius equal to m or n

,

describe an arc.

In this case the arc intersects AB in only one point, B .

T here will be but one solution.

When AO is right or obtuse, the problem is impossible.

Q. E . F .

205. S CHOLIUM.

— In Case I I I .,when the given angle is

right or obtuse,and the given side opposite the angle is

less than the other side,the problem is impossible.

PROPOS IT ION XLI . PROBLEM .

206 . Given two s id es a nd the inclu d ed a ng le of a

p a ra llelog ram, to constru ct the p a ra llelog ram.

Given—m and n the two given sides and 0 their ih

cluded angle of a parallelogram.

Requ ired— T o construct the parallelogram.

134 PLANE GEOME TRK—BOOK III.

PROPOS IT ION XLI I I . PROBLEM.

208. To circumscr ibe a circle abou t a g iven

ang le.

Given—AB C any triangle.

Required— T O circumscribe a circle about AB C.

ants— Erect DE and FG perpendicular to AB and AC,respectively , at their middle points (5T hen 0

,the point where these two perpendiculars inter

sect, is the centre required .

With 0 as a centre,and with a radius equal to AO, de

scribe the circle AB C.

T hen ABC is the required circle.

Proofi—T his is evident from 161. Q. E . F.

PROPOSIT ION XLIV. PROBLEM.

209. To inscr ibe a circle in a g iven tr iangle.

Given—ABC any triangle.

Requ ired— T O inscribe a circle in ABC.

PROBLEMS OF CONS TR UCT ION.

7135

Cons—Draw AE and BD,bisecting AA and AB respect

ively from 0,their point of intersection, draw OF

perpendicular to AB .

With 0 as a centre,and with OF as a radius

,describe

the circle FGH.

T hen AB,B C

,and AC are tangent to the circle.

Proof— For all points on AE are equally distant from

AB and AC,and all points on DB are equally distant from

AB and B C ; hence 0 is equally distant from AB,B C

,and

AC Q. E . F .

PROPOS IT ION XLV. PROBLEM .

210. At a g i ven p oin t in a g iven circumference , tod r aw a tang en t to the ci rcumference.

Given— C a point in the Ci rcumference CD.

Requ ired— T O draw a tangent to CD at C.

Cons— If the centre is not known, it must be found

by 207.

Draw the radius 00.

At C erect AB perpendicular to 00.

T hen AB is the required tangent.

Proof— For a line perpendicular to a radius

tremity is tangent to the circle

136 PLANE GEOME TRK—BOOK II I

PROPOSIT ION XLVI . PROBLEM.

211 . F rom a g iven p oin t w i thou t a g i ven circle, to

d raw a tangen t to the ci rcle .

Given— P a point without the circle AB C.

Required— T O draw from P a tangent to AB C.

Cons — Draw OP connecting the given point with the

centre.

On OP as a diameter describe a circumference intersecting the given circumference at A and B .

Draw AP and BP .

T hen AP and BP will both be tangent to AB C.

Proof.

— For AOBP , being inscribed in a semicircle, is a.

right angle. 180

Hence PB is tangent to AB C. 168. Q. E . F .

EXERC IS E S .

177. T O construct an angle equal to the sum Of two given angles .

178. T O construct an angle equal to the difference between two given

angles.

179. Given an angle, to construct its complement.

180. Given an angle, to construct its supplement.

181. T O construct an angle equal to three and one half times a given

angle.

102. Given a circle whose diameter is 10 inches and a point,P ,4 inches

from the circumference, to draw a tangent to the circle from the point P


S CHOLIUM.— If the point C falls within the segment AGB

,

angle C will be greater than angle K ; if C falls withoutthe segment ACB ,

angle C will be less than angle K ;hence the arc AGB is the locus of the vertices of all

angles equal to K whose sides pass through A and B .

PROPOS IT ION XLVII I . PROBLEM.

213. To d raw a common ta ngen t to two given ci rcles.

1 . To draw an exter ior common tangent.

Given— AC and GH two circles.

Required— T O draw an exterior common tangent.

Cons—With 0 as a centre, and with a radius equal to

OC GD,describe the circumference EF.

From D draw a tangent to BB

Draw 00 through the point of tangency , and draw DO

parallel to OC, and connect C and G.

T hen CG is the required tangent.

Proof— For, by construction ,CE is equal and parallel to

DG ; hence EDGC is a parallelogram But the angle

E is a right angle and the parallelogram is a

rectangle, and the angles C and G are right angles.

T herefore CG is tangent to both circles. Q. E . F.

PROBLEMS OF CONS TR UCTION. 139

2. To draw an interior common tangent.

Given— CC and DG two circles.

Requ ired—T o draw an interior common tangent.

Cons—With 0 as a centre,and with a radius equal to

00 DG,describe the circumference E F.

From D draw a tangent to BE

Draw OE through the point of tangency , and draw DG

parallel to OE ,and connect C and G.

T hen CG is the required tangent.

Proof— For,by construction ,DG is equal and parallel to

CE ; hence EDGC is a parallelogram But the angle E

is a right angle and the parallelogram is a rect

angle, and the angles C and G are right angles.

T herefore CG is tangent to both circles. Q. E . F .

S CHOLIUM .

—I f two circles are tangent externally , the

two interior tangents reduce to a common tangent. If

they are tangent internally , the two exterior tangents

reduce to a common tangent . If they intersect, only the

ex terior tangents are possible. I f one circle is wholly

within the other,the problem is impossible.

140 PLANE GEOME TR I’Z—B OOK I I I .

EXERC IS E S .

PROBLEMS FOR CONST RUCT ION.

T he bes t method to discover the solution of a problem in Geometry is

to suppose the problem solved , and construct a figure accordingly . If

the problem is care fully analyzed , and auxiliary hues drawn when

neces sary , the solution can generally be effected .

183 . Construct an angle of of

184. Construct an angle of of of of

185. Given the base and altitude of an isosceles triangle, to construct

the triangle.

186 . Given one Of the equal sides and one of the equal angles of an

isosceles triangle, to construct the triangle.

187. Given one side and the hypotenuse Of a right triangle, to con

struct the triangle.

188. Given the hypotenuse and an acute angle of a righ t triangle, to

construct the triangle .

189. Given the altitude and one of the equal angles of an isosceles

triangle, to construct the triangle .

190. Construct a square, having given its diagonal.

191. Construct a parallelogram,having given the diagonals and their

included angle .

192. Construct a circle, with a given radius,which shall pass through

two given poin ts .

199. Construct a circle,w ith a given radius, tangent to a given circle

and to a given straight line .

194 . Construct a circle , w ith a given radius, tangent to two given inter

secting circles .

195. Inscribe in a given circle a triangle equiangular with a given

triangle .

196 . Draw a line tangent to a given circle and parallel to a given

straight line .

197. Draw a line tangent to a circle and perpendicular to a given

straight line .

198. Draw a line tangent to a given circle and making a given angle

with a given straight line .

199. T risect a right angle.

200. Construct a parallelogram,having given a side and its two

diagonals .

142 PLANE GEOME TRK—BOOK II I.

217. Find the locus Of the vertex of a right triangle having a given

hypotenuse as its base (5

218. F ind the locus Of the middle points of all chords of equal lengththat can be drawn in a given circle (Q

219. F ind the locus of the middle points of all chords that can be

drawn through a given point, P, in the circumference of a given circle.

220. If from a given point, P, lines be drawn intersecting a given cir

cumference in B and 0, find the locus of the middle point, D, Of the

chord B C.

221. If from a given point, P,lines be drawn to any point of a given

circumference, find the locus Of the middle point, B , of the secant

PA (9

222. I f the straight line, AB ,of given length moves so

.

that its ends

constantly touch two fixed lines ,AO and B0, which are perpendicular

to each other, find the locus of the middle point P.

229. If upon a given base, AB ,a triangle ABC is constructed, having

a given vertical angle 0. find the locus of the foot of the perpendiculars

from the vertices A and B upon the opposite sides.

B O O K IV .

AREA AND RELAT IONS OF POLYGONS .

DEF INIT ION S .

214. T he Area Of a polygon is its ratio to another sur

face regarded as the unit of measure.

T he unit for measuring polygons is generally a square

whose side is a definite linear unit. T he unit of surface

is called a superficial unit.

215. E qual Poly gons are polygons which when applied

to each other coincide throughout .

216 . E qu ivalent Poly gons are polygons which have

equal areas.

Equivalent polygons may differ in form,and then would

not be equal. T hus a triangle, square, circle, etc.

,each

having the area of a square foot, would be equivalent but

not equal.

Equal polygons are always equivalent.

217. S imilar Poly gons are polygonswhich aremutually

equiangular, and which have their homologous sides pro

portional.

Homologous sides, lines, angles, etc. of similar polygons

are those which are similarly situated . Homologous lines

make equal angles with the sides.

218. S imilar Arcs , S ectors , or S egments , in different

circles,are those whose arcs subtend equal angles at the

centre.

NOT E.- T he symbol=e=is used for the words is equivalent to.

143

144 PLANE GEOME TRK—BOOK I V.

AREA OF POLYGONS .


219. Para llelog rams ha ving equa l bases and equa l

a ltitudes a re equ ivalen t .

Given—ABCD and BEGH two parallelograms havingequal bases and equal altitudes.

T o Prove D ABCD D EFGH.

Dem—Apply D BEGE to D AB CD,and

,since they have

equal bases and equal altitudes, EFGH will take the posi

S ince AD BC,and AL BK

, 79

And 39

T hen A DAL A GBE . 55

Now,since ABOL A DAL D AB CD

,

And AB CL A CBK D ABE L,or D EFGH

,

T herefore D AB CD D EFGH. Ax . 1. Q. E . D.

220. COR. 1 ,

— A triangle is eq uivalent to one half of a

parallelogramhaving an equal base and an equal altitude.

For A ABC is equal to one half Of D DEHK,having an

equal base and an equal altitude. 80

But D DEHK D DEFG,or to any other parallelogram

having an equal base and an equal altitude. 219

146 PLANE GEOME TRK—B OOK I V.

CASE I I . When the bases are incmnmensurable.

Given—AB CD and EFGH two rectangles having equal

altitudes and incommensurable bases.

T o ProveAB CD AB

Dem— T ake any unit, as EM,which is contained an

exact number of times in EF.

T hen,since AB and E F are incommensurable, EM will

be contained in AB a certain number of times,with a

remainder,KB

, which is less than the unit of measure.

Draw KL parallel to B C.

T henAKLD AK

Case I .

E FGH E F

Now,if we diminish the unit of measure indefinitely,

the remainder KB will diminish indefinitely ,

And AK will approach AB as its limit, and AKLD will

approach ABCD as its limit.

Hence we have two variables,

AKLD AK

E FGH EF

which are always equal.

And since the limits of equal variables are equal,

T henAKLD AK

,E FGII EF

AB CD

AREA OF POLYGONS . 147

223. COR . Two rectangles having equal bases are to each

other as their altitudes.

S ince either side of a rectanglemay be taken as its base.

PROPOSIT ION I I I . T HEOREM.

224. Any two rectang les a re to ea ch other as the

p rod u cts of thei r bases by thei r a lt i tud es .

Given—R and r any two rectangles having the bases B

and b and the altitudes A and a respectively .

R B X AT O Prove

r b x a

Dem— Construct the rectangle 8 having the base B and

the altitude a.

T hen the rectangles R and S , having equal bases, are to

each other as their altitudes.

HenceR i

s

S a

And the rectangles r and 8, having equal altitudes, are

to each other as their bases.

Hence f fMultiply ing,

1

: i)

; f225. S CHOLI UM.

— T he product of two lines means the

product of their numerical measures

1 48 PLANE GEOME TRK—B OOK I V.


220. I h e a rea of a rectang le is equa l to the p rod uct

of i ts base and alti tud e.

Given—R a rectangle whose altitude and base are A

and B respectively, numerically expressed in terms of the

linear unit, and let u be the unit of surface whose side is

the linear unit.

T o Prove—Area R A x B .

Dem— S ince any two rectangles are to each other as the

products of their bases by their altitudes

R A X ET hen

u 1 x 1

But is the area of R

T herefore area B A x B . Q. E . D.

A X E .

227. COR — The area of a square is equal to the square ofits side.

228. S CHOLIUM—When the sides of a rectangle are com

mensurable, the truth of this theorem

may be seen by dividing the rect

angle into squares, each equal to the

unit of measure.

T hus,if the base of the rectangle contains 8 units

,and

the altitude 4 units,the rectangle contains 32square units.

T his result may be Obtained in two ways : lst. We mayregard the rectangle as composed Of 4 rows of squares,


COR . 2. Tumparallelograms having equal bases are to eachother as their altitudes.

For, cancelling the bases in t;

COR . 3. Two parallelograms having equal altitudes are to

each other as their bases.

For,cancelling the altitudes in

f E.

p b

COR . 4, Two parallelograms having equal bases and equal

altitudes are equivalent.

For,cancelling both the bases and altitudes In

P

p

1,or P¢p .


231. T he a rea of a tr iangle is equ a l to one ha lf the

p rod uct of i ts base and a lti tud e .

Given—AB C a triangle having its altitude equal to a

and its base equal to b.

T O Prove—Area AB C= i a x b.

Dem— Construct the parallelogram ABEC.

Now, A AB O= l B ABE C. 220

But D ABE C= a x b. 229

Hence A ABO= i a x b. Q. E . D.

232. COR . 1 . Two triangles are to each other as theproducts

of their bases by their altitudes.


S QUARE S ON LINE S .


235. The squ a re d escr ibed on the hyp otenu se of a

r igh t tr iang le is equ iva len t to the sum of the squ a res

d escr ibed on the other two sid es .

Given— ABC 0. right triangle right- angled at C.

T o Prove— T he square ABED equivalent to the sum Of

the squares B CHF and ACGK .

Den — Draw CL parallel to BE ,and draw AF and CE .

S ince AB CH and AACB are right angles,ACH is a straight

line for the same reason B CG is a straight line.

In the triangles CEE and ABE,

AB BE and B C BF,sides of the same square,

And AABE = ACEE,

since each is equal to a right angle AAB C.

Hence A CBE A AEE. 55

S ince the triangle CBE has the same base and altitude

as the rectangle BELN,

T hen A CBE 5D BELN.

soUARES ON LINES . 155

PROPOS IT ION X. T HEOREM.

239. I n any tr iang le, the squ a r e of a side opp osite

an acu te ang le is equ a l to the sum of the squ a res ofthe other two sid es , min u s tw ice the p rod u ct of one

of these sides and the p roj ection of the other sid e

upon i t .

F IG. 2.

Given— B an acute angle of the triangle AB C, and draw

CD perpendicular to AB .

Dem— In Fig. 1,

AD AB DB .

In Fig. 2,

AH= DB AB .

In the right triangle ADC,

235

EP .-r . AEM IYET 2AB x DB , (2) Sq. (1)

(3) 236

Adding (2) and

m + ITO’,orM =AE + ET P— 2A3 x DB . Q. E . D.

,

SOHOLmM .— T he loth Proposition can be very readily

drawn from the 9th,by transposing the terms and re

ducing.

EXERCIS E S .

227. T he side Of an equilateral triangle is 12 inches ; find the altitude

and area of the triangle.

226 . If the hypotenuse of a right triangle is 40 inches and the base 30

inches, find the perpendicular.

156 PLANE GEOMET RK—BOOK I V.

PROPOS IT ION XI . T HEOREM.

240. I n any tr iang le, if a med ia l line be d rawnfr om the vertex to the base,

1 . The sum of the squ a res of the other two sid es is

equa l to tw ice the squ a re of ha lf the base, p lu s tw icethe squ a re of the med ia l line.

2. Th e d if'

erence of the squ a res of the other two

sid es is equ a l to tw ice the p rod u ct of the ba se an d

the p roj ect ion of the med ial line up on the base.

Given—DE the projection of the medial line upon AB .

T O Prove

l

Dem.— In A ADC, Ac

? AT ) ’ ZAD

S ince AD= DB,

F = E +W + Ae x DE . (1)

(2)

Adding (1) and we have

S ubtracting (2) from we have

PLANE GEOME TRYZ—B OOK I V.

CA DB

CE DET o Prove

Dem.—Draw AD and B C.

T hen,since the triangles

ADC and CDE have their

vertices at the same point, D,and their bases in the same

straight line, they have equal altitudes, and are to each

other as their bases.

Whence A AB O 232,3.

For a similar reason the triangles B CD and DOE are to

each other as their bases.

Whence A B CD DB§ 232. 3

A CDE DE

But the triangles ADC and B CD have the same base,

CD,and the same altitude

,since CD is parallel to AB .

Whence A ADC¢ A B CD. 232,4.

Hence the two proportions have a couplet in each the

same,and the remaining couplets will form a proportion .

CA DB“ 31.

CE DET herefore Q. E . D .

243. COR . 1 . When a line is drawn parallel to the base ofa triangle, the corresponding segments are proportional to the

two sides of the triangle.

T he result of 242may be written ,

CA OE = DB zDE .

By composition ,CA CE CE DB DE : DE

, 125

Or

Also AE AC BE BD.

PROPORT IONAL LINES . 159

244. COR . 2.

— S traight lines drawn parallel to the base ofa triangle divide the other sides proportionally.

From this theorem it is evident that

FD DA

CK KG GE EB

Dropping the ratios in parentheses, the

corresponding segments are proportional.


245. CONVERSELY—A line wh ich d i vid es two sid es

of a tr iang le p r op ort iona lly is pa ra llel to the th i rd

side .

Given—DE a line drawn so that

AC B C

DC E C

T O Prov e— DE parallel to AB .

Dem— I f DE is not parallel to AB ,draw DF parallel.

T henAC B C

242DC FC

A B CBut

,by hypothesi s,

DC E C

T henB C B C

I AX. 1FC E C

Or FC E C.

T hen DE must coincide with DE, and is parallel to AB .

Q. E . D.

160 PLANE GE OME TRY.—BOOK I V.

PROPOSIT ION XV. T HEOREM.

246. I n any tr iang le , the bisector of an ang le d ivid es

the Opp osi te s ide in to segmen ts p ropor tiona l to the

adj acent sid es .

Given— CD bisecting the angle ACE .

T o ProveDA CA

DB CB

Dem—Draw BE parallel to DC, and produce

T hen An A0,

And Am Ap .

But,by hypothesis, Am An.

T hen Ao Ap,

And

Now,since CD is parallel to BE ,

T henDA CA

DB CE

Putting GB for its equal CE ,we have

DA CAQ. E . D.

247. COR .—CONVERSELY.

— If a line is

angle of a triangle dividing the opposite side proportional to

the adjacent sides, it bisects the angle.


S IM ILAR POLYGON S .


250. T r iang les wh ich a re mu tua lly equ iangu lar

a re s im i lar .

Given— T he triangles AB C and DEF, with

AA= AD,AR = AE ,

and AC= AF.

T o Prove—A ABO and A DEF similar.

Den —Place the triangle DEF upon AB C, so that AF

coincides with AC, and A DEE will take the position of

S ince ACGH AA,GH is parallel to AB ,

AC B CAnd

GC’ H0"

OrAC BC

DE EF

In a similar manner it may be shown that

AC AB

DF DE

Hence AABC and A DEF have their homologous sides

proportional, and are similar. 217. Q. E . D .

251. COR . 1 .

—T wo triangles are similar when two angles

of one are equal respectively to two angles qf the other.

252. COR . 2. Two right triangles are similar when an acute

angle of one is equal to an acute angle of the other.

SLMILAR POLYGONS . 163

253. COR . 3.

- If two triangles are similar to a third tri

angle, they are similar to each other .

PROPOS I T ION XVI I I . T HEOREM .

254. T wo tr iang les a r e s imi la r when th ey ha ve a n

angle of on e equ a l to a n a ng le of the oth er , an d the

s id es inclu d ing thes e a ng les p rop or t iona l .

Given— T he triangles AB C and DEF , with

AC B CC= AF andé

DE EF

T o Prove—A AB C and A DEF similar.

Dem— Place the triangle DEF upon A AB C, SO that AF

coincides with AC, and A DEF will take the position Of

r A0 B 0Then

,by hypothesis,

CC HG

Hence GE is parallel to AB , 245

And ACGH = AA,and ACHG= AB . § 37

T hen A ARC and A GHC,or its equal A DEE,

are similar.

250. Q. E . D.

EXERCIS E S .

229. In the triangle AB C, AB 21 , AO= 16 , and B C= 12 ; find the

segments of AB formed by the bisector of the angle 0.

280. In the triangle AB C, AC= 30, B C z 10, and AB =: 30; find the

segments of AB formed by the bisector of the external angle at G.

231. If the sides of a triangle are a , b, and c, find the S ides Of a similar

triangle Whose base is d . e

164 PLANE GEOME TR Y.- B OOK I V.


255. Two tr iang les a re s imi la r when thei r homol

ogou s sid es a re p rop or t iona l.

Given—T he triangles AB C and DEF, with

AC AB B C

DF DE EF

T o Prove—A AB C and A DE E S imilar.

Den — T ake CG DE,and CH EF

,and draw GH.

T hen A ABO and A GHC are similar. 254

AC AB AC AB1Hence

ac GH’

DE GH52 7

AC ABhBut, by hypot eS Is,

DE DE

ABHence

GH’ Ax . 1

And A DEE = A GHO. 57

T herefore the triangles AB C and DEF are similar.

Q. E . D.

256. COR . The altitudes of two similar triangles are to eachother as any two homologous sides.

EXERCIS E .

282. In the triangle ABC, AC=24 in . and BC=30 ia . ; if a line

parallel to the base AB cuts the side AC 8 in . from the base, how far

from the vertex 0 does it cut the side B C?

166 PLANE GEOME TRK—BOOK I V:

PROPOS IT ION XXI . T HEOREM .

259. The lines d rawn from the ver tex of a tr iang le

to the base d ivid e the base an d a pa ra llel to i t p ro

p ortiona lly .

Given—DE parallel to AB ,in the triangle AB C, and CF

and CG drawn from the vertex to the base.

AF FG GB

DH E X KET O Prove

Dem.

— S ince A DHC is similar to A AFC, A E KO to

A POC,and A KE C to A GB C

, § 250

AF FG GBT hen

DH E X KE217

Dropping the ratios in parentheses, the base and parallel

are divided proportionally . Q. E . D.

260. COR — If AB is divided into equal .parts at F and G,

then DE is divided into equal parts at H and K.

PROPOS IT ION XXI I . T HEOREM .

261. If a p erp end icu la r be d rawn fr om the ver tex

of the r igh t ang le to the hyp oten u se of a r igh t tr iang le,

I . Th e tr iangles formed a re s imi la r to the wholetr ia ng le , a n d to ea ch other .

I I . The p erp end i cu la r is a mean p rop ort iona l be

tween the segmen ts of the hyp oten u se .

S IDIILAR POLYGONS . 167

I I I . E i ther leg is a mea n p r op or t iona l between the

whole hyp otenu se and the adj acen t segmen t .

Given— CD perpendicular to the hypotenuse of the right

triangle AB C.

T o Prove , 1 .

— T he triangles ADC and BDC similar to

the triangle AB C and to each other.

Dem— In the right triangles ADC and ABC, AA is

common .

Hence A ADC and A ABO are similar.

In like manner, A BDO and A ABO are similar.

Hence A ADC and A BDC are similar.

T o Prove , 2. AD DC= DC : DB .

Dem—S ince the triangles ADC and BDC are similar,th eir homologous sides are proportional

Hence AD DC DC : DB . (1)

T o Prove , 3. AB : AC= AC zAD.

Den — S ince the triangles AB C and ADC are S imilar,

their homologous sides are proportional

Hence AB : AO= AC : AD. (2)

In like manner,AB B C= B C : BD. (3) Q. E . D.

262. COR . 1.

— The sq uare cf the perpendicular is equal to

the product of the segments of the hypotenuse.

From D_

C’ = AD x DB . (4) 5 119


263. COR . 2. The square of either leg is equal to the prod

uct of the hypotenuse and the adjacent segment.

From A_

O AB x AD. (5) 119

From (6) § 119

264. COR . 3.

— T7Ie squares of the legs are proportional to

the adjacent segments of the hypotenuse.

D’ ‘

d’

5 b 6AC AR x AD AD

A . 2.IV] mg ( )AB X DB DB

x

265. I f from any point in a

circumference a perpendicu

lar be let fall,and chords be

drawn to the extremities Of

the diameter,then AB C is a

right triangle and

1 . The perpendic ular is a mean proportional between the

segments of the diameter .

2. Each chord is a mean proportional between the diameter

and the segment adjacent to that chord.

PROPOS IT ION XXI I I . T HEOREM .

266. I n any r igh t tr ia ng le the squ a re of the hyp ot

cun ee is equa l to the sum of the squa res of the legs .

Given— AB the hypotenuse Of the right triangle AB C.

T o Prov e


at D,and their bases in the same line AE

,they have the

same altitude, and are to each other as their bases.

HenceA M E 415 . (1)A CDE CE

A ABE BE.In like manner

,

A ADE DE(2) 232

,3 .

A ABB AE X BEMultiply ing (1) by

A CDE CE x DEQ. E . D .


269. Two s imi la r tr iang les a re to each other as the

squa res of thei r homologou s s id es .

Given —AC and DE,homologous sides Of the similar tri

angles ABC and DEF.

T o Prove A AB C : A DEF W if? ”

Dem— Draw the altitudes CM and EN .

S ince the triangles are similar,

§ 217

And i CM : lFN AC : DE. 256,129

Multiply ing, A AB C : A DEF A7 7" DE ’. 132. Q. E . D.

270. COR . T wo similar triangles are to each other as the

squares of any two homologous lines.

S ince A AMC and A DNE are similar,

250

§ 217

And zw . § 128

Hence A AB C : A DEF CW EN“ m DIV” . 131

S IMI LAR POLYGONS . 171

PROPOS IT ION XXVI . T HEOREM .

271. T wo p olygons a re s imi la r w hen they a re com

p osed of the same n umber of tr iang les, s imi la r eachto each , and simi la r ly p laced .

Given— A AB C similar to A FGH, A ACD to A FHK,

and

A ADE to A FE L.

T o Prov el—EB similar to LG.

Dem— S ince the corresponding triangles are similar,

AI A2.

A3= A4.

Adding, AC= AH.

In like manner, AD = AK

,and AA AF.

Also AB = AG,and AE = AL .

Hence AB CDE and FGHKL are mutuallv equiangular.

Again , since the corresponding triangles are similar,

AB B C CD DE AE217

FG GH E X KL FL

Dropping the ratios in parentheses, the homologous sides

a re proportional.

T herefore EB and LG are similar. Q. E . D.

EXERC IS E .

233 . What is the relation of two similar triangles whose altitudes are

6 and 9 feet respectively? Of two similar triangles whose bases are 8

an d 12 chains res pectively

172 PLANE GEOME TR K—B OOK I V.


272. CONVERS ELY— Two s imi la r p olygons may be

d ecomposed in to th e same n umber of tr iang les, sim

i la r each to each , a nd sim i la r ly p laced .

Given—AB CDE and FGHKL two similar polygons.

T o Prove— A AB C similar to A FGH, A ACD to A FEK,

and A ADE to A FKL .

Dem— Draw the homologous diagonals AC, AD,FYI

,

and FK .

S ince the polygons are similar,

AB = AG,and

AB B C

FG G'H

Hence A AB C and A FGH are similar.

S ince the polygons are S imilar,

AB CD AGHK . (1)And

,since the triangles AB C and N H are similar

A1 A2. (2)Subtracting (2) from A3 A4.

S ince the polygons are similar,and also AABO is sim

ilat to A FGH,

T henAC CD

FH E X

Dropping the ratio in parentheses,

A ADC and A FEK are similar. 254

In like manner A ADE and A FKL are similar. Q. E . D.

174 PLANE GE OllIE TRY.-B OOK I V.

PROPOS IT ION XXIX. T HEOREM .

275. Two s imi la r p olyg ons a r e to each oth er a s the

squ a res of a ny two homolog ou s s id es .

Given— P and P’the areas of the similar polygons EB

and LG respectively .

P A? B7) ”

F i s an “

Dem— S ince the polygons can be decomposed into thesame number Of similar triangles then

A FGH A FE K A FKL

Dropping the ratios in parentheses, the correspondingtriangles are proportional,

AndA AB O - l A ACD A ABB AAB C

A FGH A FIIK A FKL A FGH—s

‘

1AARC AB B_C

, etc, o. E . n.

P'

A FGH W GH2

T O Prove

T herefore

276. COR . Two similar polygons are to each other as the

squares of any two homologous lines of the polygons.

P ACS ince 1

P FH’

Ax . 1. Q. E . D.

RELAT ION OF LINES IN A CIRCLE . 175

RELA T ION OF L INE S IN A C IRCLE .

PROPOS IT ION XXX. T HEOREM.

277. If two chord s in tersect in a circle, their segm en ts a re recip roca lly p rop or t iona l.

Given—AB and CD any two chords intersecting at 0.

T O Prove AO C0 DO B 0.

Dem— Draw AC and BD.

T hen in the triangles AOG and BOD

AA = AD,and AC=!AB . § 179

Hence the triangles A00 and BOD are similar. 251

T herefore AO CO= DO B 0. 217. Q. E . D.

278. COR — The product of the segments of one chord is

equal to the product of the segments of the other.

From the proportion,

A0 : CO= DO B 0,

AO x BO= 00 X DO.

EXERCIS ES .

234. Prove that five times the square of the hypotenuse of a right tri

an g le are equal to four times the sum of the squares of the medial lines

from its extremities .

235. T he area

o

of a trapezoid is equal to“

the product of one of its legs

in to the distance from this leg to the middle point Of the other leg of

th e trapezoid .


PROPOS IT ION XXXI . T HEOREM.

279. If from a fix ed p oi n t w i thou t a circle a seca n t

a nd a tangen t a r e d raw n,th e ta ngen t is a mea n

.p rop ort iona l betw een the whole secan t and. i ts ex

terna l segmen t .

Given—AR a‘

secant and AC a tangent drawn from A to

the circle CDB .

T o Prove AB : AC= AC : AD.

Dem—Draw CB and CD.

T hen in the triangles AOB and ACB

[A is common,AB = AAOD. 179, 185

Hence the triangles AOB and ACB are similar. 251

T herefore AB AO= AC : AD. 217. Q. E . D.

280. COR — The square of the tangent is equal to theproductof the whole secant and i ts external segment.

From the proposition ,

Z? ” AB x AD.

EXERCIS E .

286 . Required the area of an equilateral triangle whose sides are each

16 feet.


PROPOS IT ION XXXI I I . T HEOREM .

282. I n a ny tr ia ng le the p rod u ct of two s id es is

equa l to the p rod uct of the d iameter of the ci rcum

scr ibed circle by the a lti tud e up on the th ird s ide .

Given— CD the altitude Of the triangle ABC, and CD

the diamete r of the circumscribed circle.

T o Prove AOx B C= CD x CE .

Dem—Draw DB,and ADB C is a right angle.

T hen in the right triangles EAC and BDC

AA AD.

Hence A EAC and A BDC are similar,

And

Therefore AC x B C= CD x CE . 119.

EXERCIS E S .

241. Find the longest line that can be drawn on a floor whose length

is 60 feet and wid th 40 feet .

242. F ind the shortest distance from the lower corner of a room 40

feet long, 80feet wide, and 12 feet h igh ,to the opposite upper corner .

248. F ind th e shortes t distance from the lower corner of a room 60

feet long, 40 feet wide, and 15 feet-high ,

to the Opposite upper corner,

keeping on the surface Of the room.

244. F ind the area Of a triangle whose sides are 8, 10, and 14 feet

respectively .

245. A tree 120feet high was broken Off in a stom i ; the broken end

remained on the stump, and the top struck 25 feet from the base : how

high was the stump ?

RELAT ION OF LINES IN A CIRCLE . 179

PROPOS IT ION XXXIV. T HEOREM.

288. I n any tr ia ngle the p rod u ct of a ny two s id es

is equa l to the p rod u ct of the segmen ts of the th i rd

s id e formed by th e bisector of the opp osi te ang le,

p lu s the squa r e of the bisector .

bisector Of the angle C in the triangle

ACB .

T o Prove CA x CB = AD x BD + CD”.

Dem.

— Circumscribe a circle about AB C, produce CD to

E,and draw AE .

In the two triangles CAE and CDB,

By hypothesis, 4 ACE AB CD,

And AE AB .

Hence the triangles CAE and CDB are similar,

And

Or CA x CB = CE x CD,

(DE OB ) OD,

But DE x CD AD x DB .

Substituting, CA x CB AD x DB CD".

EXERC IS E .

246. T he S ides of a triangle are 12, 18, and 24 in . respectively : find the

radius of the circumscribed circle.

180 PLANE GEOMETRK—BOOK I V.

PROPOSIT ION XXXV. T HEOREM.

284. The p rod uct of the two d iagona ls of a quad

r ila tera l inscr ibed in a circle is equa l to the sum ofthe p rod ucts of i ts opp osi te sid es .

Given—AC and DB the diagonals Of the quadrilateral

AB CD inscribed in the circle ADC.

T o Prove AC x DB = AB x DC+ AD x B C.

Dem— Draw DE ,making AADE ABB C ; then ,

addingto each AEDE

, AADB AEDO.

In the triangles EDO and ADB,ADCA ADBA . 179

And,since AEDO AADB ,

T he triangles EDO and ADB are similar. 251

Hence DC : DB E C : AB,

217

Or DB x E C= AB x DC. (1) 119

Again ,in the triangles ADE and BDC

ADAC= ADB C, 179

And,by construction

,AADE = ABDC.

T hen the triangles ADE and EDO are similar. 251

Hence AD DE = AE : B C, 217

Or DB x AE AD x B C. (2) 119

Adding (1) and we have

DB x (E C+ AE ) AB x DC+ AD x B C,

DB X AC= AB x DC+ AD X B a Q. E . D.


CD

P = CD X lo.

EXERC IS E S .


247. T he area of a rhombus is equal to one half of the product Of its

diagonals (Q

249. If the middle point of one Of the non -

parallel sides Of a trapezoid

is joined w ith the extremities of the opposite side,the triangle formed

is one half Of the trapezoid .

249. T he area of a triangle is equal to one half the product Of its

perimeter by the radius Of the inscribed circle.

250. If the, points Of bisection Of the sides Of a given triangle be

joined , the triangle thus formed will be similar to the given tri angleand equivalent to one fourth of it .

251. T he four lines join ing the middle pointsOf the four sides Of any quadrilateral form a

parallelogram.

262. T he perimeter Of the parallelogram formed by joining themiddlepoints of the four S ides Of any quadrilateral is equal to the sum of the

diagonals of the quadrilateral.

RELAT ION OF LINES IN A CIRCLE 183

258. T he d ifference of the squares Of two sides of any triangle is equal

to the difference Of the squares of the projections Of thes e sides on the

th ird side (Q

254 . If similar polygons are constructed upon the sides Of a right tri

angle, the polygon on the hypotenuse is equal to the sum of the poly

gons on the other two S ides.

255 . T he sum Of the squares Of the four sides

Of any quadr ilateral is equal to the sum Of the

squares Of the diagonals, increased by four

t imes the square Of the line joining the middlepoints Of the d iagonals.

256 . I f one Of the parallel sides Of a trapezoid is double the other,each diagonal will cut off one th ird Of the other.

257. If a line tangent to two circles cuts another line joining theircentres, the seg ments of the latter will be to each other as the diametersof the circles .

258. I f two circles are tangent to each other internally , the chords of

th e greater'

drawn from the poin t Of tangency are divided proportionally

by the circumference Of the Smaller .

259. If D is the middle point Of the hypotenuse

A C of the right triangle AB C, prove that

260. I f the perpendicular from the vertex Of a right angle upon the

h y potenuse d ivides the hypotenuse in to two parts wh ich are to each

o ther as 4 to 1,the longer leg Of the triangle is double the shorter .

261. An angle Of a triangle is acute, righ t, or Obt use according as the

sq uare of the Opposite side is less than ,equal to, or greater than ,

the sum

o f the squares Of the other two sides.

262. I f DEEG is a rectangle inscribed in

th e righ t triangle AB C, prove DE a mean

p roportional between AD and CC.

268. I f two circles are tangent to each other, externally or internally ,

an y two straight lines drawn through the point Of contact and termi

n ated both ways by the circumferences will be divided proportionallyby them.

264. T he altitude Of an equilateral triangle is equal to three times the

radius Of the inscribed circle.

184 PLANE GEOME TRK—DOOK I V.

265. T he sum Of the squares Of the segments of two chords which are

perpendicular to each other is equal to the square Of the diameter of

the circle.

266 . If from the vertical angle of a triangle a

perpendicular be drawn to the base, the prod

uct Of the sum and difference Of the two S ides

is equal to the product Of the sum and differ

ence of the segments Of the base (Q

267. Prove Prop . VIII Art. 235,by using the following diagram.

PROBLEM S OF CONS T RU CT ION .

PROPOS IT ION XXXVI I . PROBLEM.

286. T o d i vid e a stra igh t line in to any number

of equ a l pa r ts .

Given—AB any straight line.

Requ ired— T O divide AB into three equal parts.

Cons— Draw the indefinite straight line AH,making any

angle with AB .

T ake AC any length ,and make CE and EG equal to AC.

Draw GB,then draw EF and CD parallel to GB .

T hen AD DE = FB . 244. Q. E . F.

186'

PLANE GEOME TRY.—BOOK I V.

‘

Cons—T ake AD m,DE n

,and AB p .

Draw DB,then draw E C parallel to DB .

T hen AD : DE = AB : B C,

. § 242

Or m z n = s C. § 113. Q . E . F .

289. COR — I f AB is made equal to n, we shall have

m n n B C.

BC is then a third proportional to m and n.

PROPOS I T ION XL. PROBLEM.

290. To construct a mean p rop ortiona l to two g iven

lines .

Given—m and n any two straight lines.

Requ ired— T O construct amean proportional tomand n .

Cons — T ake AB m,and B C n .

Describe a semi - circumference on AC as a diameter,

and erect BD perpendicular to AC.

T hen BD is a mean proportional to m and n .

Proof— For AB BD BD B C,

265

Or § 112. Q. E . F .

291. DEFINI T ION .

— A straight line is divided into Enz

treme and Mean Ratio whenone segment is a mean pro

portional between the whole line and the other segment .

When the point Of division falls within the line, it is

said to be divided internally ; if the point falls without

the line, it is divided externally.

PROBLEMS OF CONS TR UCT ION. 187

PROPOS IT ION XLI . PROBLEM.

292. To d ivid e a g i ven stra igh t line in to ex treme

and mean ra tio.

Given—AR any S traight line.

Required— T O divide AB into extreme and mean ratio.

Cons— Draw CB perpendicular to AB and equal to one

half Of it.

With C as a centre, and with a radius equal to one

h alf Of AB ,describe the circumference GBD.

Draw AC,and produce it to D.

T ake AF equal to AG,and AH equal to AD.

T hen AB is divided into extreme and mean ratio inter

nally at F,and externally at H .

P roof— For AD AB AB AG,

Or

By division , AD AB AB AB AF : AF. 126

B ut AD AB AG, or AF, and AB AF = B F!

Hence AF : AB = B F : AF.

B y inversion, AB AF = AF : BF.

Again, by composition,

§ 125

But AD AB EH,AH= AD

,and AB AF = AH .

Hence BH AH AH : AB . Q. E. F..

‘

190 PLANE GEOME TRx—Roox I V.

PROPOSIT ION XLIV. PROBLEM.

296. To constr u ct a square equ iva len t to a g iven

para llelog ram .

Given— AO any parallelogram.

M u ired— T O construct a square equivalent to AC.

Cons — Construct E X a mean proportional to AB and

DE . 290

T hen the square LN,described on HK

,will be equal to

the parallelogram AC.

Proof.— For

,by construction

,

LE2

AB x DE .

area LMNP area AB CD.

PROPOS IT ION XLV. PROBLEM.

297. T o constr u ct a squ a re equ i va len t to a g i ven

tr iang le .

Given—AB C any triangle.

Requ ired— T O construct a square equivalent to AB C.


Cons — Construct HG a mean proportional to AB and

lDC.

T hen the squareKM described on HG will be equivalent

to the triangle AB C.

Proof— For, by construction ,

AB : KL KL i CD,

area KLMN area AB C.

PROPOS IT ION XLVI . PROBLEM.

298. T o construct a tr iang le equ i va len t to a g i ven.

p olygon .

Given— AB CDE F any polygon .

Requ ired— T O construct a triangle equivalent toAB CDEF.

Cons — Draw the diagonal BD,draw CH parallel to DB ,

and connect D and H.

T hen area A B CD area A BED. 232,4.

Hence area AB CDEF area AHDEF.

Again ,draw EH

,draw DX parallel to EH,

and connect

E and K.

T hen area A EHD area A EHK. 232,4.

Hence area AHDEF area AKEF.

192 PLANE GEOME TRY.—B OOK IV.

Draw the diagonal AE ,draw FG parallel toAE, and con

nect E and G.

T hen area A AEE area A AEG. 232,4.

Hence area AKEF area GKE .

T herefore area AB CDE F area GKE . Ax . 1. Q. E . F .

299. S CHOLIUM .—By means of 297 and 298a square

can be constructed equivalent to any given polygon .

PROPOS IT ION XLVI I . PROBLEM.

800. On a g iven stra igh t lin e as a base , to constr uct

a recta ng le equ i va len t to a g iven rectangle

OG

I?“

b L q

Given— AB CD a rectangle, and E F any straight line.

Required— T O construct on EF a rectangle equivalen t

to AB CD.

Cons— Find a fourth proportional, LM,to AB

,AD

,and

EF. 288

Construct the rectangle BEGH ,with EF and EH LN )

as its adjacent sides. 205

T hen area AB CD area BEGH.

Proof— For KP : KL PN : LM,

Or

And AB X AD = EF X EH.

Hence area AB CD area BEGH.

194 PLANE GE OME TRx—Roox I V.

PROPOS IT ION XLIX. PROBLEM.

802. T o constru ct a p a ra llelogram equ i va lent to a

g iven squ a re ,ha ving the d ifi

'

erence of i ts base an d

a lt i tu d e equa l to a g i ven lin e.

Given— AB CD any square, and KL any line.

Required— T O construct a rectangle equivalent toAB CD,

and having the difference Of its base and altitude equal

to KL.

Cons— Ou KL as a diameter describe a circumference.

Draw KM perpendicular to KL and equal to AB . Draw

MN through the centre 0.

T hen the parallelogram EFGH , constructed with MN as

a base and MP as its altitude,is equivalent to the square

AB CD.

Proof— For

Hence

Also

Hence area AB CD area EFGH.

EXERCIS ES .

By construction find x

272.

278. If a z ab (Q

274. If x2= 12. lf x = (3,

275. If 7 : x=x : 7— x.

276 . If x3= 3V§ . If x3= 2;/


PROPOSI T ION L. PROBLEM.

808. T o constru ct a squ a r e ha ving a g i ven ra t io to

a g i ven squ a r e .

E"

m ht

Given— P any square and any ratio n m.

Requ ired— T O construct a square which shall have to P

th e ratio Of n m.

Cons — Ou the line E F take EH equal to m and HF

equal to n .

On E F construct the semi - circumference EGF.

Construct GH perpendicular to EF,and draw GE and

GF .

On GE take GK equal to AB,and draw KL parallel

T hen the square Q, constructed with its equal to

GL,will have to P the ratio Of n m.

Proof — For AB GE is a right angle.

And,since GH is perpendicular to EF,

(W n

GE m

But,since KL is parallel to EF,

CL GF

GK GE

Hence 227. Q. E . F.

196 PLANE GEOME TE R—B OOK I V.

PROPOS IT ION LI . PROBLEM.

804. To constr u ct a p olygon up on a g iven lin e

s imi la r to a g iven p olygon .

Given— AB CDE any polygon , and FG any line.

Required—T O construct on FG a polygon similar to

AB CDE .

Cons— Draw the diagonals AC and AD.

Construct AG equal to AB ,and An equal to Am. § 195

T hen the triangles AB C and FGH are similar. § 251

In like manner construct A FHK similar to A ACD,and

A FKL similar to A ADE .

T hen FGHKL will be similar to AB CDE . 271. Q. E . F .

EXERCIS E S .

277. T o construct a Square equ ivalent to the sum Of two given squares

whose S ides are 4 and 5 res pectively (Q

278. T O construct a square equivalent to the d ifference Of two given

squares whose sides are 7 and 3 respectively (Q

279. T o divide a triangle into two equivalent parts by drawing a line

from the vertex to the base (Q232, Cor .

280. T o divide a tri angle into two parts wh ich shall be to each oth er

as 3 to 4 by drawing a line from the vertex to the base .

281. T o construct an isosceles triangle equivalent to a given triangle .

282. T o construct a right triangle equivalent to a given triangle .

288. T O construct a square five times a given square (3

284 . T O construct a square equivalent to a given trapez ium (Q285 . T o divide the base Of a triangle proportional to the other two

sides (Q

PLANE GEOME TRK—B OOK I V.

PROPOS IT ION LII I . PROBLEM.

con stru ct a p olygon s im i la r to a g iven

an d equ iva len t to another p olygon .

Q any two polygons.

construct a polygon similar to P and

equivalent to Q.

Cons— Construct squares equivalent to P and Q

and let m and n denote the sides of the squares equal to

P and Q respectively .

T hen construct a fourth proportional to m,n,and a

,th e

base Of P . Let c be this fourth proportional.

Upon c,homologous to a

,construct the polygon R sim

ilar to P . T hen R will be the requ ired polygon .

Proof.

— For m n a c,

And m’na

a202.

But $m’

,and Q¢ n

’.

T hen

Hence R a’r Q.

HARMONIC PROPOR T ION. 199

PROPOS IT ION LIV . PROBLEM .

808. T o find two stra ig ht li nes wh ich ha ve the

same ra tio as the a rea s of two g i ven p olygons .

Given— P and Q any two polygons.

Requ ired— T O construct two lines which have the same

ratio as the areas Of P and Q.

Cons— Reduce P and Q to triangles and then to

equivalent squares

Let m and n be the sides Of squares equivalent to P and

Q respectively .

Construct the right angle AOB take CB mand

CA n,and draw AB

,and CD perpendicular to it.

T hen area P : area Q DB : AD.

Proof— For

But

Hence area P : area Q DB : AD. Q. E . F .

HARMON IC PROPOR T ION .

809. A line is divided Harmonically when it is dividedinternally and externally in the same ratio.

T hus,if AB is divided inter

nally at C and externally at D,

A 3 B 7)

CA CB = DA DB,

then AB is d ivided harmonically .


S ince this proportion may be written

AC AD B C BD,

that is, since the ratio Of the distances Of A from C and D

is equal to the ratio Of the distances Of B from C and D,

the line CD is divided harmonically at A and B .

T he four points A,B,C,D are called harmonicpoints, and

A and B are called conjugate points, as are also C and D .

PROPOS IT ION LV. PROBLEM.

810. To d ivid e a g iven stra igh t line ha rmon ica lly

in a g iven ra t io.

Given— AB any straight line.

Required—T O divide AB harmonically in the ratio Of

m to n .

Cons— Construct the triangle ABE having the three

given lines AB ,m,and n as sides. § 202

Bisect AAEB by the line EC, and ABEF by the line

ED.193

T hen AB is divided harmonically at C and D in the

ratio Of m to n.

Proof.— For 246

And 248

Hence CA CB DA DB . 131. Q. E . F.

B O O K V .

REGULAR POLYGON S —MEA SUREMENT OF

THE CIRCLE .

812. DEFIN IT ION— A Regular Poly gon is a polygon

which is both equilateral and equiangular.

PROPOSI T ION I . T HEOREM.

818. If the ci rcumfer ence of a circle be d ivid ed

in to any n umber of equ a l p a r ts,

1 . T he chord s j oin ing the su ccess i ve p oin ts of d ivis ion form a regu la r inscr ibed p olygon .

2. T he tangen ts d raw n a t the p oin ts of d i vis ion

form a r egu lar circumscr ibed p olygon .

Given—ACE a circumference divided into any numberOf equal parts .

T O Prove , l st—AB CDEF a regular polygon .

Dem— S ince by hypothesis the arcs are equal,202

REGULAR POLYGONS . 203

T hen chord AB chord B C= chord CD,etc.

,

And are B ODBE = arc CDEFA,etc.

T hen AA= AB = AC,etc.

T herefore the polygon AB CDEF is regular.

T O Prov e, 2d .— GHKLMN a regular polygon.

Dem— In the triangles AGB ,BHC, OKD,

etc.

AB B C= CD,etc. 156

And since arc AB arc B C arc CD,etc.

,

AGAB = AGBA AHBC= AHCB ,etc. 5185

Hence A AGB A BHC A OKD,etc. g56

Whence AG AH AK,etc. , § 45 (a)

And 66

T hen GH HK KL,etc.

T herefore the polygon GHKLMN is regular. § 312. Q. E .D.

814. S CHOLIUM 1.- I t is thus seen that if an inscribed

polygon is given, a circumscribed polygon Of the same

number Of sides can be formed by drawing tangents at

the vertices Of the given polygon . And if a circumscribed

polygon is given ,an inscribed polygon of the same num

her of sides can be formed by joining the consecutive

points Of tangency .

T he following method is generally preferred

815. SCHOLIUM 2.

—1st. IfAB CD

be the inscribed polygon ,

bisect the arcs AB,B C

,CD

,etc. in

the points F , H ,L,etc.

,and draw

E

the tangents EG,GK

,KM

,etc. at

these points.

T hen,since arc EH arcHL

,etc.

,

T he polygon EGKM is regular.

204 PLANE GEOME T RK—B OOK V.

S ince OF is perpendicular to both AB and EG

the sides AB and E G are parallel.

For the same reason all the other sides Of AB CD,etc.

are parallel to the sides Of E GKM,etc. respectively .

2d. If the circumscribed polygon EGKM,etc. is given ,

Draw OE ,OG

,OK

,etc.

,and connect AB

,B C

,CD

,etc.

,to

Obtain the inscribed polygon .

816. S CHOLIUM 3.—If the chords AF

,FB

,BH

,etc. be

drawn,a regular inscribed polygon Of double the number

Of sides will be formed .

If tangents be drawn at A,B,C,etc.

,a regular circum

scribed polygon of double the number Of sides will be

formed .


31 7. A ci rcle may be ci rcumscr ibed abou t, or in

scr ibed in , any r egu la r p olygon .

Giv en—AB CDEF a regular polygon .

T O Prove , lst — T hat a circle can be circumscribed

about AB CDEF .

Den — A circumference can be described through A,B,

and C

Let 0 be the centre Of this circumference,and draw OA,

OB,and 0C.

206 PLANE GEOME TRY.- B OOK V.

822. COR .

— The angle at the centre of a regular polygon is

equal to four right angles divided by the number of sides.

For AAOB ABOC= ACOD,etc.

, 57

And AAOB ABOC, etc. 4 right angles. 19

Hence each angle at the centre equals four right angles

divided by the number Of sides.

PROPOSIT ION I I I . T HEOREM.

828. R egu la r polygons of the same number ofs id es a re simi la r .

Given—AB CDEF and GHKLMN two regular polygons

Of the same number Of sides.

T O Prov e— AB CDEF and GHKLMN similar.

Dem— S ince AB = B O==CD,etc.

,and GH =HK

KL,etc.

,

AB 2 . etc. Ax . 2.

OH HK KL

T herefore the homologous sides Of the polygons are

proportional.

T he sum of the angles Of AB CDEF equals the sum of

the angles of GHKLMN . § 98

Hence etc.

T herefore AB CDEF and GHKLMN are similar. 217

Q. E . D.


824. COR . The perimeters of regular polygons of the samenumber of sides are to each other as the radii of the circum

scribed circles,or as the radii of the inscribed circles and their

areas are to each other as the squares qf these radii .

For these radii are homologous lines Of the similar

polygons. 274, 276

PROPOSIT ION IV. PROBLEM.

825. To inscr ibe a squa re in a g i ven ci rcle.

Given—ABC any circle.

Requ ired— T O inscribe a square in AC.

Cons— Draw the diameters AC and BD perpendicular

to each Other,and draw AB

,B C

,CD

,and DA.

T hen AB CD is the required square.

Proof— For arc AB arc B C arc CD arc DA. 153

‘T hen the chords AB

,B C

, etc. are equal, 155

And the angles AB C, B CD,etc. are right angles. 180

T herefore ABCD is an inscribed square. 5313

826 . COR . The side of the inscribed square is to the rad ius

as the square root of 2 is to I .

For 235

Or AB AoV2. Ax . 2.

Hence AB : AO= V2 1 . 120. Q. E . F .

208 PLANE GEOME TRK—B OOK V.

PROPOS IT ION V. PROBLEM.

827. T o inscr ibe a r egu la r hex agon in a g iven

ci rcle .

Given—ACD any circle.

Required— T o inscribe a regular hexagon in AOB .

Chris — Draw the radius AO.

With A as a centre and with AO as a radius, describe an

arc at B,and draw AB and B 0.

T hen AB is a side of a regular inscribed hexagon .

Proof— For the triangle AOB is equilateral. and [ A03

is one third of two right angles, or

Hence the arc AB is one six th of a circumference,

And the chord AB is a side of a regular inscribed hex

agon . 313

T herefore,to inscribe a regular hexagon in a circle

,

apply the radius six times as a chord . Q. E . F .

828. COR . 1 . T he side of a regular inscribed hexagon is

equal to the radius of the circle.

829. COR . 2.

— By joining the alternate vertices of the reg

ular hexagon an equilateral triangle is inscribed in the circle.

880.- COR . 3. The side of an inscribed equilateral trian

gle is to the rad ius as the square root of 3 is to 1 .

For in the righ t triangle FDC,

W ¢ FC§ —l —W = 3551.

210 PLANE GEOME TE R—B OOK V.

T hen, since [A AAB O, the sum of the angles of the

triangle AOB equals

[ A 4 ZABO+ 21 0+

Or 54 0= and £ 0= or 115 of a circumfer

ence.

T herefore AB is the side of a regular inscribed dec

agon . § 313

Hence to inscribe a regular decagon in a circle divide

the radius in extreme and mean ratio, and apply th e

greater segment as a chord ten times. Q. E . F .

882. COIL— By joining the alternate vertices of the decagon,a regular pentagon will be inscribed in a circle.

PROPOS IT ION VI I . PROBLEM.

838. To inscr ibe a reg u la r p en ted ecagon in a g iven

circle.

Given—AD any circle.

Requ ired— T o inscribe a regular pentedecagon in AD.

Cons— Let AC be the side of a regular inscribed hex

agon ; then arc AC is i Of the circumference.

Let AB be the side of a regular inscribed decagon ; then

arc AB is 315 of the circumference.

Hence arc AO arc AB arc B C,is l 1

16 , or 1

15 of the

circumference,and the chord B C is the side of a regular

inscribed pentedecagon . Q. E . F.


884. COR . I .

— By bisecting the arcs subtended by the sides

of any polygon , another polygon of double the number of sides

can be inscribed in a circle.

T hus, we can inscribe regular polygons

From the square of 8, 16, 32, 64, etc.

From the hexagon of 12,24

,48

,96

,etc.

From the decagon of 20,40

,80

,160

,etc.

From the pentedecagon of 30,60

,120

,etc.

335. Until 1801 it was supposed that these were the only regular poly

gons that could be constructed by elementarygeometry , wh ich employs

the straight line and circle only . B ut GAUS S , an eminent German mathe

matician,in h is Disquisitimies Arithmeticae, Lipsize , 1801 , proved that it is

possible to construct regular polygons of 17 sides,of 257 Sides, and in

general of any number of sides which can be expressed by n

being an integer and 2" 1 a prime number .

PROPOS IT ION VI I I . T HEOREM .

886. The a rea Of a r eg u la r p olygon is equ a l to the

p rod uct of i ts p er imeter by one ha lf of i ts ap othem.

Given—AD a regular polygon , P its perimeter, and r the

apothem.

T o Prove—Area AD P x lr .

Dem—Draw the radn 0A, 0B , etc.

,forming the triangles

AOB,BOC, etc.

,with the common altitude r .

T he area of AOB AB x 154 ,BOO B C x gr , etc.

Adding, area AD (AB B C etc.) x gr,P x i n Q. E . D.



337. If a regu lar p olygon be inscr ibed in , or ci r

cumscr ibed abou t , a ci rcle, and the number of i ts

s id es be ind efin i tely increased ,

1 . T he limi t Of i ts apothem is the rad iu s .

2. T he limi t of i ts p er imeter is the ci rcumfer

ence .

3 . The limi t of i ts a rea i s the a r ea of the circle.

Given— AB and CD sides of two similar inscribed and

circumscribed polygons.

T o Prove , l st — T hat the limit of OF is 0A.

Dem— Draw the radius OE to the point of contact, then

OE is perpendicular to A'R

T hen 0A OF AF. 47

I f the number of sides of the inscribed polygon be

increased indefinitely , the length of each side will be

indefinitely diminished , and'

the limit of AB will be zero,

and the limit of QAB ,or AF

,will also be zero.

Hence the limit of 0A OF is zero.

T herefore the limit of OF is 0A.

T o Prove , 2d .

— T hat the limit of the perimeters of the

two polygons is the circumference of the circle.

Dent — Let p and P denote the perimeters of two similar

inscribed and circumscribed polygons respectively , and C

the circumference.


And lim. (8 s) lim.MP pm 140

But lim. EF = O. 139

Whence lim. (S s) 0,

Or lim. S lim. 8. 142

But the area of the circle is evidently greater than that

of the inscribed polygon, and less than the area of the

circumscribed polygon .

T herefore lim. S K lim. 8. Q. E . D.

PROPOS I T ION X. T HEOREM.

888. The ci rcumferences of ci rcles a re to each other

as their rad i i .

Given— C and C ' the circumferences of two circles, and

R and R ’ their radii.

T o Prove

De m.— Inscribe in the circles regular polygons of the

same number of sides,and denote the perimeters by P

and P’.

en

P’

R’

324

Whence PR’

P’R . 119

If,now

, the number of sides of each polygon be indefi

nitely increased, P will approach C as its limit, and P ’

will approach C’

as its limit


842. COR . 4.- The circ umference 0] a circle is 7cD,

or

27m.

For,from equation we have

3 71, or C=

And,since D 212

,

S ubstituting, C 27tR .

Norm—T he numerical value of r can be obtained onlymately . For practical purposes is used .

PROPOS I T ION XI . T HEOREM .

848. The a rea of a ci rcle is equa l to ha lf the

p roduct of i ts circumference by i ts rad iu s .

Given— S the area,C the circumference, and R the ra

dins of a circle.

T o Prove S = 40 x R .

Dem.

—Circumscribe a regular polygon about the circle,

and denote its area by A and its perimeter by P.

T hen A %P x R . 336

I f,now

,the number of sides of the polygon be indefi

nitely increased, A will approach S as its limit,and P will

approach 0 as its limit

T hen,by the theory of Limits

,

lim. A lim. .}P x R. 140

S 50 x B . (1) 145. Q. E . D.

218 PLANE GEOME TB K—B OOK V.

PROPOS IT ION XI I . PROBLEM.

847. To find the va lu e of the chord of one ha lf an

a re, in terms of the chord of the whole a re, when

the rad iu s of the circle is 1 .

Given—AR the chord of the arc AB ,and AD the chord

of one half the arc AB ,in a circle whose radius is 1.

Required—T o find the value of AD in terms of AB .

Cons—Draw the radius OD perpendicular to AB,then

OD bisects the chord AB and also its arc and

draw AO.

T hen in the triangle AOD,

— 2D0 x CO.

But when the radius is 1,

2 2CO.

In the right triangle AOC,

But, since AO 1,and AO= lAB ,

T hen

Clearing of fractions,

Or 200 1/4 mS ubstituting the value of 200 in

Q. E. D.

MEAS UREMENT OF THE CIRCLE 219

848. COR. 1 .

— 1f we denote AB by S and AD by s, we have

the

General Formula 8 V4 S’.

849. COR . 2.— If the radius of the circle is R

,the general

formula is

1/R(2R V4? i

PROPOSIT ION XI I I . PROBLEM .

850. T o comp ute the numer ica l va lu e of rt , app rox

ima tely .

S ince C= 27m,

When R 1, 7: 5C.

Hence the value of 7: is one half of the circumference

of a circle whose radius is 1.

By beginning with a regular inscribed hexagon the side

of wh ich is equal to the radius of the circle,and finding

the perimeters of the successive inscribed polygons of

double the number of S ides,by using the general for

mula of § 348,

8 V4 S’

,

we can find the approximate value of the circumferenceof a circle whose radius is 1.

Denote one side of a regular inscribed

Hexagon by so,

Dodecagon by 312,

Polygon of 24 sides by etc.

T hen, when R = 1,

s, 1 .

Substituting th is value of s, in the general formula,

81: V2 V I I,

220 PLANE GEOME TRK—BOOK V.

And

Or V8;

And

.

2— x’4

Or

And doubling the number of S ides seven times, which

case the polygon will have 768 sides,

.00818121.

If there are 768 sides in the perimeter, there are 384

sides in the semi- perimeter.

Hence the semi- perimeter is

384 x 8163 384 x .00818121 3 14158464.

We may regard as the approx imate value of

the semicircumference of a circle whose radius is 1,and

therefore the approx imate value of n . Q. E . D.

851. S CHOLIUM .

- Archimedes, who was born 287 B . c.,

was the first to assign an approx imate value to at . By

inscribing and circumscribing regular polygons of the

same number of sides,and then doubling the number

of sides, he proved that its value is between 3} and

or,in decimals

,between and

Clausen and Dase, of Germany , about A .D. 1846,com

pated the value of 7! independently of each other,and

their results agreed to the two hundredth decimal place .

Others have carried the value to over seven hundred deci

mal places.

T he value tq fifteen decimal places is

n


819. T he radius of any inscribed regular polygon is a mean propor

tional between its apothem a nd the rad ius of a similar circumscribed

regular polygon .

820. T he area of the ring included between two concentric circles is

equal to the area of a circle whose diameter is a chord of the outer circle

and a tangent to the inner.

821 . An equilateral polygon circumscribed about a circle is regular if

the number of its S ides is odd .

822. An equiangular polygon inscribed in a circle is regular if the

nu mber of its sides is odd .

828. An equilateral polygon inscribed in a circle is regular.

824. An equiangular polygon circumscribed about a circle is regular .

825. T he area of an inscribed regular dodecagon is equal to three

times the square of the radius. Prove geometrically .

I f R represents the radius, a the apothem,and s the side of a regular

inscribe dpolygon , prove that

826 . In an equilateral tria ngle,

s=R I/B'

, and a n QR .

827. In a square,

3 R 1/2'

,and a iR I/Z

828. In a regular hexagon ,

S = R, and a tin/3.

829. In a regular decagon ,

s tau/5“

and a Int/10 zyl’

.

880. In a regular pentagon ,

s iR l/IO and a = 1RVm881. In a regular octagon,

s R ” , and a Hi t/2 V2.

882. In a regular dodecagon ,

s ==R l/2 V3, and a a sMi l/2 V3 .

888. T he square of the side Of a regular inscribed pentagon ,

the square of the side of a regular inscribed decagon , is equal

square of the radius.

I f R represents the radius of a circle, prove that

884. T he area of an inscribed equilateral triangle is 2127373.

MEAS UREMEN T OF THE CIRCLE . 223

886. T he area of an inscribed Square is

886 . T he area of a regular inscribed hexagon is QR“V3 .

887. T he area of a regular inscribed octagon is 2R’V2.

888. T he area of a regular inscribed dodecagon is 3R’

.

889. T he area of a regular inscribed pentagon is GR’ I/10 2V6.

840. I f a circle be circumscribed about a right triangle, and on each

of its legs as a diameter a semicircle be des cribed exterior to the trian

gle, the sum of the areas of the semicircles exterior to the circumscribed

circle is equal to the area of the triangle.


841. Find the area of a circle whose diameter is 10 inches.

842. Find the area of a circle whose circumference is 60 feet.

848. Find the diameter and circumference of a circle whose area is

sq . ft.

844. T he radius of a circle is 12 in . ; what is the radius of a circle 5;times as large ?

846. T he radn of three circles are 3, 4, and 5 respectively ; what is the

radius of a circle equivalent to their sum

846. What is the rad ius of a circle whose area is one ninth of the

area of a circle whose radius is 12 inches ?

847. What is the area of a sector whose angle is in a circle whose

diameter is 24 feet ?

848. Find the length of a chord in a circle 10 inches in diameter, thatis 4 inches from the centre.

849. Find the area of a square inscribed in a circle whose radius is 12

inches .

860. Find the area of a regular pentagon inscribed in a circle whose

radius is 16 inches .

861. Find the area of a regular hexagon inscribed in a circle whose

radius is 8 inches .

862. F ind the area of a regular octagon inscribed in a circle whose

rad ius is 10 inches .

868. How many degrees in an arc whose length is equal to the length

of the rad ius

864. F ind the radius of a circle equivalent to an equ ilateral tri angle

whose S ide is 20 inches .

866. If a semicircle be des cribed on each of the three sides of a right

triangle, the semicircle on the hypotenuse is equal to the sum of the

semicircles on the other two sides .

SUPPLEMENT T O PLANE GEOMET RY .

MAXIMA AND M INIMA .

852. A Maximum magnitude, among quantities of the

same kind , is that which is the greatest.

858. A M inimum magnitude, among quantities of the

same kind, is that which is the least.

T hus, the max imum straight line inscribed in a circle

is the diameter ; and the minimum straight line that can

be drawn from a point to a straight line is the perpen

dicular.

In general a maximum or a minimum value of a math

ematical expression is the value that is greater or less than

any of the values which immediately precede or follow it .

Hence amathematical expression may have two ormore

maxima or two or more minima values.

T hus,in y x

‘16a:

a1929: 160

,

If a= 1, y= 41 ;

z = y 16,a minimum ;

cc= 3, y 25 ;

ac 4, y 32

,a maximum ;

5) y 25

,

x= 6, y 16,a minimum ;

a: 7, y 41.

But in elementary geometry , which treats only of the

line and the circle, there can be but one maximum or one

minimum value.

854. Isoperimetric figures are those which have equal

perimeters.

224

226 PLANE GE OME TRY.— S UPPLEMENT .

PROPOSIT ION I I . T HEOREM.

857. Of isop er imetr ic tr i a ng les ha ving the same

base, tha t wh ich is isosceles is the max imum .

Given—ABC and ABD two isoperimetric triangles hav

ing the same base, AB ,with AB C isosceles.

T o Prove Area AB C area ABD.

Dem— Draw the altitudes CG and DE.

T hen CG DF ; for if CG DF,

(AD DB ) (AC CB ) .

And if CG DE,then very much more would

(AD DB ) (AC CB ) .

But both of these conclusions are contrary to the hy

pothesis.

Hence CG DE.

Multiply ing both members of (1) by iAB ,

We have (tAB x CG) (tAB x DE ) .

Whence area AB C area ABD. § 231. Q. E . D.

858. COR — Of all isoperimetric triangles, that which is equilateral is themaximum.

For if the max imum triangle is not isosceles when any

side is taken as the base,its area would be increased by

making it isosceles. § 357

T herefore the maximum triangle is equilateral.

MAXIMA AND J IIINI JIIA. 227


859. Of a ll tr iang les formed w i th the same tw o

gi ven s id es , tha t in wh ich these s ides a re p erp en

d icu la r is the max imum .

Given— T he triangles AB C and ABD,with B C

and BD perpendicular to AB .

T o Prove Area ABD area AB C.

Dem— Draw.CE perpendicular to AB .

T hen

Hence

Multiplying both members of (1) by tAB ,

We have (tAB x DB ) (tAB x CE ) .

Whence'

area ABD area AB C. 231. Q.

PROPOS IT ION IV. T HEOREM .

860. Of a ll isop er imetr ic p lan e figu res , the ci rcle

is the max imum .

l st . The max imum figu re must be convex .

228 PLANE GEOME TRY.—S UPPLEMENT .

IfAB CD be a concave figure, and the re- entrant portion ,

AB C, be revolved about the line AC into the position AE C,

the figure AECD will have the same perimeter as AB CD ,

but a greater area.

Hence the maximum figure must be convex.

2d . T he stra ight line w h ich bisects the p er imete r

of a max imum p lane fig u re bisects i ts a rea a lso.

Let AB bisect the perimeter of ACBDA,then will AB

bisect its area also.

For if ACB and ADB are unequal, suppose ACB ADB ;

then if ACB be revolved about the line AB to the position

of AFB,the area of ACBFA would be greater than ACBDA.

But by hypothesis ACEDA is a maximum.

Hence ACB and ADB cannot be unequal.

T herefore the area is bisected .

3 d . The figu re ACBFA is a ci rcle.

Draw . EF perpendicular to AB from any point in the

arc ACB .

T hen from the revolution of ACB about AB as an ax is,

E0 F0,

And AAEE A AEE . 232

Now, the angles AFB and AEB must be right angles, for,

if they are not, the areas of the triangles AEB and AFB

230 PLANE GEOME TRY.—S UPPLEMENT .


862. Of a ll the p olygon s constructed w i th the same

g iven sid es , tha t is the max imum wh ich can be in

scr ibed ia a ci rcle .

Given— P a polygon constructed with the sides a,b,c,d,

and inscribed in a circle,C ; and Q another polygon con

structed with the same sides,but not inscriptible in a

circle.

T o Prove P Q.

Dem— On the sides a,b,c,d of the polygon Q, construct

circular segments equal to the segments on the correspond

ing sides of the polygon P .

T hen the figure S has the same perimeter as the circle C.

Hence area C> area S . 360

S ubtracting the circular segments from both figures, we

have

P Q. Q. E . D .

EXERCIS E S .

866 . If the lineAB is bisected at C, and D is a point on AC'

, prove that

HF + BF = 2XC§ em, and

AD x BD m m .

867. Prove that the sum of the squares on the two segments of a givenline is a min imum when the segments are equal.

868. Prove that. the rectangle of the two segmen ts into wh ich a g iven

line can be d ivided is a maximum when the given line is bisected .

MAXI JIIA AND MINIMA. 231

PROPOS IT ION VI I . T HEOREM.

868. Of a ll isop er imetr ic p olygons ha ving the same

n umber of sid es, the ma x imum p olygon is r eg u la r .

Given— AB CDEF the maximum polygon having the

given perimeter and a given number of sides.

T o Prove— T hat AB CDE F is a regular polygon .

Dem— If two of its sides,as AG and CO

,were unequal,

we could substitute for ACC the isosceles triangleABC,hav

ing the same perimeter as ACC, but a greater area. 357

In like manner the whole polygon could be increased

without changing the length Of its perimeter or the

number Of its sides.

For the polygon constructed with the same number Of

equal sides to be a max imum must be inscriptible in a

circle

T herefore it must be a regular polygon . Q. E . D.

EXERC IS E S .

869. F ind the minimum straight line between two non - intersecting

circumferences .

860. Of all triangles of a given base and area the isosceles triangle has

the greatest vertical angle .

861 . T he sum of the lines drawn from the centre of an equilateral

triangle to the three vert ices is less than the sum of the lines drawn

from any other point to these vertices .

232 PLANE GEOME T RK—S UPPLEMENT .

PROPOS IT ION VI I I . T HEOREM .

364 Of a ll p olygons h aving the same n umber ofs id es and the same a rea , the regu lar p olygon has

the min imum p er imeter .

Given—P a regular and Q any irregular polygon having

the same number of sides and the same area as P.

T o Prove—Perimeter of P perimeter of Q.

Dem— Draw the regular polygon R,having the same

perimeter and the same number of sides as Q.

T hen Q R . 363

And,since the area of P equals the area of Q, P R .

But of two regular polygons having the same number

of sides, the one Of least area has the least perimeter.

T herefore the perimeter of P is less than that of R ,or

less than that of Q. Q. E . D.

PROPOSIT ION IX. T HEOREM .

365 Of two isop er imetr ic regu lar p olygons , tha t

wh ich has the g rea ter number of sid es ha s the

g rea ter area .

SOLID GEOMETRY.

B O OK V I .

PLANE S AND POLYHEDRAL ANGLE S .

86 7. A Plane is a surface such that a straight line joining any two points in it lies wholly in the surface.

868. A plane is determined by given lines or points when

one plane, and only one, can be drawn through them.

869. A straight line is Perpendicular to a Plane when it

is perpendicular to every straight line of the plane passing

through its foot.

870. A straight line is Parallel to a Plane when it cannot

meet the plane though both be indefinitely produced .

871. Two planes are parallel when they cannot meet

though both be produced indefinitely .

LIN E S AND PLANE S .


872. If a stra ight lin e ha s tw o of i ts p oin ts in a

p lane , i t lies wholly in tha t p lane.

LINE S AND PLANES . 235

Given—AB a straight line having two points, C and D,

in the plane MN .

T o Prove—T hat AB lies wholly in the plane MN.

Dem— Connect the two pointsC and D by a straight

line,CD.

T hen CD lies wholly in the plane.

But AB coincides with CD.

Hence AB lies wholly in the plane.


878. A p la ne is d eterm in ed

1 . B y a stra ight lin e an d a p oin t w i thou t the line.

2. B y th ree poin ts n ot i n the same stra igh t line .

3 . B y two stra igh t lin es wh ich in tersect .

A. B y two p a ra llel lines .

Given— C a point without the straight line AB .

T o Prov e— T hat a plane is determined by the line AB

and the point C.

Dent — For if any plane, as MN,is passed through AB ,

it may be revolved about AB as an ax is until it contains

the point C.

Now ,if the plane is turned in either direction about

AB,it will cease to contain the point C.

T he plane is therefore determined by the given straight

line and the point without it. Q. E . D.

236 S OLID GEOME TE R—BOOK VI.

T he remaining parts of the theorem are demonstrated

like the first part. Give the proof.


874. Th e in tersect ion of two p la n es is a

line .

Given— T he planes MN and PQ intersect in the lineAB .

T o Prove— T hat AB is a straight line.

Dem.

—Draw a straight line connecting the points A

and B .

T his line will lie in MN and in PQ. 372

S ince the straight line AB lies in both planes, it mustbe their intersection. Q. E . D.


875. F rom a p oin t w i thou t a p lan e on ly one p er

p end icu la r can be d rawn to the p la ne.

238 SOLID GEOME TRK—B OOK VI .

PROPOS IT ION V. T HEOREM .

876 . F rom a g i ven p oin t in a p lan e on ly on e p er

p end icu lar can be erected to th e p lan e.

Given—AB a perpendicular to MN from the point A.

T o Prove— T hat only one perpendicular can be erected

to the plane MN .

Dem— If possible, let AC be another perpendicular to

the plane MN ,erected at the point A .

S ince AB and AC intersect, a plane may be passed

through them and its intersection AP with the

plane MN is a straight line We shall then have

two perpendiculars, AB and AC,erected from the same

point in a straight line, which is impossible. § 14

Hence only one perpendicular can be erected from the

point A in the plane MN . Q. E . D.


877. If, from a p oin t w ithou t a p lane , a p erp en

d icu la r and Obliqu e lines a re d rawn to the p la ne

1 . T h e p erp end icu la r i s shor ter than any on e ofthe Obliqu e lin es .

2. Obliqu e lin es d rawn'

from the poin t , meet ingthe p la ne a t equ a l d is tances fr om the foot of the

p erp end icu la r , are equ a l.

242 S OLID GEOME TRK— B OOK VI .

T hen,since AB is perpendicular to both MN and PQ, it

is perpendicular to CD and CE . § 369

We Should then have two perpendiculars drawn from

the same point to the same straight line, which is impos

sible. 26

Hence only one plane can be passed through C perpen

dicular to AB . Q. E . D .


888. T hrough a g iven p oin t in a stra ight lin e bu t

one p la ne p erp end icu la r to the line can be p assed .

Given—C a point in the line AB .

T o Prove—That but one plane can be passed through

C perpendicular to AB ..

Dem— Pass the plane MN through C perpendicular

to AB .

If another plane can be passed through C perpendicular

to AB,let PQ be that plane.

Draw CD and CE in the planes MN ,PQ, and DCB .

T hen,since AB is perpendicular to both MN and PQ, it

is perpendicular to CD and CE . 369

We should then have two perpendiculars drawn from

the same point in the same straight line, which is impos

sible. 14

Hence only one plane can be passed through C perpen

dicular to AB . Q. E . D.

LINES AND PLANES . 243

PROPOS IT ION X. T HEOREM .

884. If from the foot Of a p erp en d icu la r to a p la n e

a stra igh t lin e is d rawn a t r igh t a ngles to a ny lin e

Of the p lan e , a nd the p oin t of i n ter sect ion is j oin ed

w i th a ny p oin t Of the p erp en d icu la r ,th is la st line

w i ll be p erp end icu la r to the line of the p lane.

Given— AP perpendicular to the plane MN,PD drawn

perpendicular to any line B C in MN,and D

,the point Of

intersection, joined with any point A in AP .

T o Prove— AD perpendicular to B C.

Dem— T ake DB DC,and draw PB

,PC, AB ,

and AC.

S ince PD is perpendicular to B C, and BD CD,we have

PB PC. 22

T hen in the two triangles APB and APC,since PB PC

and AP is common,we have

AB =- AC. § 55

T herefore the line AD has two points, A and D,each

equally distant from B and C.

Hence AD is perpendicular to B C. 25. Q. E . D.

885. COR — The line B C is perpendicular to the plane of the

For it is perpendicular to AD and PD at their point of

intersection D. 380

244 SOLID GEOME TRK—B OOK VIZ

PROPosrrION XI . T HEOREM.

886. If on e of two p a ra llel lines is p erp en d icu la r

to a p la n e, the other is a lso p erp end icu lar to the

p lane .

Given—AB and CD parallel, and AB perpendicular

T o Prove— CD perpendicular to MN .

Den — Pass a plane through AB and CD 373, cut

ting MN in BD draw AD,and in the plane MN draw ED

perpendicular to BD.

T hen AD is perpendicular to ED, § 384

And ED is perpendicular to the plane of ABDO. 385

S ince AB and CD are parallel, and ABD is a right angle,

then CDB is a right angle. 36

Hence CD is perpendicular to BD and ED at their point

of intersection,and is therefore perpendicular to MN.

380. Q . E . D.

887. COR . 1 .

— Two lines which are perp endicular to the same

plane are parallel to each other .

Given— AB and CD perpen

dicular to the plane MN .

T o Prove— AB and CD parallel.

Dem.—I f AB and CD are

not parallel, draw ED parallel to AB .

T hen ED will be perpendicular to MN .


PROPOS IT ION XI I I . T HEOREM.

890. If a stra ight line and a p lan e a re p ara llel,

the in tersect ion s of the p lan e w i th p la’

nes p a ssed

th rough the line a re p a ra llel to tha t line and to

one another .

Given—AB parallel to the plane MN,and let

and OH be the intersections of MN with planes passed

through AB .

T o Prove—AB parallel to CD,EF

,and CH .

Dem— AB and CD lie in the same plane.

S ince AB cannot meet the plane MN,it cannot meet CD,

a line of the plane MN.

T herefore AB and CD are parallel.

In like manner,EF

,GH

,etc. are parallel to AB .

Also CD,EF

,OH

,etc. are parallel to each other.

§ 388. Q. E . D.

891. COR — If a straight line and a plane are parallel, a

parallel to the line through any point of the plane lies in the

plane.

Given— AB parallel to the

plane MN,and CD a line drawn

through any point C Of the plane

MN .

T o Prove— T hat CD lies in the

plane MN.


Dem— Pass a plane through AB and the point C 373,

intersecting the plane in a parallel to AB which

must coincide with CD,since only one line can be drawn

through the point C parallel to AB . Ax . 7. Q. E . D.

PROPOS IT ION XIV. T HEOREM .

392. Two p la n es p erp end icu la r to the same stra ight

line a re p a ra llel.

Given— T he planes MN and PQ perpendicular to the

line AB .

T o Prove—MN and PQ parallel.

Dem.

— If MN and PQ are not parallel, they will meet if

sufficiently produced .

Let C be a point in their intersection , and we Shall then

have two planes drawn through C perpendicular to AB ,

which is impossible . § 382

T herefore MN and PQ are parallel. 371. Q. E . D.

EXERC IS E S .

866 . How many planes are determined by four points in space , no

three Of which are in the same straight line ?

867. Why is a three- legged stool always stable on the floor wh ile a

four- legged table may not be ?

868. If one of two parallel lines is parallel to a plane, what is true of

the other ? Prove it .

. 869. S tate four condi tions that determine a plane .

248 S OLID GE OME TR Ill—BOOK VI .


898. If two p ara llel p la nes a re cu t by a th ird

p lane, the in tersections a r e pa ra llel.

Given— T he parallel planes MN and PQ cut by the

plane AD in the lines CD and AB respectively .


Dem—AR and CD lie in the plane AD.

And,since the planes MN and PQ cannot meet, however

far they are produced, AB and CD cannot meet if produced

indefinitely .

T herefore AB and CD are parallel. 31. Q. E . D.

894. COR — Parallel lines included between parallel planes

Given—AO and BD parallel lines included between the

parallel planes MN and PQ.

T o Prove AC BD.

Dem— Pass a plane through AC and BD 373,4) inter

secting MN and PQ in the parallel lines CD and AB respec

tively

T hen ABDC is a parallelogram,

And AC BD.


897. COR . 2. Through a given point a plane can be drawn

parallel to a given plane, and but one.

Given — A any point, and MN

any plane.

T o Prove— T hat a plane can be

drawn through A parallel to MN.

Dem.— Draw AB perpendicular

T hen through A one plane, and but one, can be passed

perpendicular to AB let PQ be that plane.

T hen PQ and MN are parallel. 392

T herefore but one plane can be drawn through A par

allel to MN. Q. E . D.

PROPOSIT ION XVI I . T HEOREM.

898. If two a ng les n ot in the same p la ne have

thei r sid es p a ra llel and ex tend ing in the same d i

rect ion ,they a r e equ a l, and thei r p lanes are p a r

a llel.

and MN the planes of the angles BAC and

EDF,and AB and AC parallel to DE and DE respectively .

T o Prove , lst—ABAC AEDF.

Den — T ake AB DE and AC DE,and draw B C

,EF,

AD,BE

,and CF!


S ince AC and DE are equal and parallel, the figure

ADFC is a parallelogram. § 83

Hence AD is equal and parallel to CF . § 79

In like manner,AD is equal and parallel to BE .

T hen BE is equal and parallel to CF . Ax . 1 . 33

Hence BE FO is a parallelogram,and B C= EF. 79

T herefore A ABO A DE E, 57

And ABAC AEDF. 45 at)"

T o Prove , 2d— PQ parallel to MN.

Dem— S ince two lines which intersect determine the

position of a plane, § 373, 3

T hen the position of PQ is determined by AB and AC,

and that of MN is determined by DE and DF .

S ince AB and AC are parallel to DE and DE,respectively,

T hen the planes MN and PQ are parallel. Q. E . D.

EXERCIS E S .

872 If two parallel lines in tersect a plane, they make equal angles

with It .

878. Find the locus of all points at a given distance from a given

plane.

874. From a given poin t without a plane construct a perpendicular to

the plane.

876. From a given point in a plane erect a perpend icular to the

plane.

876 . F ind the locus of all points in space equally distan t from anytwo given points .

877. F ind the locus

a . Of all points equally distant from two parallel planes .

b. Of all points in Space equally distant from the ex tremities of a

given straight line .

c. Of all points in space equally distant from three given points not

in the same straight line.

252 SOLID GEOME TRK—BOOK VI .


899. If two stra igh t lin es a re in tersected by three

p ara llel p lanes, their cor resp ond ing“

segments are

p rop ortiona l.

and DE two lines intersected by the par

allel planes MN,PQ, and BS in the points A, B , C, and

AB DE

B C EF

Dem—Draw AF,and pass a plane through AC and

AF 373, intersecting the planes PQ and MN in BG

and CF.

T hen DC and CF are parallel.

In like manner,GE is parallel to AD.

AB DE242

B C E F

AndAB DE

B C EF

400. COR . 1.—If two lines are cut by any number of parallel

planes, the corresponding segments are proportional.

401. COR . 2.—If any number of lines are cut by three par

allel planes, their corresponding segments are proportional.

Hence

Ax . 1. Q. E . D.

254 S OLID GEOME TRK—B OOK VI .

404. T wo dihedral Angles are equal when they can be

placed so that their faces coincide . Hence it is evident

lst. That two dihedral angles are equal when their plane

angles are eq ual.

2d . T hat the plane angles of equal dihedral angles are equal.

405. T wo dihedral angles are adjacent when they have

a common edge and a common plane between them.

T hus,the dihedral angles MDCQ and QDCE 402) are

adjacent.

406. T wo dihedral angles are vertical when the faces of

one are the extension of the faces of the other.

T hus,the dihedral angles PCDN and MDCQ 402) are

vertical angles.

407. When one plane meets another,making the adja

cent angles equal, the angles thus formed are right dihedral

angles, and the planes are perpendicular to each other.

T hus, if the plane QC 402) meets the plane MN,

making the adjacent angles MDCQ and QDCE equal, they

are right angles, and the planes are perpendicular.

408. T he Projection of a point on a plane is the foot of

the perpendicular drawn from the point to the plane.

T he projection of the point A on the plane MN is B .

T he projection of the line AB C,etc. on the plane PQ is

the line abc, etc. formed by the projection of all the points

of AB C,etc. on the plane.

256 S OLID GEOME TRY._ BOOK VI.

2d. If the angles are incommensurable.

Dem—T ake a unit of measure that is contained an

exact number of times in AEFG. If we apply this unit

to AAB C, there will be a remainder,AABO

,less than the

unit of measure.

Pass a plane through BD and B 0. § 373, 3

T hen , since the angles EFG and CEO are commensurable,we have

ACEDO ACEO

AGFHE AEFG

Now,if the unit of measure is indefinitely diminished,

the remainder AABO will be indefinitely diminished, and

AOBO will approach ACBA as its limit,and ACBDO will

approach ACDDA as its limit.

1' ACBDO

limACEO

lst part.

Hence 1m.

AGFHE AEFG

ACBDO ACBDA‘

AGEHE AGFHE

AABO

AEFG

T hen4 CBDA 4 AB C

Ax . 1 . e e . n.

AGFHE AEFG

410. SCHOLIUM.—S ince dihedral angles have the Same

ratio as their plane angles, the plane angle is taken as the

measure of the dihedral angle. S ee 177.

258 SOLID GEOME TRY.—B OOK VI .

PROPOS IT ION XXI . T HEOREM .

412. If two p lanes a re p erp end icu la r to each

other , a stra ight lin e d rawn in one of th em , p er

p end icu la r to thei r in tersection , is p erp end icu la r

to the other .

Given—T he plane PQ perpendicular to MN, and AB per

pendicular to their intersection B Q.

T o Prove—AB perpendicular to MN.

Dem— Draw B C in MN perpendicular to B Q.

S ince the plane PQ is perpendicular to MN,the plane

angle AB C is a right angle.

T herefore AB,being perpendicular to B Q and B C at B ,

is perpendicular to the plane MN . § 380. Q. E . D.

418. COR . 1.—1f two planes are perp endicular to each other

,

a perpendicular to one Of the planes from any point Of theirintersection will lie in the other plane.

Given— T he plane PQ perpendicular to MN ; at B in

their intersection draw AB perpendicular to MN.

T o Prove—T hat AB lies in PQ.

Dem.

—A line drawn in PQ perpendicular to BQ at B will

be perpendicular to MN .

T hen,if AB does not lie in the plane PQ, we shall have

two perpendiculars to the same plane at the same point,which is impossible Hence AB lies in PQ.

DIHEDRAL ANGLES . 259

414. COR . 2.- If two planes are perpendicular to each other

,

a perpendicular to one of theplanesfrom any point of the other

will lie in the other plane.

Given— T he plane PQ perpendicular to MN,and from

any point, as A of PQ, a perpendicular to MN is drawn .

T o Prove—T hat AB lies in PQ.

Dem—A line drawn from A in PQ perpendicular to B Q

will be perpend icular to MN.

T hen, if AB does not lie in the plane PQ, we shall have

two perpendiculars to the same plane from the same point,which is impossible. Hence AB lies in PQ. 375


415. If two in tersecting p lanes a re each p erp en

d icu la r to a th ird p lane, thei r in tersect ion is a lso

p erp end icu la r to tha t p lane .

Given— T he planes PQ and RS , intersecting in the line

AB, perpendicular to MN .

T o Prove— T hat AB is perpendicular to MN .

Den — For a perpendicular to MN at B will lie

PQ and RS .

I t is therefore their line of intersection.

Hence AB is perpendicular to MN . Q. E . D.

260 S OLID GEOME TRY.—B OOK VI .

PROPOS IT ION XXII I . T HEOREM.

416 . E very p oin t in the p lane wh ich bisects a d i

hed ra l ang le is equa lly d istan t from the faces ofthe angles .

Given— A any point in the plane MR which bisects the

dihedral angle PMQN.

T o Prove—A equally distant from the planes PQ and N Q.

Dem— Draw AB and AC perpendicular to the plan es

NQ and PQ respectively .

Pass a plane through AB and AC 373, making the

intersections BD,AD

,and CD.

T he plane ACB B is perpendicular to the planes NQ

and PQ. § 411

T hen ACDB is perpendicular to MQ. § 415

Hence ADB and ADC are the plane angles of the d i‘

hedral angles PMQR and NMQR . § 403

Whence AADB AADC. 404

Hence rt . A ABD rt . 4 1ACD. § 59

T herefore AB AC, 45 (a)

And A is equally distant from the planes PQ and NO.

Q. E . D.

262 S OLID GEOME TRY.—BOOK VI .


418. The ang le wh ich a stra ight lin e makes w i th

i ts own p roj ection up on a p lane is the least ang le

wh ich i t makes w i th any line of tha t p lane .

Given—AB the projection Of AP on the plane MN,and

AC any other line drawn through A in MN .

T o Prove APAB APAC.

Dem—T ake AC=AB,and draw PB and PC.

T hen,in the triangles PAB and PAC

,PA is common

,

And 377, 1

Hence PAB

.

<PAC. § 62. Q. E . D.

POLYHEDRAL ANGLE S .

419. A Polyhedral Ang le is the Opening between three

or more planes which meet at a common point.

T hus,the angle S

—ABCD,formed by

the planes ASB ,BS C, CSD, and DSA,

meet

ing in the common point S , is a polyhe

dral angle.

S is the vertex.

SA, SB , SC, and SD are the edges.

T he triangles SAB,SAD

,etc. are the

faces.

T he angles ASB ,BS C

,etc. are the face angles.

264 S OLID GE OME TRY.—B OOK VI .

PROPOSIT ION XXVI . T HEOREM.

428. T he sum of a ny two face a ng les of a tri

hed ra l ang le is g rea ter than the th i rd .

T he theorem requires proof only when the angle con

sidered is greater than each of the others.

Given— T he trihedral angle S —AB C,with the face

angle AS C greater than either ASB or BS C.

T o Prove AASB AE S C) AAS C.

Den — In the face AS C draw SD equal to SB,making

the angle CSD equal to the angle CSB ; draw AC through

D,and then draw AB and B C.

In‘

the triangles CSD and CSE,CS is common

,and by

construction SD SB and ACSD ACSB .

Hence

And

In A ACB , (AB B C) (AD DC) .

Subtracting the equals B C and DC,

T hen AB AD.

In the triangles ASB and ASD,AS is common

, SD SB,

and AB AD.

Hence AASB AASB .

266 SOLID OEOME TR Ifi—B OOK VI .

But the sum of all the angles of the triangles whose

vertex is S is equal to the sum of all the angles of the

triangles whose vertex is O. § 48

S ubtracting the angles at the bases from these equal

sums,the sum of the angles at S is less than the sum of

the angles at 0.

But the sum of the angles at 0 equals four right

angles. 19

Hence the sum of the angles at S is less than four right

angles. Q. E . D.

PROPOS IT ION XXVI II . T HEOREM.

425. If two tr ihed ra l a ngles ha ve the three fa ce

ang les Of th e one r esp ecti vely equa l to the th r ee

face a ng les of the other , the cor r esp ond ing d ihe

d ra l a ng les a re equa l .

Aasb,AESC Absc, and AAS C

Aasc.

T o Prove— T he dihedral angle SA the dihedralangle as.

Dem— Lay off the six equal distances SA,SB , S C, sa, sb,

and sc,and draw AB

,B C

,CA

, ab, be, and ca.

T hen A ASB A asb,

POLYHEDRAL ANGLES . 267

And AB ab. 45 (a)In like manner

,B C be

,and AO= ac.

T hen A ABO A abo, 57

And AA= Aa . § 45 (a)

T ake AD ad,and draw DE in the face ASB and DE in

the face ASC, perpendicular to SA ; these lines meet AB

and AC respectively , since the triangles SAB and SAC are

isosceles. Draw EF. On sa take ad AD,and construct

def as DE F was constructed .

In the triangle ADE and ads, AD ad and AA Aa.

Hence A ABB A ade. 60

T hen AE ac,and DE de.

In like manner,

AF = af, and DF = df.

Whence 55

And

Finally , 57

And AEDF Acdf.

T herefore the dihedral angle AS equals the dihedral

angle as. § 404, 1

In a similar manner, the dihedral angles SB and S C can

be proved equal to the dihedral angles SOand sc respec

tively . Q. E . D.

426. SCHOLIUM.— I f the face angles of the trihedral

angles are similarly placed , the trihedral angles may be

applied to each other and will coincide in all their parts,

and w ill therefore be equal

I f the face angles are not similarly placed , the tribe

dral angles will not coincide,but will be symmetrical

Both cases are represented in the figure.

427. COR — If the symmetrical trihedral angles are isosceles,

they are always equal.

268 S OLID GEOME TRK—B OOK VL

EXERCIS E S .


886 . If a plane meets another plane, the sum of the adjacent dihedralangles is equal to two right angles .

887. If two planes intersect each other, the opposite or vertical dihe

dral angles are equal to each other.

888. If a plane intersects two parallel planes ,

a . T he alternate- interior dihedral angles are equal.

b. T he exterior- interior d ihed ral angles are equal.

c. T he sum of the interior dihedral angles on the same side is equal

to two right angles .

889. If two dihedral angles have their faces parallel, they are either

equal or supplementary .

890. If a plane is perpendicular to a line at its middle point, any

point in the plane is equally distant from the extremities of the line,

and any po int out of the plane is unequally distant from the extremities

of the line.

891. If two planes are parallel to a third plane, they are parallel to

each other.

892. If a line is parallel to one plane and perpendicular to another,

thes e two planes are perpendicular .

898. I f two planes wh ich intersect contain two lines parallel to each

other, the intersection of the planes is parallel to the lines .

894. T he three planes wh ich bisect the dihedral angles of a tri hedral

angle meet in the same straight line.

896. T he th ree planes passed through the bisectors of the three face

angles of a trihed ral angle, and perpendicular to thes e faces respectively,

intersect in the same straight line.

896 . T he three planes passed through the edges of a trihedral angle,

perpendicular to the opposite face angles ,intersect in the same straight

line.

397. T he three planes passed through the edges of a trihed ral angle

and the bisectors of the Opposite face angles intersect in the samestraight line .

898. Suppose a polyhed ral angle formed by three, four, five equilateraltriangles . What is the sum of the face angles at the vertex

899. If the edges of one polyhed ral angle are respectively perpend ic

ular to the faces of a second polyhedral angle, then the edges of the

latter are respectively perpend icular to the faces of the former .

270 S OLID GEOME TRK—B OOK VI I .

T HE PR I SM .

488. A Prism is a polyhedron two of whose faces are

equal and parallel polygons, and whose

faces are parallelograms.

T he parallel faces are the bases ; the re

maining faces are the lateral area ; the inter

sections of the faces are the lateral edges ; the

perpendicular distance between the bases is

the altitude.

484. Prisms are named from their bases. A Triang ular

Prism is one whose bases are triangles. A Quadrangular

Prism is one whose bases are quadrilaterals, etc.

485. A Right Prism is one whose lateral

edges are perpendicular to the bases.

An Oblique Prism is one whose lateral

edges are not perpendicular to the bases.

A Regular Prism is a right prism whose

bases are regular polygons. § 312

486. If the prism S —DEF - is cut by a

plane not parallel to the base, that portion

of the prism included between the base and

the plane, as AB C—DEF,is a Truncated

Prism.

487. A Right Section of a prism is the sec

tion made by passing a plane perpendicular to the lateral

edges.

488. A Parallelopiped is a prism whose

bases are parallelograms. All the faces are

therefore parallelograms.

489. A Right Parallelop iped is a parallelo

piped whose lateral edges are perp endicular,

to its bases.

THE PRISM. 271

A Rectangular Parallelopiped is a right

parallelepiped whose bases are rectangles ;

that is, all the faces are rectangles.

440. A Cube is a rectangular parallelo

piped whose edges and faces are all equal.

PROPOS IT ION I . T HEOREM .

441. The sections of a p r ism mad e by p ara llel

p lanes a re equa l p olygons .

Given—AD and EX parallelplanes cutting the prismMN.

T o Prove— AD and FK equal polygons.

Dem.

‘- S ince AD and EX are parallel planes, AB is par

allel to FG,B C to GH

,etc. 393

Hence AB FG,B C CH

,etc.

, § 81

AAB C= AFGH, AB CD AGHK,etc. 398

Hence the polygons AD and FK are both equilateral

and equiangular.

T herefore AD and FK are equal polygons. § 215 . Q. E . D.

442. COR — Any section of a prism made by a plane par

allel to the base is equal to the base.

272 SOLID GEOME TE R—B OOK VII .


443. The la tera l a rea of a p r ism is equ a l to the

p rod uct Of the p er imeter of a r igh t section Of the

p r ism by a la tera l edge.

Given—AD a right section of the prism HP.

T o Prove—T he lateral area E X x (AB B C CD

DE AE ) .

Dem— S ince AD is a right section AB is perpen

dicular to FG.

Hence the area FHKG HK x AB . § 229

In like manner,area HLMK HK x B C,

area LNOM E X x CD,etc.

Adding, the lateral area EX x (AB B C CD DE

AE ) . Q. E . D.

444. COR . The lateral area of a right prismis equal to the

perimeter Of the base‘

multiplied by the altitude.

EXERC IS E S .

400. Find the lateral area of a righ t pentangular prism, each of the

sides of the base being 10 in . and the altitude 12 in .

401 . F ind the entire surface of a right triangular prism the sides of

whose base are 8, 12, and 16 in .,res pectively , and altitude 20 in .

274 S OLID GE OME TRK—B OOK VII .

446. COR . 1.

— T too right pr isms are equal when they have

equal bases and equal altitudes.

For the equal faces may always be S imilarly placed .

447. SCHOLIUM.

— T he demonstration above applies to

two truncated prisms, without change.


448. Any Obli qu e p r ism is equ i va lent to

p r ism whose ba se is a r igh t sect ion of the

p r ism, and whose a lt i tu d e is equa l to a

edge of the Oblique p r ism.

Given— PI a right section of the oblique prism AO.

T o Prov e— T he prism AO equal to the right prism FT,

whose altitude FQ equals the lateral edge AL.

Dean— Produce AL to Q, making FQ AL.

Pass a plane through Q parallel to F1,meeting the edges

produced in B ,S,T,and V, completing the right prism PT .

I t is evident that

T he truncated prism AI the truncated prism LT . § 447

Adding the solid F0 to both members Of this equation ,

T he Oblique prism AO the right prism FT . Q. E . D.

THE PRISM. 275


449. The opp os i te faces of a p ara llelop ip ed a re

equa l and p ara llel.

Given—AG a parallelopiped .

T o Prove— T he faces AF and DG’ equal and parallel.

Dem— AB is equal and parallel to DC. 79

In like manner,AB is equal and parallel to DH.

Hence

and the faces AF and DG are parallel. 398

T herefore the face AF = the face DG’. 90. Q. E . D.

In a similar manner we may prove AH equal and par

allel to BG’.

450. Com—Either face of a parallelop iped may be taken as

the base.

EXERCIS E S .

402. What is the locus of all points equidistant from two consecutive

faces of a polyhedron

403 . In how many ways can a polyhedral angle be formed with equi

angular tnangles ? With squares ?

404. In how many ways can a polyhedral angle be formed with equi

angu lar pentagons ? With equiangular hexagons ?

405. Can a polyhedral angle be formed by equiangular heptagons ?

406 . F ind the entire surface of a regu lar hexagonal prism,each side

of the base being 6 in . and the altitude 8 in .

407. F ind the entire surface of a regular triangular prism,each side

of the base being 4 ft. and the altitude 3 chains.

276 S OLID GEOME TRK—B OOK VII .


451. Any p a ra llelop ip ed is equ i va len t to a rest

a ng‘

u la r p a ra llelop ip ed of the same a lti tud e and

equ iva len t base.

Given—AG an oblique parallelopiped whose base is

AB CD and altitude F0.

T o Prove— AG equivalent to the right parallelopiped

UZ,having the same altitude and an equivalent base.

Dem— Produce the edges AB ,DC

,EF

,and HG.

T ake NP = E F,and pass the planes NI and PM perpen

dicular to NP,forming the right parallelopiped KQ.

T hen parallelopiped AG parallelopiped KQ. § 448

Produce the edges IX,ML

,RN

,and QP .

T ake YX RN, and pass the planes YT and XV per

pendicular to YX,forming the right parallelopiped UZ.

T hen parallelopiped KQ parallelopiped UZ. 448

But UZis a rectangular parallelopiped .

T herefore AG is equivalent to the rectangular parallelo

piped UZ.

Now,base AC¢ base KM 230

,Cor. 4

Also 387, 394

Hence AG and UZhave the same altitude and equiv

alent bases. Q . E . D,

278 S OLID GEOME TRK—BOOK VII .

PROPOS IT ION VII I .

rIBI EOREM.

458. Two rectangu la r p ara llelop ip ed s having equ al

bases are to each other a s their a lti tu d es .

CASE I . When the altitudes are commensurable.

Given—P and Q two rectangular parallelopipeds havingequal bases, and commensurable altitudes AB and CD.

P AB

Q CD

Dem—Let AE be a common unit Of measure Of AB and

CD,and suppose it is contained 4 times in AB and 3 times

in CD.

Then $11; 3. (1)

I f we pass planes through these points Of division par

allel to the bases, the parallelopiped P will be divided into

four equal parts, and the parallelopiped Q into three equal

T o Prove

parts. 442,446

T hen 5 (2)

And from (1) and (2) we have

P ABAx . 1.

Q CD

THE PR ISM. 279

CASE I I . When the altitudes are incommensurable.

G iven—P and Q two rectangular parallelopipeds havingequal bases, and incommensurable altitudes AB and CD.

P AB

Q C’D

Dem— T ake any unit Of measure that is contained an

exact number of times in CD.

S ince AB and CD are incommensurable,the unit Ofmeas

ure will be contained in AB a certain number of times,with

a remainder EB less than the unit of measure.

Pass a plane through E parallel to the base, and designate the parallelopiped whose altitude is AE by P

’.

T henP

,AE

Case I .

Q CD

Now,if we diminish the unit of measure indefinitely ,

the remainder EB will be diminished indefinitely ,

And AE will approach AB as its limit,and P

’will

approach P as its limit.

T o Prove

Hence we have two variables

which are always equal.

And,since the limits of equal variables are equal,

I

lim. lim.£ 3

.

Q on

P AB

Q C'D

T herefore

280 S OLID GE OME TR I'

.—B O0K VII .

454. COR .

— Two rectangular parallelopipeds having twoHimensions the same are to each other as their third dimensions.


455. Two rectangu la r p a ra llelop ip ed s ha ving equ al

a ltitu des a re to each otheras thei r bases.

Given— P and Q two rectangular parallelopipeds ; the

dimensions of P are a,b,e,and those of Q are a

’

,b’

,e.

T o ProveP (b ,

Q a x b

Dem—Construct the rectangular parallelopiped R whosedimensions are a, b

’

,0.

T hen the parallelopipeds P and B have the two dimensions a and c the same.

Whence P b,

R b

And the parallelopipeds R and Q have the two dimeh

sions b’and c the same.

Whence 2Z, 454

Multiplying these equations, and remembering to cancelR from both terms, we have

PQ. E . D.

Q

282 SOLID GEOME TE R—BOOK VII .


458. The volume of a rectangu la r p a ra llelop ip ed

is equ a l to the p rod uct of i ts th ree d imens ions,the

u n i t of volume being the cube whose edg e is the

linea r u n i t.

Given—P a rectangular parallelopiped whose dimen

sions are a,b,c,and U the unit of volume whose dimen

sions are 1,1,1.

T o Prove—Vol. P a x b x c.

Dem.- We have £

a x b x ca x bx c.

U 1 x 1 x 1

S ince U is the unit of volume,

T henP

vol. P .

U

T herefore vol. P a x b x c.

459. COR . l .—S ince a x b is the area of the base,

T hen the volume of a rectangular parallelopiped is equal to

the product of its base and altitude.

460. COR . 2.

— The volume of a cube is equal to the cube ofits edge.

For, if the three dimensions of a parallelopiped are each

equal to a,the solid is a cube whose edge is a.

Hence vol. cube a x a x a a’.

THE PRISM. 283

461. SCHOLIUM 1.— T he volume of a rectangular paral

lelOpiped is,more strictly speaking, equal to a x bx c times

U ; that is, the volume of a parallelop iped is the unit of volumetaken as many times as there are units in the product of itsthree dimensions.

462. SCHOLIUM 2.—When the three

dimensions of the parallelopiped are

exactly divisible by the linear unit,

the proposition is readily proved by

dividing the solid into cubes,each

of which is equal to the unit of

volume.

PROPOS IT ION XI I . T HEOREM.

463. The volume of any p ara llelop ip ed is equal

to the p roduct of i ts base by i ts a lti tud e .

Given—ABCD—E any parallelopiped whose base is

ABCD,and altitude AM.

T o Prove—Vol. ABCD—E AB CD x AM.

Den — Construct the rectangular parallelopiped AB CD—M whose base is AB CD

,and altitude AM.

Now,

AB CD—E 4+ ABCD—M. 5451

But vol. ABCD—M AB CD x AM. 459

Hence vol. AB CD—E = ABCD x AM. Ax . 1 . Q. E . D.

SOLID GEOME TR Id—BOOK VH.

PROPOS IT ION XIII . T HEOREM.

The volume of a tr ia ngu la r p r ism i s equalp roduct of i ts base and a lti tude.

Given— F0 the altitude of the triangular prismABD E .

T o Prove—Vol. ABD—E ABD x F0.

Dem—Complete the parallelopiped ABCD—E .

T hen vol. ABD —E l vol. AB CD—E . 452

But i VO]. AB CD—E lAB C'D X FO. § 463

Hence vol. ABD - E ABD x F0. 80. Q. E . D.


465. The volume of a ny p r ism is equ a l to the

p roduct of i ts base and a lti tude.

Given—ABCDE—F any prism.

286 SOLID GEOME TE R—BOOK VII .

468. Pyramids are named from their bases. A Trian

gular Pyramid is a pyramid whose base is a triangle ; a

Quadrangular Pyramid is one whose base is a quadrilateral ;a Pentangular Pyramid is one whose base is a pentagon, etc.

469. A Regular Pyramid is a pyramid whose base is a

regular polygon 312) and whose ver

tex lies in the perpendicular erected at

the centre of the base.

470. T he lateral edges of a regular

pyramid are equal 377,

Hence the lateral faces are equal isos

celes triangles

471. T he S lant He ight of a regular pyramid is the

perpendicular d istance from the vertex to any side of

the base ; as, SG.

472. A T runcated Py ramid is the portion of a pyramid

included between its base and a plane cutting all the lateral

edges.

473. A F rustum of a Pyrami d isa truncated pyramid whose bases are

parallel.

T he Altitude of a frustum is the per

pendicular distance between its bases ;

as,NP .

T he S lant Height of a frustum of a

regular pyramid is the altitude of anylateral face ; as, OH .

The lateralfaces of a frustum of a regular pyramid are equaltrapezoids.

For the faces Of the pyramid are equal and the

faces of the pyramid cut off are equal ; hence the figures

that remain are equal.

THE PYRAMID. 287


474. The la tera l a rea of a regu lar py ramid is

equa l to the p er imeter of i ts base mu ltip lied by

on e ha lf i ts sla n t heigh t .

Given — SE the slant height of the regular pyramid

S—AB CDE .

T o Prove— Lat. area S —AB C’DE (AB B C CD+DE

+AE ) x lSH.

Dem— S ince the lateral faces are equal isosceles triangles

T hen area SAE AE x lSH,

area SAB = AB x lSH,etc.

Adding these equalities, we have

Lat. area S —AB CDE (AE AB B C CD DE ) x

Q. E . D.

475. COR — The lateral area of the frustum of a regular

pyramid is equal to half the sum qf the perimeters of its basesmultiplied by the slant height of thefrustum.

EXERC IS E .

411. F ind the lateral area Of a regular triangular pyramid whose Slantheigh t is 40 ft . and the sides of the base each 30 It . Of the frustummadeby a plane bisecting the slant height.

288 SOLID GEOME TRY.—BO0K VII .

PROPOSIT ION XVI . T HEOREM.

476. If a py ramid is cu t by a p lane p ara llel to

i ts base,

1st . T he edges and a lt i tude a re d ivid ed p rop or

tiona lly .

2d . The section is a p olygon simi lar to the ba se.

Given— PK a plane parallel to the base of the pyramid

S —ABCDE ,cutting the edges in the points F, G,H,

K,and

L,and the altitude in 0.

SF S G SH SK SL

Dena— S ince the planes FK and AD are parallel,

T hen FG is parallel to AB ,GH to B C, etc.

SA SB S C SD SE SP

SF SG SH SK SL S 0

T o Prove , 2d— AB C’DE similar to FGHKL.

Dem— S ince FG is parallel to AB , GH to B C,etc.

,

T hen AFGH AAB C,

AGHK AB CD,etc. s398

Hence AB CDE and FGHKL are mutually equiangular.

Now, A SAB is similar to A SFG, A SBC to A S GH ,

etc. § 254

Hence

290 SOLID GEOME T RK—B OOK VII:


479. If the a lti tu d e of a tr ia ngu la r py ramid be

d i vid ed in to a ny number of equ a l p a rts , a nd p lanes

be p assed th rough these p oi n ts of d ivision p a r a llel

to the ba se, a nd a ser ies of p r isms be inscr ibed in ,

or ci rcumscri bed abou t , the p y ramid , and the n um

ber of d i vis ions in to wh ich the alt i tu de of th e p y r a

m id is d i vid ed be indefin i tely increased , the l imit

of the volume of the p risms w i ll be the volume ofthe py ramid .

Given—T he pyramid 0—HLM. with the altitude di

vided into equal parts, and the series of prisms A,B , and

C inscribed in,and D

,E,F,and G circumscribed about

,

the py ramid , and the number Of equal parts into which

the altitude is divided being indefinitely increased .

T o Prove— T hat the limit of the sum of the inscribed

or circumscribed prisms will be the volume of the pyra

mid .

Dem.

— Divide the altitude EH into any number of

equal parts, as four, and through these points of division

pass planes parallel to the base, forming sections in the

pyramid .

THE P YRAMID. 291

With these sections as upper bases construct the prismsA,B,C,and denote their sum by s and the lim. 8 by p .

With these sections as lower bases construct the prisms

D,E,F,G,and denote their sum by S and the lim. S by P .

Now, each inscribed prism is equivalent to the circum

scribed prism immediately above it. 466,4

Hence the difference between the sum of the circum

scribed prisms and the sum of the inscribed prisms is the

prism D.

T hen S s= D.

Now,by increasing the number of parts into which the

altitude HE is divided,we can make the volume of D as

small as we please, since we diminish the altitude without

changing the base.

T hen lim. (S s) lim. D.

But lim. (S s) P p .

And lim. D 0.

Hence P p 0.

And P = p .

T hat is, the limits of the sum of the volumes of the

inscribed and circumscribed prisms are equal.

But the volume of the pyramid lies between the in

scribed and circumscribed prisms.

T herefore the volume of the pyramid is the limit of the

inscribed and circumscribed prisms. Q. E . D.

EXERC IS E S .

412. Each face of a triangular pyramid is an equilateral trianglewhose

S ide is 6 . F ind the entire area.

413 . T he altitude of a pyramid is 16 Pa,and its base is a square 15 ft.

on a side. What is the area of a section parallel to the base, whose

distance from the vertex is 10 It ?

292 S OLID GEOME TRK— B OOK VII .


480. Two tr iangu la r py ramids having equ iva lent

bases and equa l a lti tud es a re equ iva len t .

Given— N—HLM and 0— PQR two pyramids with

equal altitudes and equivalent bases.

T o Prove N HLM -

v 0 - PQR .

Den — Place the pyramids so that their bases are in th e

same plane, and let HK be their common altitude.

Divide HK into any number of equal parts, as fou r,and through these points of division pass planes parallel

to the bases , forming sections in each pyramid .

With these sections as upper bases, construct the prisms

A,B,C in N—HLM

,and D

,E,F in 0— PQR .

T he corresponding sections of the two pyramids are

equal and,since the altitudes are equal,

and C= F.

Let v’

D + E + F.

T hen V= V’

Ax . 2.

Now,if the number Of parts into which HK is divided

is indefinitely increased , V will approach N—E LM as i ts

limit,and V

’will approach 0

—PQR as its limit.

T hen lim. V==lim. V’.

294 S OLID GE OME TRY.—B O0K VII .


489. The volume of a ny py ramid is equ a l to one

th ird of the p roduct of i ts base and a lt itude .

Given— 0—AB CDE any pyramid .

T o Prove— Vol. 0—AB CDE 1}base AB CDE x H0.

Dem— T hrough the lateral edge AO and the diagonals

AC and AD 373, 3) pass planes dividing the pyramid

into triangular pyramids.

T hen vol. 0—AB C= 2}base AB C x H0.

vol. O—AGB I base ACD x H0.

vol. 0—ADE = l base ADE x H0.

Adding these equalities, we have

vol. 0—AB CDE = 1}base AB CDE x H0. Q. E . D .

484. COR . 1 .

— Any two pyramids are to each other as the

products of their bases and altitudes.

COR . 2. Two pyramids having equal altitudes are to each

other as their bases.

COR . 3. T wo pyramids having equivalent bases are to

each other as their altitudes.

COR . 4. Two pyramids having equivalent bases and equal

altitudes are equivalent.

S CHOLIUM .

— T he volume of any polyhedron may be

obtained by dividing it into pyramids.

THE P YRAMID. 295


485. T he volume of a fr u stum of a tr iang u la r

p y ram id is equ a l to the sum of i ts bases an d a

mea n p rop or t ion a l between them,mu ltip lied by on e

th i rd of i ts a lt i tu d e .

Given .—AB C—D the frustum of a triangular pyramid.

Denote the area of the lower base by B ,the area of the

upper base by b, and the altitude by K.

T o Prove—Vol. AB C—D (B b VB x b) x M .

122

Dem— Pass a plane through AC and F 373, form

ing the triangular pyramid F—‘

AB C.

Pass a plane through DF and C 373, forming the

triangular pyramid C—DEF.

T here will remain the triangular pyramid F—ACD.

Draw FG parallel to AD, then draw GD and GC.

Now,

vol. F—ACD vol. G—ACD. 480

But the pyramid G—ACD may be regarded as having

the base AGC and its vertex at D .

T hrough FE and G pass a plane 373, forming the

prism AGH DEF ; then A AGH= A DEF, § 433

And GH is parallel to FE ,and consequently to B C. 388

Now,AGH zAGC= AH : AC

, § 232, 3

296 S OLID GEOMEZIRK—B OOK VII .

And AGO : AB C= AG AB . § 232, 3

S ince GH is parallel to B C,

AH : AC= AG zAB .

Hence AGH : AGC= AGC : AB C.

T he vol. F—AB C= B x tK.

vol. C—DEF= b x SK .

vol. F—ACD= vol. D—AG’O= VB X bX SK.

122

T herefore vol. ABC—D= (B + b VB x b) x iK .

Q. E . D.

Non e— It is seen that y B—‘

b‘

x


486. The volume of a fru stum of a ny py ram idis equ a l to the sum of i ts ba ses a n d a mean p ro

p or tiona l between them, mu ltip lied by on e th i r d ofi ts a lti tu d e.

Given— AB CDE —F the frustum of any pyramid

S —AB CDE .

Denote its bases by B and b,and its altitude by K.

T o Prov e— Vol. AB CDE —F = (B b VB x b'

) x ix:

Dem— Construct the triangular pyramid O—LMN,

having the same base and altitude as S —AB CDE .

298 S OLID GEOME TB K—BOOK VII .

Dena— T he prism AB C—DEF is composed Of the prism

GE G—DKL and the pyramids D—KLFE and C—ABE G.

Now,

vol. GHO—DKL DKL x GD. § 464

But GD zl(GD HK LC ) .

Hence vol. GE G—DKL= DKL x 3}(GD+ HK + L0) . (1 )

T he vol. D—KLFE =KLFE x tDO, § 482

i (KE + LF ) x KL x IDO, § 233

l KL x Do x e(KE + LF ) .

But tKL'

x DO area DKL.

Hence vol. D KLFE = DKL x §(KB LF ) .

In like manner,

vol. C—ABHG DKL x MAG BH ) .

Adding and we have

Vol. AB C—DEF = DKL x e[(GD HK LC) (KELF ) (AG EH )] DKL x MAD BE CF ) .

Q. E . D.

S IM ILAR POLYHEDRON S .

488. S imilar Poly hedrons are polyhedrons having the

same number of faces,similar each to each and similarly

placed , and their homologous polyhedral angles equal.

S IMILAR POLYHEDRONS . 299

T he faces,lines, and angles of similar polyhedrons

which are similarly placed are called homologous faces,lines

,and angles.


489. T he homologou s edges of simi la r p olyhed rons

a re p rop ort iona l.

Given— S —ABC and 0—DEF two similar polyhedrons.

SA SB AB

OD OE DE’

Dem— S ince the polyhedrons are similar, 0—DE F will

take the position of S —MNP,and their corresponding faces

T o Prove

will be similar. 488

Hence 217

OrSA S B AB

Q. E . D.

OD OE DE’

490. COR . 1.—Any two homologous faces of two

polyhedrons areproportional to the squares of any two

gous edges.

S ince the corresponding faces are similar,

SAB ST ; j SB C 25W SACT hen

ODE OE ’ OEF OF OED

300 S OLID GEOME TRK—BOOK VII .

491. COR . 2.— The entire surfaces of two similar polyhedrons

areproportional to the squares of any two homologous edges.

SAB SB C SA0 AB C490

ODE OEF OED DEFS ince

SAB SB C+ SAC+ ABC SAB sT s’

ODE OEF OFD DEF ODE OE ‘

surf. S—AB C : surf. 0—DEF = SF :W .


492. If a tetrahed ron is cu t by a p lane pa ra llel toone of i ts faces, the tetrahed ron cu t is s im i lar

to the first .

Given—DEF 0 a tetrahedron cut off by the planeDEF

parallel to ABC.

T o Prove—DEF—O similar to AB C 0.

Dem— T he edges AO, B 0, and CO are divided propor

tionally by the plane DEF. § 476

T hen DEO is S imilar to ABO, EFO to BOO, and FDO to

ACO. 254

T he homologous trihedral angles are equal, since theyare formed by equal face angles. § 425

T herefore DE F—0 is similar to AB C—0. § 488. Q . E ; D.

302 S OLID GEOME TE R—B OOK VI II


494. Two s imi la r p olyhed rons may be d ecomp osed

in to the same number of tetrahed rons simi la r each

to each and simi la r ly p laced .

Given—ABC— F and GHK 0 similar polyhedrons.

T o Prove—T hat they may be decomposed into the

same number of tetrahedrons similar each to each and

similarly placed .

Dem—Draw the diagonals FA,FC

,DC

,0G

,0K

,and NK.

T hen the polyhedrons will be decomposed into the

same number of tetrahedrons,similarly placed .

Let AB C—F and GE K —0 be two homologous tetra

hedrons.

T he triangles ABF and B CF are slinilar to GHO and

HKO respectively . § 272

S ince the polyhedrons are similar,the dihedral angle

FB is equal to the dihedral angle H0. § 488

T herefore AB C F is similar to GHK 0. 493

In like manner, we may prove any two homologous

tetrahedrons similar.

T herefore the polyhedrons are decomposed into the

same number of tetrahedrons similar each to each and

similarly placed . Q. E . D.

S IMILAR POLYHEDRONS . 303

PROPOS I T ION XXVI I I . T HEOREM.

495. Two polyhed rons comp osed of the same num

ber of tetrahed rons sim i la r each to each a nd s imi

la r ly p laced ar e s imi lar .

Given— ABC—F and GEK—0 two polyhedrons com

posed Of the same number of tetrahedrons similar each

to each and similarly placed .

T o Prove—AB C—F similar to GHK—0.

Dem.

— S ince the tetrahedrons are similar, their faces are

similar triangles.

Hence AB FD is similar to GHON,ACED to GKMN

, and

B CEF to HKMO. 271

S ince the tetrahedrons are similar, their trihedral angles

are equal.

Hence the polyhedral angle A is equal to the polyhedral

angle G,since each angle is composed of two equal tri

hedral angles.

In like manner any other two homologous polyhedral

angles may be proved equal.

T herefore AB C—F and GHK— 0 have equal polyhedral

angles similarly placed and the same number of faces

S imilar each to each ; hence they are similar.

§ 488. Q. E . D.

304 S OLID GEOME TRY.—B O0K VII .

PROPOS IT ION XXIX . T HEOREM.

496. Two simi la r tetrahed rons a re to each other

as the cubes of any two homologous edges .

Given—ABC—D and EFG—H two similar tetrahedrons,

and V and V’

their volumes.

T o Prove

Dent —Now, 3;AB C x DO

S ince the tetrahedrons are similar,

T henAB C E

,

And

Ac DO A?Multi 1 in these equalitiesP y g

ET “

V 34?Hence

And since any two homologous edges are proportional

to any other two,the two similar tetrahedrons are to each

other as the cubes of any two homologous edges. Q. E . D.

S CHOLIUM.

—Any two similar polyhedrons are to each

other as the cubes of their homologous edges.

For any two similar polyhedrons may be decomposedinto the same number of tetrahedrons similar each to

306 S OLID GEOME TRK—B OOK VII .

2. With squares only one is possible.

S ince each angle of a square is we may form a

polyhedral angle with

3 squares, for 3 x 90° forming a hexahedron.

Hatched ron or wbc

3. With regular pentagons only one is possible.

S ince each angle of a regular pentagon is 108°

101)we may form a polyhedral angle with

3 pentagons, for 3 x 108°

forming a dodecahedron.

S ince each angle of a regular hexagon is 120°

no polyhedral angle can be formed with three or more

hexagons.

In like manner,no polyhedra l angle can be formed with

three or more regular polygons of more than S ix sides ;hence these are the only regular polyhedrons.

T herefore only five regular polyhedrons are possible.

Q. E . D.

REGULAR POLYHEDRONS ’. 307

499. S CHOLIUM .

— T he five regular polyhedrons may be

constructed from cardboard as follows

Draw on cardboard the following diagrams.

Cut the figures out entire,and at the lines separating

adjacent polygons cut the cardboard half through .

By folding these figures the regular polyhedrons can

readily be formed .

Hexahed ron Octahedron

Icosahedron

EXERC IS E S .


Q 4. F ind the lateral area Of a right prism whose base is a regular

hexagon 8 inches on a side,and whose altitude is 6 feet.

415. Find the entire area of a right prism whose base is a rectangle 6

inches by 8 inches, and whose altitude is 8 feet.

416 . F ind the lateral area of a right pyramid whose base is a regularpentagon 10 inches on a side

,and whose Slant height is 5 yards .

417. Find the entire area of a righ t pyramid whose base is a regular

hexagon 8 inches on a side,and whose slant height is 30 inches.

308 S OLID GE OME TRY.—B O0K VI I .

419. Find the lateral area Of the frustum of a right pyramid whoseupper base is a regular heptagon 5 inches on a side

,the lower base 6

inches , and whose slant height is 18 inches .

419. F ind the entire area of the frustum Of a right pyramid whose

bases are squares, the upper base 6 ft .,the lower base 10 ft. , and sla nt

heigh t 5 yards .

420. F ind the volume of a right hexagonal prism each side of whose

base is 8 inches,and whose altitude is 20 inches .

421. F ind the volume Of the frustum Of a square pyramid whose alti

tude is 24 inches, the S ides Of the upper base 6 inches , and of the lower

base 8 inches .

422. T here are two similar triangular pyramids whose homologous

sides are 30 and 40 inches respectively ; required the relation of thei r

volumes. What is the relation Of their areas‘

I‘

423. F ind the edge Of a cubical bin that will contain 1200 bushels of

grain .

424. T he lateral edge of a regular hexagonal pyramid is 10 ft . , and the

sides of its base 6 ft. ; find the lateral area,the entire area, and the

volume.

425. Find the lateral area and volume Of the frustum of a square

pyramid whose altitude is 18 ft. , the S ides of the lower base 6 ft. , and

of the upper base 4 ft.

426 . F ind the volume Of a truncated prism whose lateral edges are

12, 16 , and 20 ft . respectively , and the sides Of a right section are 6,8,

and 10 ft. respectively .

427. How far from the vertex must a square pyramid whose altitude

is 12 ft. and base 6 ft . on a side he cut by a plane parallel to the base

to divide it into two equ ivalent parts ?

428. Find the edge of a cube whose contents are numerically 10 times

its lateral surface .

429. T wo bins Of similar form contain 360 and 1215 bushels of wheat

respectively . If the first bin is 4 ft. long, what is the length of the

second

430. T he lateral edge of a frustum of a regular hexagonal pyramid is

12,and the sides of its bases are 12 and 6 , respectively ; find its lateral

area and volume.

431 . A square pyramid 40 ft. high ,whose base is 12 ft. on a S ide, is cut

by a plane parallel to the base 20 ft . from the vertex ; required the area

of the upper section .

B O OK V I I I .

THE CYLINDER , CONE , AND S PHERE .

T HE CYLINDER .

500. A Cy lindrical S urface is a curved surface gener

ated by a moving straight line which constantly touches a

given curve and in all positions is parallel to a given

fixed straight line not in the plane of the

curve.

T hus,if the line GH moves so that it

constantly touches the curve DHE and in

every position is parallel to its original

position , the surface generated is a cylin

drical surface.

T he moving line GH is the generatrix.

T he curve DHE which it touches is the directrix.

Any straight line, as CD or GB,is an element of the surface.

In this general definition of a cylindrical surface the

directrix may be any curve. T he number of kinds of

cylindrical surfaces is therefore unlimited . Hereafter we

shall assume it a circle,as this is the only curve whose

properties are discussed in elementary geometry .

501. A Cy linder is a solid bounded by a cylindrical

surface and two parallel planes.

T he parallel planes are the bases.

T he cylindrical surface is the lateral area .

T he perpendicular distance between the bases is the

altitude.

310

THE CYLINDER . 311

COR . It follows from this definition that The elementsof a cylinder are equal and parallel. § 394

502. A Right Cylinder is a

cylinder whose elements are

perpendicular to its bases ; as A.

An Oblique Cylinder is one

whose elements are oblique to

its bases ; as B .

A Circular cylinder is one whose bases are equal parallelcircles.

T he Axis of a circular cylinder is the line joining thecentres of its bases ; as

NOT E —A right circular cylinder is called a cylinder of revolution,since

it may be generated by the revolution of a rectangle about one of its

sides as an axis.

503. S imilar Cylinders are cylinders whose axes are pro

portional to the radii Of their bases.

504. A prism is inscr ibed in a cylinder when its bases are

inscribed in the bases of the cylinder. T he cylinder then

circumscribes the prism.

A cylinder is inscribed in a pr ismwhen the bases of thecylinder are inscribed in the bases of the prism. T he

prism then circumscribes the cylinder.

505. A Section of a cylinder is a plane figure formed by.

cutting the cylinder with a plane in any direction.

A Right Section is a section perpendicular to the axis.

An Axial Section is a section formed by a plane passing

through the axis.

A plane is tangent to a cylinder when it passes through

an element of the curved surface without cutting the sur

face. T he element through which it passes is called the

element of contact.

312 SOLID GEOME TE R—B OOK VIII .


506. A section of a cy lind er ma de by a p lane p ass

ing through an elemen t is a p a ra llelog r am.

Given—AB CD a section of the cylinder AC made by a

plane passed through the element AD.


Dem— Pass a plane through AD,and it will intersect

the circumference in B .

Draw B C parallel to AD. B C will then be an element

Of the surface and will also lie in the plane

AC. 31

Now,since B C lies in both surfaces

,it must be their in

tersection .

T he line DC is parallel to AB . 393

T herefore AB CD is a parallelogram. § 74. Q. E . D.

507. COR . 1.

— Every section of a right cylinder made by a

plane perpendicular to its base is a rectangle.

COR . 2. The greatest section Of a cylinder made by a plane

passing through an element is the axial section.

COR . 3. The fig ure formed by joining the corresponding

extremities of two elementswi th straight lines is a parallelogram.

COR . 4. The elements of a cylinder are equal.

314 S OLID GEOME TR Y.—B OOK VI II .


509. The la tera l a rea of a circu la r cy lind er is

equa l to the p er imeter of a r ight section , mu ltip lied

by an element .

Given— S the lateral area, C the perimeter of a right

section , and E an element,of the circular cylinder AB .

T o Prove S = C X E .

Dem—Inscribe in the cylinder 9. prism whose base is a

regular polygon .

Denote the lateral area by S’

,and the perimeter of a

right section by P .

T hen S’

P x E . 5443

I f the number Of faces of the prism be indefinitely in

creased,

T hen S’will approach S as itS limit

,and P x E will

approach C x E as its limit.

We Shall then have two equal variables.

Hence lim. S’= lim. P x E . 5140

Or S = C x E . Q . E . D.

510. COR . 1.

— 7he lateral area qf a right cylinder is equal

to the circumference Of the basemultiplied by the altitude.

T his may be formulated thus, S = 27IR x H in

which R is the radius of the base and H the altitude.

THE CYLINDER . 315

511. COR. 2. The total area Of a cylinder is

T = 27tR(H + B ) ,in which T denotes the total area

,B the radius of the base,

and H the altitude.

For S = 27IR x H and the bases 27tR’. § 344

Adding, T 27tR(H B ) .

PROPOSIT ION IV. T HEOREM.

512. The volume of a ci r cu la r cy linder is equal to

the p roduct of i ts ba se an d a lt i tud e.

Given— V the volume, B the area of the base,and H the

altitude of the circular cylinder AD.

V= B X H.

Dem— Inscribe in the cylinder a prismwhose base is a

regular polygon.

Denote its volume by V’

,and the area Of the base by B

’

.

T hen V’

B’

x H . § 465

I f the number of faces of the prism be indefinitelyincreased ,

T hen I" will approach V as its limit, and B’x H will

approach B x H as its limit.

We shall then have two equal variables.

Hence lim. V’

lim. B’

x H.

T herefore V= B x H.

316 S OLID GEOME TRY.—BOOK VI II .

513. COR— The volume of a circular cylinder is

V aR'x H,

in which V denotes the volume, R the radius of the base,

and H the altitude.


514. The la tera l a reas , and a lso the en ti re a reas , oftwo simi la r r igh t ci rcu la r cy lind ers a re to each

other as the squa res of the i r a lt i tu des , or as the

squa res of the rad i i of thei r bases , and thei r volumes

a re to each other as the cubes of their a lti tud es or

of their rad i i .

Given—S and s the lateral areas, T and t the total areas,

V and v the volumes, H and h the altitudes,and R and r

the radii of the bases, of any two similar right circular

cylinders.

T o Prov eS T

8 t

V R’

(Ian

v 75

Den —S ince the cylinders similar

lowing proportions

503,125

318 SOLID GEOME TRK—BOOK VI II.

T he conical surface of a cone is the lateral area .

T he perpendicular distance SO from the vertex to the

base of a cone is the altitude.

A Circular Cone is a cone whose base is a circle. T he

straight line drawn from the vertex of a circular cone to

the centre of the base is the axis of the cone.

517. A Right Circular Cone

is a circular cone whose axis

is perpendicular to its base ;

as A.

AnOblique CircularGone

is a circular cone whose axis

is Oblique to its base ; as B .

NOT E—A right circular cone is called a cone of revolut ion,since it may

be generated by the revolution of a right triangle about one of its legs as

its axis.

518. A T runcated Cone is the portion

of a cone included between its base and a

plane cutting its convex surface ; as ABDC.

519. A F rustum of 8.Gone is a truncated

cone whose bases are parallel ; as GHFE .

T he Altitude of a frustum of a cone is the

perpendicular distance between its bases ;

T he Slant Height of a frustum Of a right

circular cone is the portion of an element included between

its bases ; as E C .

520. Similar Cones of Revolution are circular cones whose

axes are proportional to the radii of their bases. T heymay be generated by S imilar right triangles.

THE CONE . 319

521 . A pyramid is inscribed in a cone when its base is

inscribed in the base of the cone and its vertex coincides

w ith the vertex Of the cone. T he cone then circumscri bes

the pyramid .

522. A pyramid is circumscribed about a cone when its

base is circumscribed about the base of the cone and its

vertex coincides with the vertex of the cone. T he cone is

then inscribed in the pyramid.


523. E very section of a cone mad e by a p lane pass

ing th rough i ts ver tex is a tr iang le .

Giv en— CDE a section Of a cone made by a plane pass

ing through the vertex C.

T o Prove— CDE a triangle.

Dem— Pass a plane through C,D,and E .

Draw the straight lines CD and CE ; they are elements

of the convex surface, § 515

And CD and CE lie in the plane CDE . § 372

Hence CD and CE lie in the convex surface and also in

the plane, and are therefore their intersections.

Also DE is a straight line. 374

T herefore CDE is a triangle. § 41. Q. E . D.

320 SOLID GEOME TE R—B OOK VII I.

PROPOS IT ION VI I . T HEOREM .

524. E very sect ion of a ci rcu la r cone made by a

p lane pa ra llel to the base is a ci rcle .

Given— BF a section of the circular cone S —ADB, par

allel to the base ADB .

T o Prov e— B CF a circle.

Dem— Pass planes through SP and any elements

SD,etc cutting the base in PA , PD,

etc.,and the parallel

section in OE,00

,etc.

T hen OE is parallel to PA,and 00 to PD. 393

Hence A SPA is similar to A SOE ,and A SPD to

257

Whence E 0 CO§ 217

AP DP

But AP = DP ; hence E 0 C0. 149,8

Hence all points in the curve E CF are equally distant

from 0.

T herefore E CF is a circle. Q. E . D.

EXE RCIS E S .

448. Sections of a cone parallel to the base are proportional to the

square of the distances from the vertex .

449. In cones w ith equal bases and equal altitudes , sections parallel to

the bases and at equal distances from them are equal.

322 SOLID GEOME TRK—B OOK VI II .

527. COR . 2.— I7ie total area Of a cone of revolution is

T = nR(L R) ,

in which T denotes the total area,B the radius of the base

,

and L the slant height.

For S aR x L.

T he base nR’.

Adding, T = nR(L R) .


528. The volume of a ci rcu la r con e is equ a l to one

th i rd of the p rod uct of i ts base and i ts a lt i tud e.

Given—V the volume, B the area Of the base, and H

the altitude of the circular cone.

T o Prove V= tB x H .

Dem— Inscribe in the cone a pyramid whose base is a

regular polygon .

Denote its volume by V’

,and the area of its base by B

’.

T hen V’tB

’x H. § 483

If the number of the faces of the pyramid be indefi

nitely increased ,

T hen V’will approach Vas its limit, and B

’willapproach

B as its limit .

THE CONE . 323

We Shall then have two variables which always

equal.

Hence lim. V’

lim. tB’x H ,

Or V= lB x H .

529. COR . The volume of a circular cone is

V= lnR2x H

,

in which R denotes the radius of the base.

PROPOS IT ION X . T HEOREM.

530. The la tera l a reas , a nd a lso the en t i re a reas, oftwo s imi la r cones of revolu tion a re to each other as

the squ a res of thei r sla n t heigh ts , or as the squa res

of th ei r a lti tud es , or as the squ a res of the ra d i i

of thei r bases ; and thei r volumes a re to each Otheras th e cubes of thei r sla n t heigh ts , or as the cu bes

of the i r a tt i tudes, or as the cubes of the rad i i ofthei r bases .

Given— S and s the lateral areas, T and t the total areas,

Vand v the volumes, L and l the slant heights, H and h

the altitudes, and R and r the radii of the bases of two

similar cones of revolution .

S T L’

H

s t l’ h’

V L’ H 3R3

v l“‘ h3 r’i

T o Prove

324 SOLID GEOME TRY.- BOOK VIII

Dem—S ince the cones are similar,

L R H L + R25

l r h l+ r

T hen

S nRL RX

II2

R2

s 7trl r h2

r“

T

t 7tr(r l) r r l P

V lnR’H R2

H L3 H

3R‘

2 .

v inr’h e h r r

§ 5 9 Q E D

PROPOS IT ION XI . T HEOREM .

531 . Th e la tera l a rea of a fr u stum of a con e ofr evolu tion i s equ a l to on e ha lf of the sum of the

ci rcumferences of i ts ba ses mu lt ip lied by i ts slant

heigh t .

Given—S the lateral area, C and c the circumferences of

326 SOLID GEOME TRK—B OOK VI II .

PROPOS IT ION XI I . T HEOREM .

534. Th e volume of a fru stum of a ci rcu la r con e

is equ a l to the sum of i ts ba ses a nd a mea n p rop or

t iona l between i ts ba ses,mu lt ip li ed by on e thi rd i ts

a lt i tu d e .

Giv en V the volume,B and b the areas of the bases

and H the altitude of a frustum of a right circular

cone .

T o Prov e— I' = (B + b V B x b) x lH .

De m. —Inscribe in the frustum the frustum Of a pyram id

whose base is a regular polygon .

Denote its volume by I"

,and the areas Of its bases by

B’

and b'

.

T hen I’ = (B

’

b’

VET —67x ) x l II . § 4se

I f the number Of the faces Of the frustum Of the py ra

mid be indefinitely increased ,

T hen V’will approach V aS its limit

,and B

’

and b’

w ill

approach B and b,respectively , as their limits.

\Ve shall then have two variables which are always

equaL

Hence lim. I lim. (B’

b'

VI); x b’

) x tII , § 140

Or I' = (B + b+ VB x III . Q . E . D.

THE CONE . 327

535 . T he volume of a frustum of a circular cone is

V= h as“

r”

Br)H ,

in which R and r are the radii of the bases.

For B and b nr’.

T hen

S ubstituting in the formula of § 534,

V= 571022

r’

Rr)H . Q. E . D.

T HE S PHE RE .

536. A S phere is a solid bounded by a surface all

points Of which are equally dis

tant from a point within , called

the centre.

A sphere mav be generated by

the revolution of a semicircle

AB C about its diameter AC as

an ax is.

A radius of a sphere is a straight

line drawn from the centre to the

surface ; as 0B .

A diameter is a straight line drawn through the centre,

having its extremities in the surface .

From this defin ition it is ev ident that in the same

sphere or equal spheres the rad ii and diameters are equal.

53 7. A great circle is a section made by a plane passing

through the centre ; as B GD .

A sma ll circle is a section made by a plane which does

not pass through the centre as EHF .

T he semicircumference B GD is equal to the semi- cir

cumference AB C,since the rad ii 0B and 0A are equal,

and is thus one half of the circumference of a great

circle .

328 S OLID GEOME TRK—B OOK VI II .

538. A Spherical Segment is a portion Of the sphere

included between two parallel

planes ; as A,B,and C.

T he sections of the sphere

made by the two parallel planes

are the bases of the segment .

T he perpendicular distance be

tween the planes is the altitude of

the segment .

A spherical segment of one base

is a Spherical segment formed

when one of the parallel planes is tangent to the Sphere.

539. T he surface of a spherical segment is a zone.

T he bases of the zone are the circumferences Of the circles

which bound it.

T he altitude of the zone is the altitude of the correspond

ing segment.

A zone of one base is a zone formed when one of the

parallel planes is tangent to the sphere.

540. A Spherical Sector is a volume generated by the

revolution of a circular sector

about the d iameter ; as O—AB CD.

A spherical sector, in general,

is bounded by three curved sur

faces ; namely , the two conical

surfaces generated by the radii

0A and 00,and the zone gener

ated by the arc AC. T he zone is

called the base of the spherical

sector.

00 may coincide with OH,in which case the spherical

sector is bounded by one conical surface and a zone of one

base ; as O—EGF .

330 S OLID GEOME TRK— B OOK VIII .

542. COR . 1 .— If AB meets MN,

the surface generated is a

conical surface whose altitude is CD and the radius of whosebase is BD.

543. COR . 2.

—If AB is parallel to MN ,the surface generated

18 a cylindrical surface whose radius is BDand altitude CD.


544. The a rea of the su rfa ce of a sp her e i s equ a l

to i ts d iameter mu lt ip li ed by the ci rcumference ofa g rea t ci rcle .

Given— 0 the centre of the semicircumference

with the radius 0A equal to B .

T o Prove S ur. sphere AE x 27m.

Dem— Inscribe in the semicircle a regular semi - poly

gon ,and revolve both Of them about the diameter AE .

Draw OH, OK,0L

,and GM perpendicular to the chords

AB,B C

,CD

,and DE ; these perpendiculars are equal.

and bisect the chords

T hen area AB AG x 27t0H . § 541

area B C= G0 x 27toH.

area CD= OF x 271011 .

area DE FE x 27t0H.

Adding these equations,

area ABCDE AE x 27tOH.

THE S PHERE . 331

I f the number of sides of the semi- polygon be indefi

nitely increased ,

T hen the area ABCDE will approach the surface Of the

sphere, and OH will approach R as its limit.

We shall then have two variables which are always

equaL

Hence lim. area ABCDE = lim. AE x 27tOH,

Or sur . sphere AE x 27m.

545. COR . 1 . The area of the surface of a sphere is

S 47i R’or 7tD

’

,

in which S denotes the surface and R the radius.

546. COR. 2.

— The areas of the surfaces of two spheres are

to each other as the squares of their radii , or as the squares oftheir diameters.

For let S and S’represent the areas of two sphereswhose

radii are B and R’

.

T hen S : S'

47tR’ 47i R

"R2R"

,

And S S’

7tD27tD

"D’D”

.

547. COR . 3.

— The area of a zone is equal to the circumference of a great circlemultiplied by its altitude.

For the area of the zone generated by B C equals

GO x zym.

548. COR . 4.

—Zones on the same sphere, or on equal spheres,are to each other as their altitudes.

For let Zand Z’ represent the areas of two zones whose

altitudes are A and A ’

,on a sphere whose radius is R .

T hen § 547

549. COR . 5.— The area of a zone is to the area of the sur

face of a sphere as the altitude of a zone is to the diameter ofthe sphere.

For Z: S = A x 27tR s 27tR= A : D.

332 S OLID GEOME TRY.—B OOK VI II.

PROPOS IT ION XV. T HEOREM .

550. If an isosceles tr iang le be revolved abou t an

a x i s in the same p la n e, wh ich p asses th r ough i ts

ver tex w i thou t in tersect ing i ts su rfa ce , the volume

g en era ted is equ a l to the a rea g en era ted by the

base , mu ltip lied by one th i rd of the a lt i tud e.

Given— T he isosceles triangle AB C revolved about the

ax is FG,and AD its altitude.

T o Prove—Vol. ABC= area CB x tAD.

Dem— Produce CB to C,and draw BE and CF perpen

dicular to AC .

Now,vol. ACG' vol. ACF vol. G'CF

,

lnCF“x AF + M OP

”x G

'F, § 529

x (AF + CF ) i n CF’x AG

,

tn CF (CF x AG) .

But CF x AO= AD x CG,each being equal to twice

the area of the triangle ACG. § 231

Hence vol. ACG’ M OP x AD x CG’.

But nCF x CG is the area generated by CC . § 541

Hence vol. ACG area CG x tAD,

Also vol. AB C area B C x tAD.

S ubtracting these two equations,

vol. ABC= area CB x lAD. Q. E . D .

334 S OLID GEOME TRK—B OOK VIII .

552. COR. 1. The volume of a sphere is

V and V {fl tDfl

in which V denotes the volume of the sphere, R the radius,and D the diameter.

S ince V x QR finR’

,545

V= an”x {D {ram § 545

553. COR. 2.

— 7he volumes of two spheres are to each other as

the cubes of their radii or as the cubes of their diameters.

For V z V'

§ 552

V : § 552

554. COR. 3. The volume of a spherical sector is equa l to

the area of the zone which forms its base, multiplied by one

third of the radius of the sphere.

By a process of reasoning entirely analogous to that

employed in the theorem, it may be proved that

vol. sph . sector OBB zone BD x fill.

555. COR . 4. The volume of a spherical sector is equal to

two thirds of the area of a great circle,multiplied by the altitudeof the zone.

For area zone x H. § 547

T hen vol. x H x SR inR’ x H. § 554

556. COR . 5. The volumes of spherical sectors on the same

or equal spheres are to each other as the zones which form their

bases, or as the altitudes of these zones.

For

V : V'= i 7tR’ x H z i nR

’x

EXERCIS E S .

450. F ind the surface Of a Sphere whose radius is 23 in .

451. Find the volume of a sphere whose diameter is 18 in .

452. F ind the volume of a Spherical sector the altitude of whose base

is 8 ih . ,the diameter of the sphere being 18 in .

THE SPHERE . 335

PROPOSIT ION XVI I . T HEOREM.

557. To find the volume of a sp her ica l segmen t .

Given—O the centre of a sphere whose radius is R,

And AB r,

0B h’

,

DC= B C= H,

OC= h,

VOI. AB CD V.

T o Prove—T he volume of the spherical segment generated by the revolution of ABCD about MN as an axis equal

to 7t(h h’

)R’lm:(h

’

Dem—Draw OA and OD.

Now,seg. ABCD sect. AOD cone DOC cone AOB .

Or V= i nR’H tnf’h i nr

’h’. 555

,528

But H h h’

, (1)

(2)

T hen V= inUz h’

)R’Inh(R

’ h“ ) inh’

(R’

Or V= inR’h 37tR

’h’

+ iaR’h inh

’i aR

’h’

ink“

.

Collecting, V= 7i(h h’

)R’ln(h

’ Q. E . n.

T his formula is convenient when the distances of the

bases of the segment from the centre of the sphere are

given.

336 SOLID GEOME TRK—B OOK VIII .

558. COR . 1. The volume of a spherical segment is equal tothe half s um of its bases multiplied by its altitude, plus the

volume of a sphere of which that altitude is the diameter.

Factoring § 557, we have,

V= (h (h2

hh’

hung.

Squaring H h’ 2hh’

h” .

T hen hh’= 4 h H

And h’

h h’

h” h’ i (h’

h” H’

) h”

,

W h" ) w :§(R

’r”

R’

r’

) QH’

,

SR’

§(r’

r"

) QH’.

Substituting in

V= H[3R’31s .w r

"

) per]

V= }H (7tr’

7i r”

) fiuH’.

559. COR . 2.

— The volume of a spherical segment ofbase is

inr’H t

’.

EXERCIS E S .

ORIGINAL EXAMPLES .

458. Find the lateral area and volume of a cylinder whose altitude is

40 in . and radius of the base 10 in .

454. Find the lateral area and volume of a cone whose alti tude is 8

yards and radius of the base 4 ft .

455. F ind the entire area Of the frustum Of a cone whose slant height

is 12 ft ., the radius of the upper base 3 ft . , and of the lower base 6 ft .

456. Find the volume of the frustum of a cone whose altitude is 12in .

,

the radius of the upper base 8 ih . , and of the lower base 12 in .

457. Find the surface and volume Of a sphere whose radius is 16 in .

458. F ind the area Of a zone whose altitude is 4 in . , on a sphere whose

diamete r is 20 in .

459. F ind the volume of a spherical sector. the altitude of the zone

which forms its base being 6 ih . ,on a Sphere whose radius is 15 in .

338 S OLID GEOME TB ra— B ooa'

VIII .

ORIGINAL THEOREMS .

477. What is the re lation Of the surfaces of a cube and an inscribed

sphere ?

473 . What is the relation Of the volumes Of a cube and an inscribed

sphere?

470. What is the relation of the surfaces of a cube and a circumscribed

430. What is the relation of the volumes of a cube and a circumscribedSphere

461. What is the relation of the volumes Of two cylinders generated

by successively revolving a rectangle about its two adjacent sides?

432. Prove that the surface of a sphere is equal to the lateral area of a

circumscribed cylinder .

433 . Prove that the surface Of a Sphere is to the entire area of a cir

cumscribed cylinder, including its bases, as 2 to 3.

464. Prove that the volumes of a cone Of revolution ,a sphere, and a

cylinder Of revolution are in the proportion of 1,2, 3 if the bases of the

cone and cylinder are each equal to a great circle of the sphere and their

altitudes are each equal to the diameter Of the sphere .

435. Prove that the sum Of the squares of three chords at right angles

to one another from any point in the surface of a sphere is equal to the

square of the diameter .

466. T he entire area of a cylinder of revolution whose altitude and

diameter are equal is a mean proportional between the surface of the

circumscribed sphere, and the entire area of a cone of revolution whose

slant height and diameter are equal, and wh ich is inscribed in the same

Sphere. T he same is true Of the volumes of thes e three bodies .

437. T he entire area of a cylinder of revolution whose altitude and

diameter are equal is a mean proportional between the surface of the

inscribe l sphere, and the entire area of a cone of revolution whose

Slan t height and diameter are equal, and which is circumscribed about

the same sphere. T he same is true Of the volumes of these three

bodies .

466 . A sphere ZR inches in diameter is bored through the centre witha 2r - inch auger ; derive a formula for find ing the volume of the part

remaining.

469. If an equilateral triangle is revolved about its altitude, find the

relation of the solids generated by the triangle, the circumscribed circle,

and the inscribed circle .

490. How far must a person ascend above the earth to see one fourth

Of its surface ?

B OOK IX .

S PHERICAL GEOME T RY .

S PHER ICAL S U RFA CE S .

560. A S pherical Ang le is the angle between two in

tersecting arcs of great circles.

T he arcs are the sides of the angle. T he point at which

the two arcs meet is called the vertex of the angle. T he

measure Of a spherical angle is the same as the measure

Of the dihedral angle formed by the planes of its sides.

56 1. A S pherical Poly gon is

a portion of the surface of a

sphere hounded by three or

more arcs of great circles ; as

AB CD,or EFG.

T he diagonal of a spherical

polygon is the arc of a great

circle connecting any two vertices

which are not consecutive.

562. A S pherical T riangle is a spherical polygon of

three sides.

Spherical triangles are equilateral, isosceles, etc.,in the

same manner as plane triangles.

563. A Lune is the portion of the surface of a Sphere

bounded by the semi- circumferences of two great circles.

A S phericalW edg e is the solid bounded by a lune and

the planes Of its bounding arcs.

340 S OLID GEOME TRK—B OOK IX.

564 A S pherical Pyramid is the solid bounded by a

S pherical polygon and the planes of its sides ; as O—ABCD

,

565. A Pole of a Circle is a point on the surface of a

sphere equally distant from every point in the circum

ference of the circle.


566. E very section of a sp here mad e by a p lane is

a ci rcle .

Given - ADB a section of a sphere made by a plane.

T O Prove— ABED a circle .

Dem— Draw OP perpendicular to the plane of ADB .

T ake E and D any two points in the perimeter of ADB,and draw OE and OD.

T hen OE OD. 536

Hence 378

But E and D are any two points in the perimeter ADB .

Hence ADB is a circle whose centre is C. Q. E . o.

56 7. COR . 1 .

— A diameter drawn perpendicular to the plane

of a small circle passes through its centre.

568. COR . 2.

—All great circles on the same sphere are equal.

569. COR . 3.

— Every great circle divides a sphere into two

equal parts.

SFE RRICAL S URFACES . 343

T herefore AC B C> AB .

In like manner

Ac+ AB > B C,and AB + B C> AC. E . n.

577. COR . 1 .

— Any side of a spheri cal polygon is

the sum of the other sides.

Draw the diagonal CB .

T hen AB AC+ CB . 576

But CB CD DB .

Hence AB AC+ CD DB .

578. COR . 2. The shortest distancefromonepoint to another

on the surface of a sphere is measured on the arc of a great

circle.

For, suppose the two points to be connected by the arc

of a small circle. Divide this are into any number Of

equal parts, and through these points of division draw

arcs of great circles, thus forming a Spherical polygon.

T he sum of these arcs will be greater than the arc of a

great circle joining the two given points But if

the number of divisions be indefinitely increased , the

limit of the sum of these arcs will be the arc of the

small circle join ing the two given points, and will be

greater than the arc Of a great circle joining the same

points.

EXERC IS E S .

491. One and only one plane can be passed through a given point on

a sphere tangent to that Sphere.

492. Sections of a sphere at equal d istances from the centre of a sphere

344 S OLID GEOIlIE TR Y.— B OOK IX.

PROPOS IT ION IV. T HEOREM .

579. The sum of the s id es of a sp her ica l p olygon

is less tha n the ci rcumfer ence of a g rea t circle.

Given— AB CD a Spherical polygon on the Sphere whose

centre is 0.

T o Prov e— AB B C CD DA the circumference

of a great circle.

Dem— Draw the radn AO,B0

,C0

,and DO.

T hen AAOB ABOC+ Acoo ADOA 4 rt. angles.

424

But the arcs AB ,BC, CD,

and DA are the measures of

the angles AOB ,BOC

,COD

,and DOA . 177

Hence the arcs AB B C CD DA 4 rt. angles.

But four right angles is the measure Of a circumference.

T herefore AB B C CD DA the circumference of

a great circle. Q. E . r) .

EXERC IS E S .

493. T he sections of a sphere are proportional to the products Of the

segments of the d iameter drawn perpendicular to them.

494. F ind the locus Of the cen tres of Spherical sections made by planes

that contain a given straight line .

495. Find the locus of the centres Of spherical sections wh ich pass

through a given point .

346 SOLID GEOME TRK—BOOK IX.

581. Con. 1.—The poles of a great circle are at equal dis

tances fromthe circumference, and those of a small circle are

at un qual distances.

582. COR . 2.— l polar distance of a great circle is a

quadrant.

PROPOSIT ION VI . T HEOREM .

588. If a p oin t on the su rface of a sp here is a

qua d ran t’

s d istance from each of two p oin ts in the

a rc of a grea t ci rcle, i t is the p ole of tha t a re.

P

Given—P a point on the surface of a sphere CD,at a

quadrant’s distance from each of the points A and B .

T o Prove—P the pole of the arc of a great circle AB .

Dem— Draw the radii OA,OB

,and OP .

T hen,since the arcs PA and PB are quadrants, the angles

POA and POB are right angles.

T hen P0 is perpendicular to the plane AOB . 380

T herefore P is a pole of the arc AB . 580. Q. E . D.

584. COR— If AB is perpendicular to the arcCAB , it passes

through the pole D.

Norm—T his theorem is not always true. If the two points lie at the

extremities of a diameter, any number of great circles may be drawn

through them, but P is the pole of only one of them.

SPHERICAL S URFAess 347

PROPOSIT ION VI I . T HEOREM.

585. A spher ica l angle is measu red by the a rc ofa grea t circle d escr ibed w i th i ts vertex a s a p ole,

included between i ts sid es p rod uced if n ecessary .

Given—APE a spherical angle on the sphere whose

centre is O,and AB the arc Of a great circle described

from P as a. pole.

T o Prove—T hat AB is the measure of AAPB .

Dem—Draw the tangents PD and PE,and the radn AO,

B0,and PO. T hen ADFB ‘AAOB . 169

, 403

S ince AB is the arc of a great circle described from P as

a pole, PA and PB are quadrants 582) and AO and BO

are perpendicular to PO.

T hen AAOB is the plane angle of the dihedral angle

P0.

Hence AAOB ADFB = AAPB . 403,1

But AAOB is measured by the arc AB . 177

T herefore AB is the measure of AAPB . Q. E . D.

586. COR. 1.— A

’

spherical angle is equal to the plane angle

formed by the tangents to the arcs at their point of intersection.

587. COR. 2. T he opposite or cortical angles formed by twoarcs of great circles intersecting each other are equal.

348 SOLID GEOME TRK—BOOK IX.


588. If from the ver tices of a sp her ica l tr iangle as

p oles , arcs be d escr ibed forming a second spherical

tr iang le , the vertices of the second tr iangle are re

sp ect ively p oles of the sides of the first .

Given—ABC a spherical triangle, and DEF a second tri

angle described from A,B,and C as poles.

T o Prove—D,E

,and F respectively poles of the arcs

B C,AC, and AB .

Dem—S ince B is a pole of the arc DE,the distance from

D to B is a quadrant. § 582

S ince 0 is a pole of the arc DE,the distance from D to

C is a quadrant. § 582

T hen D is a quadrant’s distance from the points B

and 0.

Hence D is a pole of the arc B C. § 583

In like manner F may be shown to be the pole Of the

arc AB ,and E the pole Of the arc AC. Q. E. D.

589. SCHOLIUM.

—T he triangle ABC may be described

with D, E , and F as poles, as DEF is described with A,

B,and C as poles. T riangles so related are called Polar

T riangles.

350 SOLID GEOME TRYZ—BOOK IX.

PROPOS IT ION X . T HEOREM .

592. The sum of the a ngles of a sp her ica l tr ia ng le

is less than six , and grea ter tha n two, r igh t ang les .

Given—ABC a spherical triangle.

T o Prove—A B C 540° and 180°

Dem— lst,

A

B

C

A B C

A 180°—EF.

B 180°—DE.

180°—DE .

Hence A B C= 540°—(EF + DE + DE ) .

But EF DF DE 579

T herefore A B C> 180° Q. E . D.

593. COR .

— A spherical trianglemay have one, two, or three

right angles, or one,two, or three obtuse angles.

DEE—A spherical triangle having

One right angle is called a rectangular triangle,

T wo right angles is called a bi- rectang ular triangle,

T hree right angles is called a tri - rectangular triangle.

352 S OLID GEOME T RK— B OOK IX.

598. N o isosceles symmetrical triangles are equal.

they may be made to coincide throughout.

PROPOS IT ION XI . T HEOREM.

599. Two symmetr ica l sp her ica l tr iangles ar e

equ iva len t .

Given—ABC and DEF two symmetrical triangles havingAB DE

,AO= DE

,and B C= FE .

T o Prove A AB C¢ A DEF.

Dena— S uppose P and Q the poles of small circles which

pass through A,B,C and D,

E,F .

T hen these circles will be equal.

For the chords AC, AB ,and B C will equal DE, DE ,

and

FE respectively . 156

Hence plane A AB C= plane A DEF, § 57

And the circumscribed circles are equal.

T hen the arcs PA,PB

,PC

, QD, QE , QF are all equal.

565

Hence the triangles ACP and DFQ are mutually equi

lateral,and also isosceles.

T hen A ACP A DFQ,

A ABP A DE Q,

354 SOLID GEOME TRY.—BOOK IX.

T o Prove A GKH.

Dem—Construct AHKL symmetricalwith A GHK then

all the parts of AHKL are equal respectively to the parts

597

A AB C= AHKL, First part.

§ 599

A AB C¢A GHK. Q. E . D.

PROPOS IT ION XI I I . T HEOREM.

601. Two sp her ica l tr iangles on the same

a re equa l or equ iva len t when two ang les a nd an

includ ed s ide of the on e a re resp ect i vely equa l to

two angles and an inclu ded sid e of the other .

Given—DEF and PQR two Spherical triangles on equal

Spheres, with AD= AP ,AE = AQ, and A L.

T o Prove—A DEF A PQR ; or A DEE¢ A PQR .

Dem— Construct the polar triangles Of DEF and PQR .

T hen and Af= Ar . § 590

Hence A def= q r,or q r . 600

T hen their polar triangles will be equal, 590

And A DEE = A PQR,or A DEF=O=A PQR Q. E . D .

602. S CHOLIUM.

—T his proposition can be proved by

direct superposition, as in § 56. Compare also § 600.

E QUAL AND S YMME TRICAL TRIANGLES . 355


603. If two spher ica l tr iangles on the same sp here

a re mu tua lly equ ila tera l, they a re mu tua lly equ i

angu la r .

Given—AB C and DEF two mutually equilateral spheri

cal tri angles ou equal spheres.

T o Prove—T he triangles AB C and DEF mutually equi

angular.

Dem.- Draw the radn 0A,

OB,CC

,C’D

, OE ,and OF.

T hen the homologous face angles of the trihedral'

angles

O—ABC and O'

- DEF are equal. 177

Hence dihedral angle OA dihedral angle 0'D. 425

But the spherical angles BAC and EDE are equal to the

dihedral angles OA and OD. § 56O

T herefore ABAC AEDF.

In like manner, AABC ADEF,

And AAOB ADFE .

Hence the triangles ABC and DEF are mutually equi

angular. Q. E . D.

604. COR . Twomutually eq uilateral spherical trianglesthe same sphere are equal or equivalent.

T he triangles are equal if their parts are arranged in

the same order,and symmetrical and equivalent if their

parts are arranged in an inverse order.

356 S OLID GEOME T RK—B OOK IX.


605. If two sp her ica l tr iang les on the same sphere

a re mu tu a lly equ iang u la r , they a re mu tu a lly equ i

la tera l.

Given — AB C and DEF two mutually equiangular

spherical triangles on equal spheres.

T o Prove— AB C and DEF mutually equ ilateral.

Dem—Construct the polar triangles of AB C and DEF.

T hen,since AB C and DEF are mutually equiangular,

A'B'

C’and D’E

’F’are mutually equilateral. § 590

Hence A'

B'

C’and D

’E

’F

’are mutually equiangular.

604

T hen their polar triangles AB C and DEF are mutually

equilateral. § 590. Q. E . D.

606. COR . Two mutually equiangular spherical tri angles

on the same sphere are equal or equivalent.

EXERCIS E S .

498. If the sides of a polar triangle are and 163° respec

tively , how many degrees are there in each angle of its polar triangle?

497. If the angles of a spheri cal triangle are and 70° te

Spectively , how many degrees are there in each side of its polar tri angle ?

498. How far above the earth’s surface must a person ascend to see i

of its surface ?

358 S OLID GE OME TRK—B OOK IX.

PROPOS IT ION XVI I . T HEOREM .

609. If two a ng les of a spher ica l tr iang le a re equal,the sides opp osi te them a re equa l .

Given— ABC a spherical triangle, with AB AC.

T o Prove AB AC.

Dem— Construct the polar triangle Of ABC.

T hen DEF will have DE DF,

And AE AF.

T hen in the polar triangle of DEF,or ABC

,

AB= AC. § 590. Q. E . D.

EXERCIS E S .

503. An equiangular Spherical triangle is also equilateral.

504. An equ ilateral spherical triangle is also equiangular .

505. A circle may be inscribed in a spherical triangle .

506. E ither angle of a Spherical triangle is greater than the difference

between 180° and the sum of the other two angles .

507. T he intersection of two spheres is a circle whose plane is perpen

d icular to the straight line joining the centres of the spheres, and whosecentre is in that line .

508. I f the arc of a great circle is drawn perpendicular to the arc Of

another great circle at its middle point, a point in the perpendicular

will be equally d istant from the extremities Of the second are .

509. If two straight lines are tangent to a Sphere at the same point,th eir plane is tangent to the sphere.

E QUAL AND S YMME TRICAL TRIANGLES . 359


610. I n any Sp her ica l tr ia ngle the g rea ter side

lies opp osi te the g reater ang le.

Given—AB C a spherical triangle, with AB AC.

T o v a AC AB .

Dem—Draw the arc of a great circle BD,making

ADB C= A0.

T hen BD DC.

But AD BD AB .

Hence AD DC AB,

Or AC> AB . Q. E . D.

6 11. COR — I a any spherical triangle the greater angle lies

opposite the greater side.

Given—AB C a spherical triangle, with the side AC>AB .

T o Prove AB AC.

Dem.

—I f AB AC,AC would equal AB . 576

I f AB AC, AC would be less than AB . 610

But each of these conclusions is contrary to the hypothe

sis that AO> AB .

T herefore AB AC.

360 SOLID GEOME TRK—B OOK IX.

MEA S UREMENT OF S PHER ICAL POLYGON S .


6 12. If the a rcs bf g‘

rea t ci rcles in tersect on the

su rface of a hemisph ere , the sum of the opp os i te

sp her i ca l tr ia ng les wh ich they form is equ i va len t

to a lu ne w hose ang le is the a ng le formed by th e

ci rcumfer ences .

Given—AB F and CBD two great circles intersecting on

the hemisphere ACED.

T o Prove— A AB C+ A BDF¢ lune BDEFB .

Dem.

—Produce the arcs ABE and CBD until they meet

in E .

T he semicircumference ABF = the semicircumference

S ubtracting the common are BF,

T hen arc AB arc E F.

In like manner it may be proved that are

and are AO= are FD.

T herefore A ABO¢ A DEF .

But A BDE - l A DEE = lune BDEFB .

Hence A AB C+ A BDF¢ lune BDEFB . Q. E . D.

362 S OLID GEOME TRYZ—B OOK IX.

CASE II . When the angles are incommensurable.

Given— AB CDA and FBEDF two lunes, whose angles

ABC and FBE are incommensurable.

T o Prove _AB CDA AABOFBEDF AFBE

Dem— T ake any angle FBH as a unit Of measure

which is contained an exact number of times in AFBE .

T hen,since AAB C and AFBE are incommensurable,

AFBH will be contained in AABO a certain number of

times,with a remainder KB C which is less than the unit

of measure.

Produce BK to D.

S ince the angles ABK and FBE are commensurable, we

have,by Case I .

,

ABKDA AABK

FBEDF AFBE

Now,if we diminish the unit Of measure indefinitely,

the remainder KCB will d iminish indefinitely ,

And AABK will approach AABC as its limit,

And ABKDA will approach AB CDA as its limit.

Hence we have two variables which are always equal.

And, S ince the limits of equal variables are equal,

MEAS UREMEN T OF SPHERIC’AL POLYG'ONS . 363

ABKDA1

. AABK1m

FBEDF AEBE

OrAB CDA AAB C

FBEDF AEBE

614. COR . 1.- The surface of a lune is to the surface of a

sphere as the angle of the him is to four right angles.

615. COR . 2.- The area of a lune is

L 2A X T,

in which L denotes the area of the lune, A its angle (expressedin right angles) , and T the area of a tri - rectangular triangle.

Q. E . D.

T he surface of a sphere is § 594

HL A

ence

8T 4)

Or L 2A x T .

6 16. COR . 3. The volume of a spherical wedge is to the

volume of a sphere as the angle of the wedge is to four r ight

angles.

T his may be demonstrated by a process of reasoningentirely analogous to that employed in the theorem.

6 17. COR . 4. The volume of a spherical wedge is equal to

the volume of a trirectangular pyramid multiplied by twice the

angle of the wedge.

EXERCIS E S .

510. T wo lunes on the same Sphere or on equal Spheres are to each

other as their angles .

511. Find the area of a tri - rectangular triangle on a sphere whose

d iameter is 24 inches .

512. T he angle of a lune is 30°on a sphere whose radius is 48 ft. find

the surface of the lune .

513 . Find the volume of a tri -rectangular spherical pyramid , the

d iameter of the Sphere being 8 in .

364 S OLID GEOME TRK—BOOK IX.


618. M o a rea of a sp her ica l tr iangle is equ a l to

i ts sp her ica l ex cess mu lt ip lied by a tr i - recta ng‘

u la r

tr iangle .

Given—ABC a spherical triangle on a sphere whose

centre is 0.

T o Prove—Area ABC= (A B C 2) x T .

Dem— Produce the S ides till they meet the great circle

T hen A DAB A FAO a lune whose angle is A. 612

But a lune whose angle is A 2A x T . § 615

In like manner A HBF + A EBK 2B x T,

And also A GCK A HCD= 20 x T .

Adding these six triangles, we observe that their sum

exceeds the surface of a hemisphere, or 4T , by twice the

area of AB C.

Hence 2 x A AB C= (2A 2B 2C ) x T—4T,

Or A ABO= (A B c 2) x T . Q. E . n.

EXERC IS E .

514. F ind the area of a spherical triangle whose angles are

and on a sphere whose radius is 60 It . Of one whose angles are

each

366 SOLID GEOME TRYZ—B OOK IX.

PROPOS IT ION XXI I I . T HEOREM.

620. If from any p oin t on a hemi sp here two a rcs

of a grea t ci rcle a re d rawn p erp end icu lar to the cir

cumference of the hemisp here, and oblique a rcs are

d rawn , then ,

1 . The shorter of the two p erp end icu la r a rcs is the

shortest a rc tha t can be d rawn from the g iven p oint

to the ci rcumference ; and the longer of th e two p er

p end icu la r a rcs is the longest a re tha t can be d rawn

from the g iven p oin t to the ci rcumference .

2. Any two obli que a r cs wh ich cu t off equ a l d is

tances from the foot of the p erp end icu la r a re equa l.

3 . The obli qu e a re wh ich cu ts off the g rea ter d is

tance from the foot of the p erp end icu la r is the

g rea ter a rc.

Given— D a point on a hemisphere whose centre is 0,DC and DH two arcs perpendicular to the circumfer

ence ACB,DE and DF oblique arcs cutting off equal dis

tances from C,and DG cutting off a greater distance from

C than DE ,

T o Prove , lst—DC the Shortest distance from D to the

circumference ACB,and DH the longest distance.

Dem—Let DF be any Oblique are drawn from D to

AOB .

MEAS UREMENT OF SPHERIC'AL POLYGONS . 367

T hen in the triangle DCF, DC is less than DE. 610

Hence DC is the shortest distance to AGB .

T o prove DH the longest distance from D to the circum

ference AOB,

Let FDK be any other great circle than CDH, passing

through D.

T hen arc PDK arc CDH,

And are FD are CD. lst part.

S ubtracting, arc DK arc DH.

T o Prove , 2d are DE = arc DF.

Dem— T he two triangles DCB and BCF have two S ides

and the included angle of each respectively equal, and

are therefore symmetrical. § 6OO

Hence arc DE arc DF. § 597

T o Prove, 3d are DC are DE .

Dem— T ake DE= DE .

T hen in the triangle DFG' we have

are DC are DF,or its equal DE . 610. Q . E . D.

EXERCIS E S .

515. Find the area of a spherical triangle whose angles are

and respectively , on a sphere whose radius is 36 in .

516 . Find the volume of a triangular spherical pyramid the angles of

whose base are and res pectively , and the diameter of thesphere is 24 ft .

517. F ind the area Of a spherical polygon whose angles are

and on a sphere whose radius is 12 ft .

518. F ind the area of a spherical quadrilateral, the angles beingand the d iameter of the sphere being 12 ft.

519. Find the volume of a spherical segment, the radius of the spherebeing 20 ft. , and the planes of the bases being 8 and 12 ft. from the

centre of the sphere .

520. What is the volume of a sphericalwedge the angle of whose baseis if the volume of the sphere is 144 cu . it .

368 S OLID GEOME TRK—B OOK IX.


621. If the s ides of a sp her ica l a ng le be p roduced

u n ti l they meet, a nd thu s form a lune, then ,

l st . If the inclu d ed a ngle is acu te, the longest a rc

of a grea t ci rcle tha t can be d rawn betw een i ts sid es ,

p erp end icu la r to ei ther s id e, is the measu re of the

ang le .

2d . If the includ ed ang le is obtu se,the shor test

a rc of a grea t circle tha t ca n be d rawn between i ts

s id es, p erp end icu la r to ei ther side, is the measu re

of the ang le .

CASE I . When the angle is acute.

Given—BAC an acute spherical angle on a sphere whose

centre is O, with its S ides produced forming the lune

ACFBA,and DE its measure.

T o Prove—DE the longest arc of a great circle that can

be drawn between ADF and AEF perpendicular to ADF .

Dem— S ince DE is the measure of AA,A is the pole

of the arc DE,and DE is perpendicular to ADF and

585

Produce DE ,and it will pass through the pole P Of the

arc ADF which lies without the sides, since AA

is acute.

70 S OLID GE OME TRK—B OOK 1X.

S ince AEP is a right angle, PEB is also a right angle.

T hen arc PE arc PB . 610

Now adding PE and PB to PD and PC,respectively ,

arc DE arc B C.

T herefore DE is the shortest are that can be drawn‘

between ADF and AEF perpendicular to ADF. Q. E . D.

EXERC IS E S .


521. T he area Of a spherical triangle, each of whose angles is is

equal to the area of a great circle .

522. A Sphere can be circumscribed about any tetrahedron.

Suggestion—All points equally distant fromA, B ,

and C lie in the perpendicular EN erected at the

centre of the circle passing through A , B , and C.

In like manner all points equally distant from B , C,

and D lie in the perpendicular MF erected at the

centre Of the circle passing through B , C, and D.A

Also EN and FM lie in the plane perpendicular to

B C at its middle point , etc .

523 . A sphere can be inscribed in any tetrahedron .

524. T he four perpend iculars erected at the centres of the faces of a

tetrahedron meet in the same point .

525. T he six planes perpend icular to the six edges of a tetrahedron at

their middle point will in tersect in a common point.

526 . T he six planes bisecting the six dihedral angles of a tetrahedron

intersect in a common point .

527. T he volume of a spherical wedge is to the volume of the sphere

as the angle of the lune which forms its base is to four ri ght angles .

528. T he volume of a Spherical wedge is equal to twice the angle

of the lune which forms its base multiplied by the volume of a tr i

rectangular pyramid .

529. T he volume of a Spherical pyramid is equal to the spherical ex

ces s Of its base multiplied by the volume of a trirectangular pyramid .

530. If the arcs of great circles be drawn from any point on a sphere

to the extremities Of another are of a great circle, they will be greater

than the sum Of two other arcs S imi larly drawn but included by them.

S UPPLEMENT T O S OLID GEOME TRY .

T HE W EDGE AND PR ISMOID .

622. A W edge is a solid bounded by a rectangle, two

trapezoids, and two triangles.

T he rectangle (ABCD) is the back.

T he trapezoids (ABFE and DCFE )are the faces.

T he triangles (ADE and B CF ) arethe ends.

T he line EF (where the faces meet) is the edge.

623. T here are three distinct forms of the wedge

l st. When the'

length Of the edge is equal to the length

Of the back .

2d . When it is less.

3d . When it is greater.

624. CASE I .

—When the length of the edge is equal to the

length qf the back.

I t is evident that this form of the wedge is a right tri

angular prism whose base is the triangle ADE and whose

altitude is EF.

Hence its volume A ADE x EF.

EXERC IS E S .

531. Find the volume of a wedge whose back is 10 in . by 6 in . ,the

altitude of the end being 12 in.

532. Find the volume of a wedge whose back is 20ft. by 12ft. and theend an equilateral triangle.

372 S OLID GEOME TRY.— S UPPLEMENT .

625. CASE I I .—When the length of the edge is less than the

length of the back.

T hrough F , the middle point of the edge, pass the plane

FCB perpendicular to the back ,dividing the wedge into

two equal parts, only one of which is represented in the

figure.

T hrough E pass the plane EGH parallel to FCB formingthe right triangular prism GHB C—F , and the quadrangu

lar pyramid AHGD—E .

Denote AB by L,EF by l, AD by b, EK by h, and the

volume by V.

T hen the prism GHB C—F = if bhl,

And the pyramid AHGD—E = b(L Olh.

Hence V lbhl bh(L l) ,

tbhl §t gbhl,

lbhl gt ,

Or V= §bh(l 2L) .

By multiply ing this expression by 2, letting l represent

the entire length of the edge, and L the entire length Of

the back,we obtain the following

RULE — Add the length of the edge to twice the length of the

back, and multiply this sum by one sixth of the product of thewidth and altitude ; the final product will be the required

volume.

EXERCISES . 375

Multiplying these two equations,

4MN = BL B l bL bl.

Substituting this value in

V= geb(BL bl 4MN ) .

But BL is the area of the lower base, bl is the area of

the upper base, and MN is the area of the middle section .

Hence,for finding the volume of a prismoid we Obtain

the following

RULE —Multiply the sum of the lower base, the upper base,

and four times themiddle section by one sixth of the altitude.

I t is readily shown that th is rule will apply to findingthe volume of a pyramid, cone, prism,cylinder, frustum of a

pyramid, frustum of a cone,etc.

EXERC IS E S .


534. Find the volume of a rectangular prismoid, one of whose bases

is 16 by 20 ft . , the other 8 by 12 ft . , and the altitude 24 It .

535. How many bushels of lime in a kiln whose lower base is 10 by12 ft . , the upper base 16 by 18 ft .

,and the altitude 16 ft . , if a bushel

contains 1193 cu . ft. ?

536 . What is the volume of a stick of hewn timber whose ends are

20and 24 inches and 10by 14 inches, its length being 30 ft .

537. Find the volume of a square pyramid whose base is 6 ft . on a S ide

and altitude 24 ft .

538. F ind the volume of a cone whose altitude is 48 ft. and the radius

Of the base 6 ft.

539. Find the volume of the frustum of a triang ular pyramid the

sides of whose lower base are 8 in . and the sides of the upper base 4 in . ,

its altitude being 18 in .

540. Find the volume Of the frustum of a cone,the radius of the

lower base being 16 ln .,the radius of the upper base 4 in . , and the alti

tude 12 in .

541. F ind the volume of a prismoid one of whose bases is 24 by 36 ft. ,

the other 20by 22 ft . , its altitude being 4 rods .

MISCELLANEOUS PROPOS IT IONS .

BOOK I .

542. If one angle of a triangle is equal to three times another angle of

the triangle, the triangle can be divided into two isosceles triangles .

543. T he sum of the three med ial lines of a triangle is les s than the

perimeter and greater than the semi- perimeter .

544. If two of the medial lines of a triangle are equal, the triangle is

isosceles .

545. T he Sum of the medial lines of a triangle is greater than three

fourths of the perimeter of the triangle.

546 . T he lines joining the middle points of the opposite S ides of a

quadrilate ral bisect each other.

547. T he bisectors of the angles of a trapezium form a second trape

zium,the opposite angles of wh ich are supplementary .

548. T he lines joining themiddle points of the sides of a square, takenin order, form a square .

549. If a diagonal of a quadrilateral bisects two angles , the quadri

lateral has two pairs Of equal S ides.

550. In a right triangle a perpendicular let fall from the vertex of the

right angle upon the hypotenuse cuts Off two triangles mutually equi

angular to the original triangle.

(n 2l180°

n

551. Prove that in a regular n-

gon each angle equals

552. What fractional part Of each interior angle does each exteri or

angle Of a regular n -

gon equal ?

553. If the diagonals Of a trapezoid are equal, the trapezoid is

isosceles.

554. In any triangle, the S ides of the vertical angle being unequal, the

med ian line drawn from the vertical angle lies between the bisectrix of

the vertical angle and the longer S ide.

555. In any triangle, the S ides of the vertical angle being unequal, its

bisectri x lies between the med ian line and the perpendicular from the

vertical angle to the base.

556. T he triangle forme d by joining points in the sides Of an equi

lateral triangle equidistant from the vertices, taken in order,is an equi

lateral triangle .

375

378 MIS CELLANEOUS PROPOS I T IONS .

meeting At?produced in E . and the circumference in D. so that DE shall

be equal to the radius of the circle, then the angle E will be cqual to one

th ird of the angle ACB .

Nora—If BE could be drawn under thes e conditions with the straight

line and the circumference only , ACB being any given angle, then the

trisartion of an angle in general would be possible by elementary geometry ; but th is has never been done.

578. If a circle is inscribed in a triangle, the distance from the vertex

of any angle to the po ints of contact of its sides is equal to the semi

perimeter minus the side ly ing opposite to this angle.

574. If two circles are tangent to each other internally at the point B ,

and a chord AC of the larger is tangent to the smaller at D,then BD

bisects the angle AB C.

575. If the opposite angles of a quadrilateral be supplementary , the

quadrilateral can be inscribed in a circle.

576 . T he perpend iculars from the vertices of a triangle to the Opposite

sides are the bisectors of the angles of the triangle formed by join ing thefeet of the perpendiculars.

577. If two straight lines are drawn through the point of contact of

two circles tangent externally , the chords of the intercepted ares are

parallel.

578. If AB is the common chord of two intersecting circles. and AC

and AD are the diameters d rawn from A, prove that the line CD passes

th rough 8 .

579. T he line join ing the centre Of a square des cribed on the hypote

nuse of a right triangle w ith the vertex of the right angle bisects the

right angle of the triangle .

580. If two circles in tersect each other, the longest common secant

that can be d rawn through either point Of intersection is parallel to

the line joining the centres of the circles .

581. Prove that the bisectors of the four angles of any quadrilateral

in tersect in four points, all of wh ich lie on the circumference of the

same circle .

CONST RUCT IONS .

582. Draw two lines that shall divide a given right angle into three

equal parts .

583 . Draw a line tangent to a given circle and perpendicular to a given

straigh t line .

MIS CELLANE OUS PROPOS I T IONS . 379

584. Construct a circle of given radius tangent to a given circle and

pass ing through a given point .

585. In a given isosceles triangle, draw a line that shall cut off a trape

zoid whose base is the base of the given triangle and whose three other

sides shall be equal to each other.

586 . T hrough a given point draw a straight line making equal anglesWith two given straigh t lines.

587. Draw a circle that shall be tangent to a given circle and also to

a given line at a given point.

588. In the prolongation of any diameter of a given circle find a point

such that a tangent from it to the circumference shall be equal to the

d iameter of the circle.

589. Construct a parallelogram whose area and perimeter are respec

tively equal to the area and perimeter of a given triangle.

590. T o find a point w ithin a given triangle such that the three

straight lines drawn from it to the vertices of the triangle shall makethree equal angles with each other .

BOOK IV .

591. Any chord of a circle is a mean proportional between its projection on the diameter drawn from one of its extremities and the diameter

itself.

592. I f on a diameter of a circle two points are taken equally distant

from the centre,the sum of the squares of the distances of any point of

the circumference from these two points is constant .

598. T he square described on the sum of two lines is equivalent to the

sum of the squares described on the lines ,increased by twice the rect

angle Of the lines .

594. T he square descri bed on the d ifference of two lines is equivalent

to the sum of the squares described on the lines, diminished by twice the

rectangle of the lines .

595 T he rectangle contained by the sum and difference of two lines

is equ ivalent to the d ifference of their squares.

596 . I f two fixed parallel tangents are cut by a variable tangent, the

rectangle of the segments of the latter is constant.

597. T he square of a side of a regular inscribed decagon together

w ith the square of the radius equals the square of a side of a regular

inscribed pentagon .

380 MI S’CELLANE OUS PROPOS I T IONS .

598. If two circles touch each other , secants drawn through their

point of contact and terminating in the two circumferences are divided

proportionally at the point of contact.

566 . T he rectangle of two sidel of any triangle equals the rectangle

of the perpendicular to the third side and the diameter of th e circum

600. I f two circles are tangent externally , the portion of their common tangent included between the points of contact is amean proportional between the diameters of the circles .

601. In any isosceles triangle the square of one of the equal sides is

equal to the square of any straight line drawn from the vertex to the

base, plus the product of the segments of the base.

CONST RUCT IONS .

602. Inscribe a square in a given triangle.

603. In a given triangle inscribe a rectangle whose sides are in a

given ratio .

604. Construct a square equivalent to a hexagon.

605. Construct a triangle, having given one angle, the side Opposite,

and the area.

606 . Divide a trapezoid into two equivalent parts by drawing a line

parallel to its bases.

607 Construct a square. having given the difference between the

diagonal of the square and its side.

BOOK V .

608. Prove that but three equal regular polygons may be used tocover a plane surface.

609. T he area of a regular inscribed octagon is equal to that of a rect

angle whose adjacent sides are equal to the sides of the inscribed and

circumscribed squares.

610. T he squares of the diagonals of a trapez ium are together double

the squares of the two lines joining the bisectors of the opposite

sides .

611. T he rectangle of the segments of chords passing through a

common point is a constant quantity .

612. T he rad ius of any inscribed regular polygon is a mean propertional between its apothem and the radius of a similar circumscribedregular polygon .

FORMULIE OF MENSURAT ION .

ABBREVIAT IONS .—b base ; h altitude ; A area ;p perim

eter ; r radius ; d diameter ; 0 circumference ; s slant

height ; V= volume ; m middle section ; 6 apothem ; a arc ;

E spherical excess ; T area of a tri - rectangular triangle.

T he numbers refer to the page .

POLYGONS .

Rectangle, 148, A bh.

Parallelogram, 149, A bh.

T riangle, 150, A lbh.

When a , b, and c represent the three sides and 8 their half

mm, 181,

T rapezoid , 151 , A }(b b’ )h.

Regular polygon , 211 , A ipe.

C IRCLE .

Circumference of a circle , 216, c ! d 2m'.

Area of a circle, 216, 217, A i cr m" i nd“ .

POLYHEDRONS .

Prism, 284, V bh.

Lateral area, 272, A ph.

Parallelopiped, 282, V bh.

Pyramid , 294, V i bh.

Lateral area, 287,Frustum Of pyramid , 295, V i b(bLateral area of frustum, 287, A “ P p

’

)s.

Pri smoid, 373, V }h(b b’ 4m) .

CYLINDER , CONE ,

Right circular cylinder, 315, V bh m’ h.

Lateral area, 314, A ch 21rrh.

R ight circular cone, 322, V i bh §m‘h.

Lateral area , 321, A ==r 408 m .

Frustum of right circular cone, 326, V 3 r h(r’ + r

” +

Lateral area, 324 , A s(r r’

)r .

382

FORM ULE OF MENS URAT ION. 383

Surface of sphere, 330, A cd 4m" = 1rd“ .

Surface of a lune , 363 , A 2a T .

Surface of sph . triangle , 364, A TE .

S urface of aph . polygon , 365 , A TE .

Surface of a zone, 331 , A 2m'h.

Volume of sphere , 333, V §Ar {M5find

”.

Volume sph . sector , 334, V fim‘ h lbr .

Volume sph . segment,336, V r

”

) §vrh’.

EXPRE S S ION S INV OLV ING 7!

‘R’ 31 41593. i n

1

fir i0°

U S EF UL ROOT S .

1/ I0

V;13

2

173

AREA S OF REGU LAR POLYGON S WHOS ES IDE IS 1.

S ides. Area. Name . S ides . Area.

T riangle 3 Octagon 8

Square 4 Nonagon 9

Pentagon 5 Decagon 10 7.694208.

Hexagon 6 2598076 . Undecagon 11

Heptagon 7 Dodecagon 12

VOLUME S OF REGULAR POLYHEDRONS WHOS ES IDE IS 1.

Name . No. of Faces.

T etrahedron 4

Hexahedron 6

Octahedron 8

Dodecahedron 12

Icosahedron 20

ANSWERS TO NUMERICAL EXERCISES.

Pag e 21.—E r: . 1 . 2. 8. 4.

Pag e 25 .—Ex. 5 . 6 . 7 . gof a

rt . angle, 3 of a rt . angle, of a rt . angle ; 8. gof a rt . angle,

1» of a rt . angle , of a rt . angle ; 9. 60°

10.

Pag e 27 .— Ex. 55

°13 . 14. 50

°and

Pag e 39 .—Ex. 24. 29° and 25 . 30

°and

Pag e 52 .—Ex. 30. 31 . 32. 60

°33 .

Pag e 5 5 .—Ex. 35 .

Pag e 6 26.—Ex. 42. 14792 43 .

Pag e 69 .—Ex. 55 . 56 . 57. 110.

Pag e 87 .—Ex. 105 . 6,m ; 108. 4 : 2 a: y ;

109. 110. 111 . 2y z z + y 2b zw+ b;

1 12. a zc = l zb ; 1 13 . 8and 20.

Page 115 .— E.r . 145 . 1156 . 25

°147 . 160

°

148. Angles 50°and arcs 120

°and 149. Angles 120

°

and arcs and 140°150. 140

°and

Pag e 117 — Ex. 151 . 152. 153 . 154.

and

Pag e 118 .—Ex. 155 . A B C and D

156 . D E = and F =

Pag e 154 .— Ex. 224. 3A. 32P . ; 225 . 2A. 112P . ; 226 . 10A .

Pag e 15 5 .—Ex . 227. Alt . ft area 623 5+ sq .

it 228. Perp . in .

do bdPag e 16 3 —Ex. 229. 230. 231 .

a a

Pag e 16 4 .— Baz. 232. 20 in .

Pag e FH .—Em. 233, 4 to 9, 4 to 9.

Pag e 178 .— Ex. 243 . 81+fi.

244. 391 9+ sq . ft . ; 245 . a.

GLOSSARY

T HIS glossary contains the etymology and meaning of manyof the common terms of Geometry . Only the words of greatest

value to the student have been selected .

KEY.—L.

= Latin ,Gr. Greek , d im. diminutive.

A cute . [L. acutus sharp ] Sharp . An acute ang le is an

angle less than a right angle .

A dj acen t . [L. ad to, j acere to lie ] Ly ing near to .

A lti tu d e . [L . altus high .] Height .

A naly sis . [Gr . mm again ,luein to loosen .] T he re

solving of problems by reducing them to equations.

A ng le . [L . angulus an angle ] T he amount of divergence

Of two lines which-meet at a common point .

A nt eced ent . [L. ante before , cedere to go .] T he firstterm of a ratio .

A x iom. [Gr . axioma that which is assumed .] A self- evi

dent truth .

B isect . [L. bi two, seam cut .] T o divide into two

equal parts .

Cen tre . [L. centmm centre ] A point within a circle

equally distant from every point of the circumference .

Chord . [Gr . chorde a string ] A straight line joining theextremities of an arc.

Circle . [L. circulus,dim. of czrcus circle ; or Gr . kirkos

circle ] A plane figure bounded by a curved line every point

of which is equally distant from a point within ,called the

centre .

Circumference . [L. circum around , ferre to beam]T he bounding line of a circle .

C ircumscrib e . [L . circum around , scribere to write, to

draw.] T o write or draw around .

Commensurable . [L. com together, mensumre to

measure ] Having a common measure .

388 CLOSSARY.

Complement . [L. comp lementam,from com with , p lere

to fill.] T he difference between an are or angle and 90°or a

right angle .

Concave . [L. can with , cavus hollow .] Hollow or

curved .

Concentr ic. [L. can together, eardrum centre ] Hav

ing the same centre .

Cone . [Gr . konos a cone, a peak .] A round body whose

base is a circle , and whose convex surface tapers uniformly to

a point called the vertex .

Consequent . [L. can together, sequi to follow .] T hesecond term of a ratio .

Constant . [L. can together, stare to stand .] A quan

tity of fixed value .

Conv erse . [L. can together, W e to turn .] A propositiou in which the hypothesis and the conclusion of the direct

proposition are reversed .

Conv ex . [L. convexus vaulted ] Rising into a spherical

form.

Corollary . [L. corollarium a gift , money paid for flowers,

from corolla ,dim. of corona a crown .] A truth easily deduced

from a proposition .

Curv ilinear . [L . curvus curved, Iinea line .] Bounded

by curved lines.

Cy lind er . [Gr . kylindros, from kyliein to roll.] A round

body with uniform diameter and parallel and equal bases.

Decagon . [Gr . delta ten , gonia an angle ] A polygon

of ten sides.

Deg ree . [L. de down , gradus a step ] One three hun

dred and sixtieth part of a circumference .

Diagonal . [Gr . dict through , gonia an angle ] A line

of a polygon connecting two angles not consecutive .

D iameter . [Gr . dia through ,matron a measure ] A

straight line drawn through the centre of a circle and termi

nating on both sides in the circumference .

Dih edral . [Gr . di two, hedra a base ] Having two

plane faces .

Dimens ion . [Gr . (ii apart , meliri to measure ] A

measure in a single line ; as length , width , etc.

390 GLOSSARY.

Isosoeles . [Gr . i308 equal, ehelos leg ] Having two

legs or sides equal.

Lemma . [Gr . lemma a thing taken for granted ] An

auxiliary proposition .

Locus . [L. locus a place ] In general, a point , line, or

surface which contain s all points having a common property ,

and no others .

Lune. [L. luna the moon .] A portion of the surface of a

sphere included between two semi - circumferences of great

circles.

Max imum [L.mar imus greatest , pl. maxima , superlative

of magma great .] T he greatest quantity or value attainable

in a given case .

M ed ian . [L. med ias the middle ] Running through the

middle .

M inimum. [L. minimus least , pl. minima ,superlative of

parvus little ] T he least quantity or value attainable in a

given case .

Mix t ilinear . [L. miscere to mix ,linea a line.] Con

taining straight and curved lines.

Nonagon . [L. norms n inth ,Gr . gam

'

a an angle ] A

polygon of nine sides . See Enneagon .

Obtuse . [L. obt'usus blunt , from ob upon ,tundo

,tusum

to strike ] Not sharp . An obtuse angle is one that is greater

than a right angle .

Oct ag on . [Gr . octa eight , gonia an angle ] A polygon

of eight sides .

Oct ah ed ron . [Gr . octa eight , hedra a base ] A poly

hedron bounded by eight planes.

Parallel. [Gr . parallelos parallel, from para beside,

allelon one another ] Extending in the same direction .

Parallelog ram. [Gr . parallelos parallel, gramma

line .] A quadrilateral whose opposite sides are parallel.

Parallelop ip ed . [Gr . parallelelos parallel, epi upon ,

pedon the ground] A prismwhose base is a parallelogram.

Pentag on . [Gr . pente five, gonia an angle ] A poly

gon of five sides .

Pen t ed ecag on . [Gr . perue five, delea ten , gonia an

angle ] A polygon of fifteen sides.

GLOSSARY. 391

Perimmer . [Gr . per i around,metron ameasure ] T he

distance around a figure .

Perpend icu lar . [L. perpendiculum a plumb- line, from

per through , pendere to hang ] A line or plane fallingat right angles upon another line or plane .

[Gr . rr pi .] T he in itial letter Of the Gr . periphe reia

circumference . Used to designate the ratio of the circumference of a circle to its diameter .

Poly g on . [Gr . polus many , gonia an angle ] A plane

figure bounded by straight lines.

Poly h ed ron . [Gr . polus many , hedra a base ] A solid

bounded by planes .

Postu late . [L. postulatum a demand] T o assume with

out proof.

P r ism. [Gr . pr isma something sawed ] A solid whose bases

are equal and parallel polygons.

Prismoid . [Gr . p risma something sawed , eidos form.]A body that approaches the form of a prism .

Proj ect ion . [L. pro forth, j acere to throw .] T he act

of throwing forward .

Py rami d . [Gr . puramis a pyramid] A solid bounded by

a polygon and triangles having a common vertex .

Quad ran t . [L. quadrans a fourth part .] T he quarter of a

circle , or of the circumference of a circle .

Quadr ilateral . [L. quatuor four, latus, lateris a side ]A polygon Of four sides.

R ad ius . [L. radius a spoke of a wheel.] A line drawn

from the centre of a circle to the circumference.

R at io (ra’shi - o or ri

’sho ) . [L . ratus to reckon .] T he

relation which one quantity has to another of the same

kind .

R ecip rocal . [L. reciprocas return ing ] Mutually inter

changeable . Unity divided by the quantity .

R ectang le . [L. rectus right , angulus an angle ] A

parallelogram whose angles are right angles

R ect ilinear . [L. rectus right , linea a line .] A plane

figure bounded by straight lines.

Rhombu s . [Gr . rhombu s,from rhombein to turn round .]

An Oblique- angled equilateral parallelogram.

392 GLOSSARY.

S calene . [Gr. shalenos unequal ] Unequal. A scalene

triangle is one in which no two of the sides are equal.

S cholium. [Gr . whole a learned discussion .] A remark

upon one or more propositions .

S ecant . [L. secure to cut .] A line that cuts another .

S ector . [L. w e re to cut .] A part of a circle compre

hended between two rad ii and the included are.

S egment . [L. secure to cut .] A part cut off from a figure

by a line or plane .

S emicircle . [L. semi half,

circular a circle ] T he halfof a circle .

S ph ere . [Gr . spha ira a ball.] A solid bounded by a curved

surface every point of which is equally d istant from a point

within called the centre .

S ub tend . [L. sub under, tendere to stretch .] T o ex

tend under , as the chord which subtends an arc.

S uperposi t ion . [L. super over , ponere to place ] Theact of superposing ; being placed upon something .

S upp lemen t . [L. sub under, p lere= to fill.] T he differ

ence between an are or angle and 180° or two right angles.

S ynthes is . [Gr . sun with ,tithenai to place ] A put

ting together .

T ang ent . [L. tangere to touch ] A right line wh ich

touches a curve .

T etrah ed ron . [Gr . tetra four, hedra a base ] A poly

hedron bounded by four bases

T ransverse ]. [L. trans across, vertere to tur n.] A

line which intersects any system of other lines .

T rap ez ium. [Gr . trapezia a table, from tetra four , pour

foot .] A quad rilateral none of whose sides are parallel.

T rape zoid . [Gr . trapez ia a table , eidos form .] A quad

rilateral two of whose sides are parallel.

T r ig on . [Gr . tr i three, go nia an angle ] A triangle.

T runca te . [L. truncare to cut T o cut off ; as a trun

cated cone .

U nd ecag on . [L. undecim eleven , Gr . gonia an angle ]A polygon of eleven sides .

Zon e . [Gr . zone a girdle ] A portion of the surface of a

sphere included between two parallel planes.

394 B IOGRAPHICAL NOTES .

lived at Alexandria, probably in the 4th century A. D. T he t e

puted inventor of Algebra .

Euclid Lived at Alexandria about 300 B . c. An

eminent writer on Geometry . His principal work is the Ele

ments of Geometry, in thirteen books, parts of which have been

used as a text- book for elementary geometry down‘

to the

present time . It is said of him that when Ptolemy asked

if there was not some easier way of learning Geometry ,he

replied ,

“There is no royal road to Geometry .

”T heorem VI II .

,

Book IV. , is the 47th proposition of the first book of Euclid’s

Elements. See Note, 4236 .

Euler Leonhard . Born 1707 ; died 1783 . A Swiss

scholar, and one of the most eminent mathematicians of mod

ern times . T heorem XI I .,Book IV .

,is due to Euler .

Hippocrates (hip -

pok’re- tez) . Born about 460 B . c. ; died

about 377 B . c. An eminent Greek mathematician . T he first

to effect the quadrature of a cu rvilinear figure by finding a

rectilinear one equal to it . He was also the author of the first

elementary text- book on Geometry . Exercise 344 is due to

H ippocrates.

K ep ler J ohann . Born 1571 ; died 1630. A Ger

man mathematician and physicist , celebrated for his work on

Astronomy . He was the first to introduce the idea of infinity

into the treatment of Geometry . He regards the circle as com

posed of an infinite number of triangles, the cone as composed

of an infinite number of pyramids, etc .

Leg endre (le A d r ien M arie . Born at T oulouse,

1752; died at Paris, 1833 . A celebrated French mathematician .

T he author of the most celebrated work on Geometry written

during the last century . I t is called Elénwnts de Geometr ic, and

was published in 1794.

Le ib n it z Gott fried W ilh elm. Born 1646 ; d ied

1716 . A celebrated German philosopher and mathemat ician .

T he founder of the I nfinitesimal Calculus.

N ew ton , S ir Isaac . Born at Woolsthorpe, 1642 ; died at

Kensington ,1727 A famous English mathematician and

natural philosopher . In 1664 he discovered the Binomial Theo

rem,and

,about 1665 , the Difl

'

erential Calculus. H is greatest

work is the Principia .

B IOGRAPHICAL NOTES . 395

Pappu s Lived about the close of the4th century .

An Alexandrian geometer . His chief mathematical work was

called The Colled ion.

Pascal Bla ise . Born at Clermont - Ferrand , 1623 ;

died at Paris, 1662. A celebrated French geometrician , philoso

pher , and physicist . At the age of sixteen he ach ieved renown

by his T reatise on Conic Sections. Later in life he undertook and

solved many of the most difficult problems of mathematics .

P lato Born at E gina, 429 or 427 B . c. ; died at

Athens, 347 B . c. An eminent Greek philosopher and mathe

matician . He made mathematics the basis of his teaching .

Conic Seaionswere first studied in his school. H e gave a simpleand elegant method for the dup lication of the cube and for the tr i

section of an ang le. Plato was the first to limit Plane Geometry

to the useBf the S traight Line and the Ci rcumference.

Ptolemy (tol’e-mi) , C lau d iu s P tolemmu s . Born 87 A. n . ;

died 165 A. D. One of the greatest astronomers, geographers,

and mathematicians of the late Greeks. T he Ptolemaic system

of astronomy , wh ich places the earth at the centre of the uni

verse, is named in his honor . In Geometry Ptolemy has been

ranked fourth among the ancients, after Euclid , Apollon ius,

and Archimedes. He fixed the approximate value of 7r at

or , in decimals,

T heoremXXV . ,Book IV .

,is due to him.

Py thagoras (pi - thag’o- ras) . Born about 582B . C . died about

500 B . c. A famous Greek philosopher and mathematician .

He is supposed to have discovered the following theorems

only three planefigures can fill up the space around a point the sum

of the angles of a p lane triang le equals two right ang les (See T heorem XVI I . , Book the circle is greater than any other p lane

figure of equal area (See é360) ; and the celebrated proposition

of the squa re on the hypotenuse (See T heorem VI I I .,Book

T h ales Born at Miletus about 640 B . c. ; died about

546 B . C. A famous Greek philosopher , astronomer , and geom

eter ; one of the seven wise men of Greece . He introduced

Geometry into Greece from Egypt . He is said to have de

termined the distance of vessels from the shore by Geometry ;

measured the height of the pyramids bymeans of their shadows ;

and discovered that all angles in a semicircle are right angles

(See Q T heorem I .,Book I I I .

,is due to T hales.

I ND E X .

[T he numbers refer to the page ]

Assnm arxoxs, 18 Conical surface, 317

Acute angle, 15 Conjugate points, 200Acute -angled triangle, 40 Consequent, 74

Adjacent angles , 15 Constant quantity , 83

Alternate angles, 31 Continued proportion, 75

Alternation, 75 Converse, 16

Altitude, 41, 54,270, 318, 328 Convex , 61

Angle, 14 Corollary , 16

Angle at the centre, 205 Couplets, 74

Antecedent, 74 Cube. 269, 271

Apothem,205 curve, 14

Arc, 88 Curved surface, 16

Area, 143 Curvilinear figure, 16Axial section, 311 Cylinder, 310

Axioms, 16 , 17, 89 Cylindrical surface, 310

Axis of cone, 318

DEGREE,110

Bass ,41

, 53, 54, 317 Diagonal, 54, 61, 269

Ri - rectangular triangle, 350 Diameter, 88, 327Bisectors, 41 Dihedral, 253

Broken line, 14 Directri x,310

,317

Division, 75

CENT RAL ANGLE, 89 Dodecahedron, 269

Centre, 88, 205, 327

Chord , 88 Evens, 253, 269

Circle, 88, 327 Element, 310, 317

Circumference, 88 Equal polygons, 143

Circumscribed , 89, 311, 319 Equiang ular, 61

Commensurable, 81 Equilateral, 40, 61

Complement, 15, 111 Equivalent polygons, 143

Composition , 75 Excess, spherical, 351

Concave, 61 Exterior angles, 31, 41

Conclusion , 16 Extreme and mean ratio, 186

Cone,317 Extremes, 74.

396

398 INDEX.

Rectangular triangle, 350

Rectilinear figure, 16Reductio ad absurdum

, 30

lie- entrant, 61

Regular polygon , 202

polyhedron , 305

pyramid, 286

Rhomboid , 53

Rhombus , 53

R ight angle, 15

R ight- angled triangle, 40

R ight cylinder, 311

Right section,

SCALENE , 40Scholium. 16

S ecant , 31, 89

Sector, 89

Segment, 89

Semicircle, 88

Semicircumference, 88

S imilar arcs , 143

cones , 318

cylinders, 311

polygons, 143

polyhedrons, 298

S lant height, 286, 318

Small circle, 327

Solid , 13 UNrr, 143, 269

Sphere . 327

Spherical angle, 3 9 VARIABLE QUANT ITY, 83exces s, 351 Vertex , 14, 269, 285

polygon, 339 Vertical angle, 15

pyramid , 340 Volume, 269

sector, 328

segment. 328 WEDGE , 371

surface, 339

triangle, 339 ZONE, 328

Spherical wedge, 3 9

Square , 53

S traigh t line, 14

Superposition, 20

Supplement , 15, 111

Surface,13

,16

Symbols. 17

Symmetrical, 263 , 351

T ABLm, 382, 383

T angent, 88, 311

T etragon ,61

T etrahedron , 269, 305

T heorem. 16

T heory of limits, 83T h ird proportional, 75

T ransversal, 31

T rapez ium, 53

T rapezoid , 53

T riangle, 40

T rigon , 61

T rihedral, 263

T ri- rectangular triangle,

T risect, 140

T runcated cone, 318

pyramid , 286

prism,270

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