exercises in euclid - forgotten books

KEY

EXERC I SES IN EUCLID .

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HONORARY FELLOW OF ST J OHN’S COLLEGE , CAMBR ID GE .

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AT THE UN IVER SITY PRE SS .

THE Keys already issued to som e of the A uthor’s work s

have b een found very useful by affording assistanc e to

private students,and by saving the lab our and tim e of

tea chers ; and this ha s led to the issue of the present

volum e . C a re ha s b een taken,a s in the form er Keys, to

present the solutions in a sim ple na tura l m anner , in order

to m eet the difficulties which a re m ost likely to arise,

and to render the work inte l lig ib le and instruc tive .

Novem ber , 1880.

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H 1 to 15

16 to 2627 to 3132

35 to 4546 to 481 to 1112 to 141 to 15

16 to 1920 to 2223 to 3031

32 to 3435 to 371 to 45 U) 9

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4 to 67 to 1819 to I)1 to 1213 to 211 to 481 to 141 to 371 to 161 to I)

MI SCELLAN EOU S

1 1 to 16

C ON TEN TS

PAGE

EXERC ISES,I N EU CL I D .

I . 1 to 15 .

1 . LET AB be the given straight line on which the isosceles triangleis to be constructed ; let D E be the straight line to which each side is to b eequal . With centre A

,and radius equal to D E

,describe a circle ; with

centre B and radius equal to D E,describe another circle ; let these circles

intersect at 0. Join A G and BG then ABO will be the triangle required .

2 . The given point and the vertex of the constructed triangle both fallon the circumference of the small circle .

3. Let AB and GD be two straight lines which bisect each other at rightangles at the point 0 so that A O is equal to OB , G0 is equal to OD , andthe angles at 0 are right angles . In GD take any point E ,

and join EA andEB then EA shall be equal to EB.

For AO is equal to B0 by hypothesis ; E0 is c ommon to th e two triangles A OE and BOE and the angle A GE is equal to the angle BOEby A xiom 1 1 . Therefore EA is equal to EB

,by I . 4.

Similarly it may be shewn that any point in AB is equally distant fromG and D.

4 . The angles AB0 and A GB are equal by I . 5 . Hence the angles DB0and D GB are equal by A xiom 7 . Therefore the sides DB and DC are equalby I. 6.

5 . The angle DBA is half the angle ABG, by construction . The angleBAD is equal to half the angle ABG,

by hypothesis. Therefore the angleDBA is equal to the angle BAD . Therefore BD is equal to AD by I . 6 .

6 . It is shewn in the demonstration of I. 5 , that the angle BGFis equalto the angle GBG ; therefore EH is equal to OH , by I . 6 . A lso it is shewnthat F0 is equal to GB . Therefore EH i s equal to GH by A xiom 3 .

7 . AFis equal to AG , by construction ; AH is common to the twotriangles FAH and GAH and EH is equal to GH , by Exercise 6 : thereforeth e angle FAH is equal to the angle GAH by I . 8 .

8 . AB is equal to AD , by hypothesis ; A G i s common to the two triangles BA G, D A G ; and the angle BA G is equal to the angle D A G,

byhypothesis : therefore the base BC is equal to the base D O,

and the angleA CE i s equal to the angle A GD by I. 4.

T o EX . EUC

2 EXERCISES IN EUCLID .

9 . The angle A GB is equal to the angle BDA by I. 8 ; and then the twotriangles A GB and BDA are equal in all respects by I . 4 ; so that the angleABC is equal to the angle BA D . Therefore A O is equal to B0 by I. 6 .

10. Let ABCD be the rhombus, so that AB ,BC , CD,

D A are all equal.Join BD . Then in the two triangles BAD , BGD the base BD is commonand the two sides BA

,A D are equal to the two sides BC ,

CD , each to eachtherefore the angle BAD is equal to the angle BGD , by I. 8 . Sim ilarly itmay be shewn that the angle ABC is equal to the angle A D C .

11 . Let ABCD be the rhombus , so that AB,BC

,GD , DA , are all equal .

Join BD . Then in the two triangles ABD,GED , the side AB is equal to the

side GB, th e side BD is common , and the base A D is equal to the base GDtherefore the angle ABD is equal to the angle CBD by I . 8 . Thus the angleABC is bisected by BD . Similarly it may be shewn that the angle ADC isbisected by BD ; and also that the angles BA D andBGD are bisected by A G.

12 . Let there be two isosceles triangles A GB, ADB on the same baseAB , and on opp osi te sides of it. Join GD then CD shall bisect AB at rightangles .In the two triangles A CD , BGD the side CD is common ; A C i s equal

to BC , by hypothesis ; and the base AD is equal to the base BD,by

hypothesis therefore the angle A GD is equal to the angle BGD by I . 8 .

Let AB and CD intersect at E . Then in the triangles A GE,BGE the

side GE is common ; the side A G is equal to the side BC ; and the angleA GE has been shewn equal to the angle BGE : therefore the triangles areequal in all respects by I . 4 . Thus AB is equal to BE and the angle ABCis equal to the angle BEG, so that each of them is a right angle .Next let the two isosceles triangles A GB, ADB be on the same base AB ,

and on the sam e side of it . Join CD and produce it to meet AB at E . Itmay be shewn as before that AB is equal to BE , and that the angles atE are right angles.

13. Let AB be the given straight line , C and D the two given points.Join CD and bisect it at E . From E draw a straight line at right angles toCD , meeting AB at F. Join OF, DE. Then CFshall be equal to D E.

For in the two triangles GEF and D EF, the side EF is common , CEi s equal to D E , and the right angle GEF is equal to the right angle D EFtherefore CFis equal to BE, by I . 4.

The problem is impossible when the two points G and D are situated onthe same perpendicular to the given straight line AB

,and at unequal dis

tances from that straight line .

14. Let AB be the given straight line , C and D the two given points.From D draw D E perpendicular to AB , and produce D E through E toa point F such that EFis equal to ED . Join CF

,and produce it to meet

AB at G join D G then CG and D G shall be the required straight lines .For ED i s equal to EF, and EG is common to th e two triangles ED G

and EFG ; the right angles GEE and GED are equal therefore by I . 4 thetriangles EEG and DEG are equal in a ll respects , so that the angle EGEi s equal to the angle D GE .

The problem is impossible wh en the two points C and D are equallydistant from the straight line AB , and not on the same perpendicular to AB .

EXERCISES IN EUCLID . 3

If C and D are equally distant from AB , and on the same perpendicular , thenany point in AB may be taken for the point G .

15 . Let the angle BA G be bisected by AD,a nd the angle BAG by AE

the angle D A E shall be a right angle .Since the angle BAD is half the angle BA G, and the angle BAE is

half the angle BA G,the two angles BAD and BAE together are half the

two angles BA G and BAG together. But the angles BAG and BAG togetherare equal to two right angles

,by I . 13 therefore the angles BAD and BA E

together are equal to a right angle .

16 . Let the four straight lines AE , BE , GE, DE meet at th e point E ,

and make the angle A EB equal to the angle GED , and the angle BEG equalto the angle DEA : then shall AE and EG be in one straight line , and alsoBE and ED in one straight line .By I . 15 , Cor . 2

,the four angles AEB, BEG, GED , DEA are together

equal to four right,

angles ; but the two angles ABB and BEG are equalto the two angles GED and D EA ; therefore the angles AEB and BEG aretogether equal to two right angles ; therefore A E and E G are in one straigh tline by I . 14. Similarly it may be shewn that BE and ED are in onestraight line.

I . 16 to 26.

17 . The angle BD A is greater than the angle CAD ,by I . 16 the angle

CAD is equal to the angle BAD , by hypothesis : therefore the angle BDA isgreater than the angle BA D . Therefore the side BA is greater than the sideBD , by I. 19. Similarly it may be shewn that GA i s greater than CD .

18. Take any point G in BC , and j oin A G . The angle A GG is greaterthan the angle ABC ,

by I . 16 and the angle A GB is greater than the angleA GB, by I . 16 . T herefore the angles ABC and A GB are together less thanthe angles A GC and AGB together ; therefore the angles ABC and A GB aretogether less than two right angles by I. 13.

19 . Join BD . The angle ABD is greater than the angle A DB,and the

angle BBC is greater than the angle BBC , by I . 18 ; therefore the wholeangle ABC is greater than the whole angle A D C . Similarly by joining AGwe can shew that the angle D CB is greater than the angle D AB.

20. Let ABCD be the square ; on B0 take any point E ; j oin AE andproduce it to meet D C produced at F then shall AF be greater than A C .

The angle D CA i s greater than the angle OFA by I. 16. The angle A GF,

which is greater than a right angle , is greater than the angle D CA , which i sless than a right angle. Therefore the angle A CF is greater th an the angleAFC . Therefore AF is greater than A G, by I. 19 .

21 . Let AB be the given straight line,0 the given point without it .

From 0 draw 00 perpendicular to AB then 0G shall be shorter than anyother straight line 0D drawn from 0 to A B.

For the angle C CD is a right angle ; therefore the angle OD O is less thana right angle, by I. 17 therefore 0G i s less than 0D ,

by I. 1 9 .

Next, let OE be a straight line drawn from 0 to AB ,and more remote

from 0G than CD is : 0D shall be less than OE .


For the '

angle OD O i s greater than the angle OEC ,by

~ I . 16 ; and theobtuse angle ODE is greater than the acute angle OD C : therefore the angleODE i s greater than the angle OED . Therefore OD is less than OE

,by

I. 19 .

Lastly, from C on the straigh t line AB take CF equal to CD,and on the

other side of G j oin OF. Then OF is equal to OD by I . 4 . And no otherstraight line can be dr awn from O to AB equal to OD

,besides OF.

For if this straight line were nearer to 00 than OD o'

r OF is,it would

be less than OD , and if it were more remote from 00 than OD or OF is, itwould be greater than OD .

22 . Let ABC be a triangle , and 0 any point. Join OA , OB, 00. Then0A and OB are together greater than AB ,

OB and 00 are together greaterthan BG

,and 00 and 0A are together greater than CA ,

by I. 20. Hence '

twice the sum of 0A , OB , and OG is greater than the sum of AB , BC, andGA ; therefore the sum of 0A ,

OB,and 0 0 is greater than half the sum of

AB, BC , and GA .

23. Let ABCD be a quadrilateral figure . D raw the di agonals A G andBD . Then AB and BC are greater than A C

,and AD and DC are greater

than A C therefore the four sides AB,BC, GD , DA are greater than twi ce

A G. Similarly it may be shewn that the four sides are greater than twiceBD . Hence twice the sum of the four sides is greater than twice the sum ofthe diagonals ; therefore the sum of the four sides is greater than the sum ofthe diagonals .

24. Let ABC be a triangle,and D the middle point of the base BC join

AD : then AB and A G together shall be greater than twice AD.

Produce AD to a point E so that D E may be equal to AD ; join BE .

In the two triangles ADC and EDB the two sides AD , DC are equal to thetwo sides ED , DB each to each ; and the angle ADC is equal to the angleEDE , by I. 1 5 therefore A C is equal to BE

,by I . 4 . The two sides AB

,BE

are greater than AE , by I. 20 ; therefore the two sides AB , AC are greaterthan AE , that is greater than twice AD .

25 . Let ABC be a triangle in which the angle G is equal to the sumof the angles A and B . A t the point C in the straight line A G make theangle A GG equal to the angle A , and let GG meet AB at D . Then thetriangle A GD is isosceles, by I . 6 . A lso as the angle A GB i s equal tothe sum of the angles A and B , and the angle A GG i s equal to the angle A ,

the angle BCG is equal to the angle B : therefore th e triangle BGD i sisosceles

,by I . 6 .

26 . It is shewn in the preceding Exercise that A GD i s an isoscelestriangle

,having AD equal to CD ; also that BGD is an isosceles triangle

having BD equal to CD . Hence , as AD and BD are each equal to CD,the

point D is the middle point ofAB , and AB is equal to twice GD .

27 . Let AB be the given base ; at the point A make the angle DABequal to the given angle . From AD cut off AE equal to the given sum

of the sides,j oin EB. A t the point B make the angle EBFequal to the

angle A EB,and on the same side of EB ; let BF meet AE at G : then A GB

will be the triangle required .

For,since the angle BBC is equal to the angle BEG

,the sides EC and

BC are equal,by I. 6 . Therefore the sum of AG and GB is equal to the sum

EXERC ISES I N EUCLID . 5

of A G and CE , that is to the given sum of the sides. A l so the base AB andthe angle BA G have the required values.

28 . From any point D in the straight line bisecting the angle A ofa triangle draw D E perpendicular to the side AB , and D E perpendicularto the side AC. Then in the triangles D AE and D AF

,the side D A i s

common ; the angle D AE is equal to the angle D AF,by hypothesis ; and the

right angles DEA and DEA are equal :.

therefore D E i s equal to BE by I . 26 .

29 . Let AB be the given straight line in which the point is to be found .

Let CD and EFbe the other two straight lines , and let them meet , producedif necessary, at 0 . Through O draw a straight line b isec ting the anglebetween CD and EF and let AB , produced if necessary, meet this straightline at K then K will be such a point as is requir ed : for the perpendicularsfrom K on the straight lines CD and EFmay be shewn to be equal in themanner of the preceding Exercise .Two straight lines can be drawn thr ough 0 bisecting angles formed by

the given straight lines , so that in general two solutions of the problem canb e obtained , but there will be only one solution if AB is parallel to either ofthe bisecting straight lines.If CD and EFare parallel the construction fails . We must then draw a

straight line KL parallel to CD and EF,and midway between them : the

intersection of this straight line‘with AB,produced if necessary

,will be the

required point . But if KL is parallel to AB there will be no solution .

30. Suppose A the given point through which the straight line is to bedrawn , and B and G the other given points from which perpendiculars areto be drawn . Join BC, and bisect it at D ; j oin AD : this shall be therequired straight line .For draw BE and CF perpendicular to AD

,produced if necessary. Th en

in the triangles BDE , CDF the sides BD and CD are equal ; the anglesBD E and GDF are equal, by I . 1 5 ; and the right angles BED and GEDare equal : therefore BE is equal to CF

,by I . 26.

31 . In the triangles A DB,A DE the side AD is common ; the ang les

BAD and EA D are equal by hypothesis ; and the right angles ADB andAD E are equal : therefore BB is equal to ED , by I . 26 .

32 . Bisec t the angle BAG by the straight line AD ; from P draw PGperpendicular to AD, and produce PG both ways to meet AB at E ,

andAC at F : then AE will be equal to AF.

For in the two triangles A GE,A GE the side AG is common the angles

EA G and FAG are equal,by construction ; and the right angles EGA and

FGA are equal : therefore AE is equal to AF,by I . 26 .

33. Let ABC be a triangle having the angle B a right angle,and let

D EFbe a triangle having the angle E a right angle ; also let A C be equal toBE

, and AB equal to D E : then shall the triangles ABC and DEF be equalin all respects .Produce GB to G , so that BG may be equal to BE and join A G . Then

the angle ABG is a right angle,by I . 1 3, and is therefore equal to the angle

D EF; also the sides AB , BG are equal to the sides D E , EF each to each :therefore the triangles ABG,

D EFare equal in a ll respects, so that AG is

6 EXERC ISES IN EUCLID .

equal to BE. But AC is equal to D E by hypothesis ; therefore A C is equalto AG, and the angle A GG i s equal to the angle A GG. Therefore the twotriangles ABC , ABG are equal in all respects, by I. 26 . But the trianglesABG, DEFwere shewn to be equal in a ll respects ; therefore the trianglesABG, DEFare equal in all respects.

I . 27 to 31 .

34. Let ABC be a triangle having the sides AB and A C equal . D rawany straight line parallel to BC, meeting AB at D , and A C at E then theangle AD E shall be equal to the angle AED .

For the angle AD E is equal l

to the angle ABC ,and the angle AEP

i s equal to the angle A GB, by I . 29. But the angle ABC is equal tothe angle A OB

,by I. 5 . Therefore the angle AD E is equal to the angle

AED .

35 . Let the straight lines A and B meet at K, let the straight lines Gand D meet at L , and let the straight lines A and D meet at M. The acuteangle at K is equal to the acute angle at M, by I. 29 ; and the acute angleat M is equal to the acute angle at L also , by I . 29. Therefore the acuteangle at K i s equal to the acute angle at L .

36 . Let the straight line AB be terminated by two parallel straightlines. Let G be the middle point of AB ; through C draw any straight lineD CE , terminated by the same parallel straight lines a s AB , so that AD isparallel to EB. Then will ED be bisected a t G.

For in the two triangles A CD,BGE the two sides A C, BC are equal by

hypothesis ; the angles A CD , BGE are equal by I . 1 5 ; and the angles CA D ,

GBE are equal by I . 29 therefore the triangles are equal in all respects byI . 26, so that CD is equal to CE .

37 . Let 0 be a point equi distant from two parallel straight lines ;through 0 draw one straight line AOB terminated by the parallels , and alsoanother straight line COD terminated by the parallels, so that A and C areon one of the parallels

,and B and D on the other : then will A C be equal

to BD .

Since the given straight lines are parallel, a straight line can be drawnthrough 0 to meet the parallels at right angles , and this straight line will bebisected at 0 because 0 is equidistant from the parallels, by hypothesis .Therefore by Exercise 36 the straight lines AB and DC are bisected at0. Thus in the two triangles A OG, BOD the two sides A O, 00 are equalto the two sides BO

,OD each to each ; and the angle A OG is equal to

the angle BOD ,by I . 15 : therefore AC is equal toBD , by I . 4 .

38 . Let ABC be a triangle : produce BA to D ; suppose that AEbisects the angle D A C

,and that it is parallel to BC : then will ABC be

an isosceles triangle.For since AE i s parallel to BC the angle D AE is equal to the angle

ABC , by I. 29 , and also the angle GAE i s equal to the angle A GB,by

I. 29. But the angle D AE is equal to the angle EA G, by hypothesis ;therefore the angle ABC is equal to the angle A GB : therefore the sideAB is equal to the side A G, by I . 6.

EXERCISES I N EUC LID . 7

39 . Take any point E in DC, and at the point E make the angle CEFequal to the given angle . Through A dr aw a straight line parallel to FE

,

and meeting CD at B then B is the required point .For the angle ABC is equal to the angleFE C by I . 29 and therefore the

angle ABC is equal to the given angle .

40. Let ABG be a triangle ; let a straight line be dr awn bisectingthe angle A

,and meeting BC at D. From D draw a straight line parallel

to AB,meeting A C at F, and also a straight line parallel to AC , meeting

AB at E then D E shall be equal to DE .

For in the triangles ABD,AFD the side AD is common ; the angle

EAD is equal to the angleFAD,by construction ; the angle ED A is equal

to the angle D AF, and the angle FDA to the angle D AE , by I . 29 ; sothat the angle ED A i s equal to the angle FD A ; hence the two trianglesare equal in all respects by I. 26 . Thus D E is equal to D E.

41 . The angle FEG is equal to the angle E CB, by I. 29 ; the angleECB is equal to the angle E CF, by hypothesis ; therefore the angle FECis equal to the angle E CF. Therefore EF is equal to EC , by I . 6 . A gain,the angleFGC is equal to the ang le GCD ,

by I , 29 ; the angle GGD is equalto the angle FGG, by hypothesis ; therefore the angle FC G is equal tothe angle FGC . Therefore FG is equal to FC

,by I . 6 . And i t has been

shewn thatFE is equal to FC therefore EFis equal to FG .

42 . Bisect the angle ABC by a straight line meeting A C at E throughE draw a straight line parallel to CB meeting AB at D : then D shallbe the point required .

For the angle D EB i s equal to the, angle EBC ,by I . 29 ; the angle

DBE is equal to the angle EBC, by construction ; therefore the angle D EBis equal to the angle D EE therefore DB is equal to D E , by I . 6 . And

the angle D EA i s equal to the angle BCA by I . 29 , and is therefore aright angle ; so that D E is perpendicular to AC.

43 . Bisect the angle ABC by a straight line , meeting A C at E ;through E draw a straight line parallel to BC, meeting AB at D. Thenwill BD , D E , EG be a ll equal.For the angle D EB is equal to the angle BBC , by I . 29 ; the

angle EBC i s equal to the angle DBE by construction ; therefore theangle D EB i s equal to the angle D BB. Therefore DB is equal to D E ,

by I . 6 . A gain , the angle A DE is equal to the angle ABC , and the angleAED is equal to the angle A GB, by I. 29 ; also the angle ABC isequal to the angle A GB : therefore the angle ADE is equal to the angleAED . Therefore AD is equal to A E , by I. 6 . But the whole AB isequal to the whole AC ; therefore DB is equal to EC . Thus BD , D E,

EC

are a ll equal.

44. From A draw a straight line bisecting the angle BAC , and meetingBC at F. Then the triangles BAF and CAF are equal in all respects

,

by I. 4 ; so that the angles AFB, AFC are equal , and therefore eachof them is a right angle. Therefore AF is parallel to ED

,by I. 28 . The

angle AED is equal to the angle GAF,and the angle ED A is equal to

the angle BAF, by I. 29 .

‘ But the angle BAF is equal to the angle GAF,

by construction ; therefore the angle AED is equal to the ang le ADE :

therefore AB is equal to AD,by I. 6.

8 EXERCISES IN EUC LID .

I . 32 .

45 . Let ABC be a triangle having the sides AB and A C equal . FrofnB draw a perpendicular to AC meeting AG at D , and from C draw aperpendicular to AB meeting it at E . Then each Of the angles CBD ,

BGEwill be equal to half the angle A .

Bisect the angle A by a straight line meeting the base BC at F.

Then the triangles BAF, CAF are equal in all respects, by I . 4 ; andtherefore the

‘

angle AFB is a right angle . Then in the triangles BAF,

BGE the angle ABC is common ; the right angle AFB is equal to th eright angle CEB therefore the third angle BAF is equal to the third angleBGE , by I . 32 . Similarly the angle CAF is equal to the angle CBD .

46. A C i s equal to CE , and BC is equal to CD , by Construction .

The angle A CE is equal to the angle BCD,each being one third of two

right angles,by I . 32 ; to each of these add the angle A GB : therefore the

whole angle A GD is equal to the whole angle BGE . Thus in the twotriangles A CD , ECB the two sides A C, CD are equal to the two sides E C ,

CB each to each ; and the included angles are equal : therefore AD is equalto BE . Similarly AD is equal to FC

47 . The figure has eight equal sides, and eight equal angles : the in

ter ior angles of the fi gh re together with four right angles are equal tosixteen right angles therefore all the interior angles of the figure are equalto twelve right angles , by I . 32

,C orollary Hence each angle is twelve

eighths of a right angle,that is, a right angle and a half.

48 . Let A and(B be the two g iven points, CD the given straight line .

A t the point C make the angle EGD equal to the angle of an equilateraltriangle ; and at the point D make the angleFD C

,

also equal to the angleof an equilateral triangle

,and on the same side of CD as the angle EGD .

_

Through A draw a straight line parallel to E C ,meeting

h

CD , produced ifnecessary , at G : through B draw a straight line parallel to FD meetingCD, produced if necessary, at H. Produce GA and HB to meet at K :

then GHK is the equilateral triangle required .

For since the angle at G is a third of two right angles, and so also isthe angle at H, the angle at K is also a third of two right angles , by I . 32 .

Hence the triangle GEK i s equia ‘ngular,and therefore also equilateral

by I . 6 .

49 . Let ABC be a triangle, having AB equal to A C. Bisect the anglesB and G by straight lines meeting at D . Produce CB to any point E .

The angle DBE i s equal to the two angles BBC ,DCB by I. 32 ; but

DCB is half the angle A GB,and is therefore equal to half the angle

ABC,and is therefore equal to the angle ABD . Therefore the angle

DBE is equal to the two angles ABD and ED G. Take away the commonangle ABD therefore the angle ABE is equal to the angle BD C.

50. The angle A GB is equal to the angle ABC , and the angle A GDis equal to the angle A D C , by I . 5 . Therefore the angles A GB and A GDtogether are equal to the angles ABC and AD C together . But the anglesA GB, A CD , ABC , AD C are together equal to two right angles, by I . 32 .

Therefore the angles A GB and A GD are together equal to one right angle.


5 1 . Produce AB to any point H,and A C to any point K ; bisect

the angle CBH by the straight line BD ,and bisect the angle ECK by the

straight li ne GD : then the angle BBC , together with half the angle BA G,

will make up a right angle.Bisect the angle ABC by the straight line BE , and bisect the angle

A GB by the straight line CE then the angles EBD and EGD will beright angles. For th e angle EBC is half the angle ABC , and the angleCBD is half the angle CBH therefore the angles EBC and CBD aretogether half the angles ABC and CBH , that is equal to a right angle,by I. 13 : thus EBD is a right angle . Sim ilarly EGD is a right angle .The angles BEG, EBO, E CB are together equal to two right angles , byI . 32 ; that is BEG together wi th

'

h alf ABC and half A GB are equal totwo right angles. Join ED . Then BED and EDB are together equal toa right angle , by

‘

l . 32 ; GED and ED G are together equal to a rightangle

,by I. 32 : therefore BEG and BD C are together equal to two right

angles. Thus EEG and BD C are equal to BEG together with half ABCand half A GB. Therefore ‘

BD C is equal to half ABC and half A GB.

Therefore BBC together wi th half BA G is equal to half BAG ,half ABC ,

and half A GB, that is equal to half two right angles , that is equal to aright angle

52 . Let ABC be a triangle . Suppose the angle ABG gr eater than thesum of the other two angles ; then twice the angle ABC is greater than thesum of the angles ABC , BCA ,

CAB,that is greater than two right angles

therefore the angle ABC is greater than a right angle. A gain , suppose theangle ABC equal to the sum of the other two angles ; then twice the angleABC is equal to the sum of the angles ABC , BCA ,

CAB,that is equal to two

right angles therefore the angle ABC is a right angle. Lastly , suppose theangle ABC less than the sum of the other two angles ; then twice the angleABC is less than the sum of the angles ABC , BOA , CAB,

that is less thantwo right angles therefore the angle ABC is les s than a right angle.

53. Construct an equi lateral triangle ABG. Bisect the angle A by astraight line AD, and bisect the angle 0 by a straight line CD . Then AD C.

will be such a triangle as is required.

For the angle DAC, being half the angle of an equilateral triangle, is onesixth of two right angles ; So also is the angle D CA ; therefore the angleADC is four sixths of two right angles

,by I . 32 . Thus the angle ADC

is four times each of the angles DAC, D GA .

54. Since BC is bisected at E the two sides AE , EC are equal to thetwo sidesFE , EB each to each ; the angle AEG is equal to the angleFEB,

by I . 15 therefore the triangles AEG and FEB are equal in all respects , sothat the angle A CE is equal to the angle FBE .

In a similar way by comparing the triangles GGA and HGB , we see thatthe angle GA G is equal to the angle HBG.

Therefore the anglesFBE , EBG,GBH are together equal to the angles of

the triangle ABC , that is to two right angles , by I . 32 . Therefore the anglesABE and ABH are together equal to two right angles : therefore HB and BFare in the same straight line, by I. 14.

5 5 . Take any straight line AB. A t the point A make the angle BADequal to a right angle. Bisect the angle BAD by the straight line AE ; and


’

bisect the angle D AE by the straight line AF ; I. 9 . Then the angle BAF isthree fourths of a right angle . At the point B make th e angle AEG equal tothe angle BAR -and on the same side of AB ; let AF and BG meet at C :then ABC will be the triangle required.

For the two angles ABC and'

BA G are by construction together equal tothree halves of a right angle , therefore the angle A GB is half a right angle byI 32 . Thus the half of the angle A GB is a fourth of a right angle

,and is

therefore equal to one third of each of the angles BA G and ABC.

56 . On AB measure off AD equal to the given straight line . A t thepoint D draw DO making with AD the angle A D Q equal to half the givenangle

,and meeting A C at Q . A t Q draw QP , on the same side of QD as

DA is , making the angle D QP equal to the angle QDP ,and meeting AB at P .

Then P Q is equal to PD , by I . 6 so that AP and PO together are equal toAD , that is to the given straight line . And the angle APO is equal to thesum of the angles PDQ and P QD , and is therefore equal to the given ang le.

57 . Let ABG be a triangle having th e sides AB and A C equal . From Bdraw a straight line making the angle DBG equal to one third of‘

th e angleABG,

on the other side of BC , and meeting A C produced at D . From G

draw a straight line making the angle EGB equal to one third of the angleA GB, on the other side of CB,

and meeting AB produced at E . Let BD and

GE intersect at F.

The triangle BFG has obviously the angles EGFand“

CBF equal , andis therefore isosceles by I . 6 .

The angle BFE i s equal to the sum of the angles BCF and GBE,by I. 32

,

and is therefore equal to two thirds of the angle ABC . The angle BEGis equal to the difference of the angles ABC and BGE

,by I. 32 , and is there

fore equal to two thirds of the angle ABC . Therefore the angles BEE andBEFare equal , and the triangle BEE i s i sosceles , by I. 6 . Similarly thetriangle CFD is isosceles.

58 . The angle AEG is equal to the sum of the angles - ECB and BBCby I. 32 , and so also is the angle DEB therefore the angles ECB and EBCare together half the angles AEG and D EB. The angles E CFand EBFaretogether half the angles EGA and EBD , by construction. Hence the anglesE CB, EBC , EGF, EBE are together half the angles AEG, D EB,

EGA ,

EBD . Take the former sum from two right angles, and the remainder is theangle BFC take the latter sum from two right angles

,and the remainder is

h alf EA G and EDB therefore the angle BFC is half the sum of the anglesEA G, EDB.

59 . Let ABC be a triangle, having the angle A GB a right angle. A t thepoint C draw a straight line CD making the angle A GD equal to the angleGAB

,and meeting AB at D. Since the angle A GB i s a right angle it is

equal to the sum of the two angles CAB and CBA , by I . 32 ; the angleA CD i s equal to the angle CAB,

by construction : therefore the remainingangle BGD i s equal to the angle CBA . Because the angles CAD and A GDare equal

,the sides AD and GD are equal , by I . 6 ; and because the angles

OBB and BGD are equal, the sides BD and CD are equal , by I. 6 . HenceAD

,BD ,

and CD are all equal ; so tha t D is the middle point of AB, and CDis equal to half of AB.

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1 2 EXERCISES IN EUCLID .

not be less than the perpendicular from A on D E ; and if this conditioni s satisfied there will be two intersections of the circle with DE , and thustwo solutions app a rently : but it will be found on examination that thereis only one distinct solution.

66. Take a straight line AD equal to the given difference of the sides .

A t the point D draw a straight line D E making with AD, produc ed thr ough

D , an angle equal to half a right angle . With centre A ,and radius equal

to the given hypotenuse, describe a circle cutting D E at B . From B drawBC perpendicular to AD produced. Then A GB will be such a triangle asis required .

For A GB is a right angle by construction ; and the hypotenuse AB isof the required length . A lso since the angle BCD is a right angle

,and the

angle BD C is half a right angle,the angle DBG is half a right angle,

and is therefore equal to the angle BD C therefore BC is equal to CD .

Thus the difference of the sides is equal to AD,that is to the given difference .

In order that the construction may be possible the given hypotenusemust be greater than the given difference of the sides.

67 . Let AB be the given hypotenuse . Bisect AB at D ; from D drawa straight line at right angles to AB , and on it take D E equal to the givenperpendicular. Through E drawFEG parallel to AB . From centre D , withrad ius equal to AD

,describe a circle cuttingFG at C. Join CA and CB :

then A GB shall be the triangle required .

For the angle A GD is equal to the angle CAD,and the angle BGD is

equal to the angle CBD , by I . 5 ; therefore the whole angle A GB is equalto the sum of the angles CAB and GBA therefore the angle A GB is a rightangle

,by I . 32 . Thus A GB is a right - angled triangle having the given

hypotenuse .From C draw CH perpendicular to AB ; then will CH b e equal to ED .

The angles EDH and GED are together equal to two right angles ; thereforeGH is. parallel to ED ,

by I . 28 ; therefore the angle ED G i s equal to theangle HCD , by I. 29 . Therefore the two triangles ED G and HGD are equalin all respects

,by I . 26 ; so that OH is equal to ED . Thus the perpendicular

from the right angle on the hypotenuse has the given length .

68 . Let D E be the given perimeter, and FGH the given angle . D rawGK at right angles to GH

,and on the same side of it as GF. A t the point

D dr aw DL ,making the angle LDE equal to half the angle FGH at the

point E dr aw EM on the same side of D E as DL ,making the angle MED

equal to half the angle FGK. Let DL and ME meet at 0 . D raw GA ,

meeting D E at‘

A , and -making the angle D GA equal to th e angle CD A ;draw CB

,meeting ED at B

,and making the angle ECB equal to the angle

CEB then ABC will be the required triangle.For the angle GAB is equal to the sum of the angles A CD , AD C ; that

is to twice the angle A D C ; that is to the angle FGH. Similarly the angleGBA is equal to the angleFGK. Thus the two angles GAB and GBA aretogether equal to a right angle ; and therefore the angle A GB is a right angleby I . 32 .

The side AG is equal to AD , and the side BC to BE, by I. 6 . Thereforethe sum of the sides A G, BC, and AB is equal to D E , that is to the givenperimeter.

EXERC ISES IN EUC LID . 13

69. Let BAG be a right angle. On AB describe an equilateral triangleADB bisect the angle BA D by the straight line AE then will the anglesBAE , EAD , D A G be all equal.For the angle BA G i s a right angle ; the ang le BAD i s one thi rd of

two right angles,that is two thirds of one right ang le, by I. 32 ; therefore

the angle CAD is one third of a right angle . And as the angle BAD isbisected by AE the

,

angle BAE is equal to the angle EA D , each beingone third of a right angle . Hence the angles BAE , EAD ,

DAG are allequal , each being one third of a right angle .

70. Let AB be the given straight line. On AB describe an equilateraltriangle ABG. Bisect the angle GAB by the straight line AD ,

and bisect theangle GBA by the straight line BD . Th rough D draw a straight lineparallel to GA , meeting AB at E ; and through D draw a straight line D Eparallel to CB , meeting AB at F. Then will AE ,

EF,FB all be equal.

Since D E is parallel to CA the angle ED A i s equal to the angle D A G,

by I . 29 ; but the angle D AE i s equal to the angle D A G, by constructiontherefore the angles ED A and D AE are equal : therefore A E i s equal toED , by I. 6 . Similarly it may be shewn that BF is equal to FD.

Because D E is parallel to CA the angle D EFis equal to the angle GAB,

by I . 29 . Similarly the angle DFE is equal to the angle GBA . Thereforethe angle EDE is equal to the angle A GB

,by I . 32 . Thus the triangle EDE

is equiangular ; and therefore it is equilateral by I. 6 .

Now AE was shewn to be equal to ED ; therefore AE is equal to EF.

Simi larly BF is equal to FE . Thus AE,EF, FB are all equal .

71 . Let LM and PO be the parallel straight lines ; and A the givenpoint . Suppose A to be between the parallel straight lines . Through Adraw a straight line perpendicular to one of the parallel straight lines , andtherefore also perpendi cular to the other by I . 29 . Let thi s straight lin emeet LM at B

,and PQ at C. From B on LM take BD equal to A C and

from G on PQ take CE equal to AB , and on the sam e side of BC as‘

BDis. Join AD and AE these will be the required straight lines .For the triangles BAD

,GEA are equal in all respects by I . 4 ; so that

AD is equal to AE , and the angle GAE i s equal to the angle BDA . But theangles BAD andBD A are together equal to a right angle

,by I. 32 : therefore

the angles BAD and GAE are together equal to a right angle. Therefore theangle EA D i s a right angle

,by I . 13 .

If A is not between the parallel straight lines,GE and BD must be taken

on opposi te sides of BC .

72 . Let ABC be the g iven triangle , and D E the given perimeter. A t

the point D m ake the angle L DE equal to half the angle ABC ; and at thepoint E , on the same side

'

of ED , make the angle MED equal to half theangle A GB. Let DL and EM meet a t F. From F draw FG ,

meeting D E atG , making the angle DFG equal to the angle FD G ; and from F draw EH ,

meeting ED at H, making the angle EFE equal to the angle FEH . ThenFGH will be the triangle required.

For FG i s equal to D G , and EH is equal to HE ,by I. 6 . Therefore the

sum of the sidesFG, GH , HE is equal to D E ,the given perimeter.

Also the angle FGH is equal to the sum of the angles ED G , DFG ,by

I . 32 ; that is to twice the angleFD G , that is to the angle ABC . Similarlythe angle FHG is equal to the angle A GB. Therefore the angle GFH isequal to the angle BA G, by I. 32.

14 EXERCISES I N EUCLID .

I . 33,‘

34

73. Let ABCD be a quadrilateral having AB parallel to D C ,and AD

equal to BC . Suppose AB less than DC ; from A draw a straight lineparallel to BC

,meeting DC at E . Then ABGE is a parallelogram . There

fore the angle ABG is equal to the angle AEG,and the side AE is equal to

the side BC, I . 34. Therefore AD is equal to AE ,and the angle A D E is equal

to the angle AED . Therefore the angles ABC and A DE are equal to th e

fingles AEG and ABD ; that is they are together equal to two right angles,y I. 13 .

In the same manner it may be shewn that the angles BAD and BCD aretogether equal to two right angles .

74. Suppose that AB and GD are equal but not parallel , and that theangle ABD is equal to the angle CDB. Produce AB and CD to meet at E .

Since the angles ABD and CDB are equal EB i s equal to ED . Thereforealso EA is equal to E C ; and the angle EGA i s equal to the angle EA G.

The two angles EBD and ED B are together equal to the two angles EA Gand EGA , by I . 32 . Therefore the angle EBD is equal to the angle EA G.

Therefore BD is parallel to A C, by I. 28 .

75 . Let ABC be a triangle ; let E be any point in A C, and D any pointin BC then AD and BE will not bisect each other.Let AD and BE intersect at F.

If possible suppose that AF i s equal to FD , and BF equal to FE . Thenthe triangles AFE and DFB are equal in all respects, by I . 4 , so that theangle EAF1 8 equal to the angle BDF. Therefore A E i s parallel to BD , byI. 27 . But this is impossible, since BD and AE , when produced, meet at G.

76 . Let ABCD be a quadrilateral , having AB equal to DC, and ADequal to BC the figure shall be a parallelogram .

Join A C ; then in the two triangles ABC and CD A , the sides BA , A G

are equal to the Sides DC,CA each to each ; and the base BC is equal to

the base DA : therefore the angle BA G is equal to the angle D GA . Therefore AB is parallel to DC, by I. 27 .

Similarly it may be shewn that AD is parallel to BG.

77 . Let ABCD be a quadrilateral such that the angle A is equal to theangle C , and the angle B equal to the angle D : then the figure shall bea parallelogram ,

The angle A is equal to the angle G, and the angle B is equal to the .

angle D ; therefore the two angles A and B together are equal to the twoangles G and D together . But the four angles A , B , G, D together areequal to four right angles, by I . 32 ; since the quadrilateral m ay be dividedinto two triangles by drawing AC or BD . Thus the angles A and B are together equal to two right angles . Therefore AD is parallel to BC, by I . 28 .

Simi larly AB is parallel to DC.

78 . Let ABCD be a parallelogram . D raw the diagonals A G and BDintersecting at E : then AC and BD shall be bisected at E .

In the triangles AED and CEB the sides AD and BC are equal , by I. 34 ;the angles AD E and GBE are equal , by I . 29 ; and the angles D AE and

BGE are equal, by I . 29. Therefore the triangles are equal in all respectsby I . 26 . Thus AB is equal to CE , and DE is equal to BE .

EXERCISES IN EUCLID . 1 5

79. Let ABCD be a quadrilateral figure , and let A G and BD intersectat E ; suppose that A G and BD are bisected at E : then ABCD will be aparallelogram .

In the triangles AED and CEB the sides AE , ED are equal to the SidesGE , EB each to each, by supposition ,

the angle ABD is equal to the angleCEB,

by I. 15 ; therefore the triangles are equal in all respects , by I . 4 .

Thus the angle ADE ' i s equal to the angle CBE therefore AD is parallel toCB

,by I . 27 .

Similarly AB is parallel to D C.

‘

80. Let ABCD b e a parallelogram , and suppose that the straight lineA G bisects the angles at A and G then the four sides of the parallelogramwill be equal.For in the triangles A GB and A GD the side A G is common , the angle

A GB is equal to the angle A GD ,and the angle BA G is equal to the angle

D A G ; therefore the side BC is equal to the side DC, and the side BA tothe side DA

,by I. 26. But AB is equal to DC, and AD is equal to BC, by

I. 34. Therefore the four sides AB,BC, GD , DA are all equal.

8 1 . Let AB and GD be the parallel straight lines,and O the given

point . In AB take any point E ,and from E as centre , with radius equal

to the given length,describe a circle meeting CD at F ; j oin EF. Through

O draw a straight line parallel to EF,meeting AB at G

,and CD at H.

Then EGHE is a parallelogram , by construction ; therefore GH is equalto EF, by I . 34 thus GE is equal to the given length .

82 . Let ABCD be a parallelogram ; let straight lines bisecting theangles A and B meet at E then A BB will be a right angle .The angles EAB and EBA are together half of the angles DAB and

ABC together. But the angles DAB and ABC are together equal to tworight angles , by I. 29 . Therefore the angles EAB and EBA are togetherequal to a right angle. Therefore the angle ABB is a right angle , by I. 32 .

83 . Let ABCD be a parallelogram . Suppose that the straight lineswhich bisect the angles A and G are not coincident : then they shall beparallel.Let the straight line which bisects the angle A meet BG at E ; and let

the straight line which bisects the angle G meet DA at F. Then, by I . 29 ,the angle BEA i s equal to the angle D AE , that i s to the half of the angleD AB, that is to the half of the angle D CB, by I . 34 . Thus the angle BEAis equal to the angle BGF. Therefore EA i s parallel to CF

,by I . 28 .

84. Let ABCD be a parallelogram ,and suppose that the diagonals A G

and BD are equal then all the angles of the parallelogram will be equal.In the two triangles ABC and BAD the side AB is comm on AD is

equal to BC , by I. 34 ; and AC is equal to BD by supposition : therefore theangle ABC is equal to the angle BAD . Similarly it may be shewn that anyother two adjacent angles are equal ; so that a ll the four angles are equal .

85 . Let AB and GD be the given straight lines ; suppose that therequired point is to be at a distance equal to E from AB , and at a distanceequal to F from CD.


D raw a straight line parallel to AB,and at a distance E from it ;

also draw a straight line parallel to CD, and a t a distance F from it ; letthe two straight lines thus drawn meet at 0 : then 0 will be the requiredpoint.For the distance of 0 from AB will be equal to E,

and the distanceof 0 from CD will be equal to F, by I . 34.

Two straight lines can be drawn parallel to AB , and at the r equireddistance from it, namely, one on each side of it ; and in like manner twostraight lines can

'

be drawn parallel to CD,and at the requi red di stance

from it : hencefour points can be found which will satisfy the conditions ofthe problem, assuming that AB and GD are not parallel .

86 . Let AB and GD be the two given straight lines in which th erequired straight line is to be terminated . Let E be a straight line to whi chthe requi red straight line is to be equal

,and F that to whi ch the required

straig ht lin e is to be parallel.From A draw a straight line parallel to F, by I . 31 and cut off from it

AG equal to E ,by I . 3. Through G draw a straight line parallel to AB,

and let it meet CD at H. Through H draw a straight line parallel to AG ,

and let it meet AB at K. Then HK is the required straight line .For EX i s equal to A G,

by I. 34 ; so that HK i s of the required length :and it is parallel to AG

,and therefore to F, by I. 30.

87. Let AEB,BFG, GGD be the three equilateral triangles : then wil l

EFbe equal to A C, and GFbe equal to BD .

In the two triangles ABC and EBE the two sides AB , BC are equal tothe two sides EB,

BF each to each . The angle FBG is equal to the angleABE ,

each being the angle of an equilateral triangle ; to each of them addthe angle ABE ; therefore the angle ABC is equal to the angle EBE. Hencethe triangles ABC and EBE are equal in all respects, by I. 4 so that A Gis equal to EF.

Similarly it may be shewn that BD i s equal to CF.

88 . Let ABCD be a parallelogram , and let ABEFbe another parallelogram having BE equal to BC, but the angle ABE greater than the angleABG : then will the diagonal BF be less than the diagonal BD .

The two angles ABC , BCD are together equal to two right angles, andso also are the two angles ABE , BEF, by I. 29 therefore the two anglesABC ,

BCD are together equal to the two ABE , BEF: but the angle ABEi s greater than the angle ABC, by hypothesis ; therefore the angle BEF isless than the angle BGD .

In the two triangles BCD ,BEF the two sides BC, CD are equal to the

two sides BE , EF each to each ; but the angle BCD is greater than theangle BEF: therefore the base BB is greater than the base BF, by I. 24.

89.Let AD

, BE ,CF be the perpendiculars from A , B , C respectively

on the straight line : the sum of AD and CF shall be equal to twice BE .

The straight line D E produced does not pass between A and G supposethat it cuts A G produced through G. Through E dr aw a straight lineparallel to AC

,and let it meet AD at G , and CF produced, through F, at H.

Thus GE is equal to AB ,and EH i s equal to BC, by I. 34 therefore GE 18

equal to EH .


In the two triangles GED , HEE the angle GED is equal to the angleHEE, by I . 15 ; the angle D GE is equal to the angle FHE

,by I . 29 ; and

the side GE was shewn equal to the side HE : hence these triangles areequal in a ll respects

,by I. 26 so that GD is equal to HF. Therefore A D

and CF together are equal to AG and CH together , that is equal to twiceBE

,by I. 34.

90. Let ABCD be the parallelogram , let 0 be the point of intersectionof the diagonals AC and BD . Then by Exercise 78 the diagonals A C andBD are bisected at 0 ; By Exercise 89 the sum of the perpendicula1 s fromA and C on any straight line outside the parallelogram is twice theperpendicular from 0 ; and also the sum of the perpendiculars from B andD is twice the perpendicular from 0 . Hence the sum of the perpendicula 1 sfrom A and C 1s equal to the sum of the perpendiculars from B and D .

91 . Let ABCD EF be the six- sided figure . Then AB is by suppositionequal and parallel to ED ; therefore ABDE i s a parallelogram,

by I . 33 .

Therefore A D passes thr ough the m i ddle point of BE , by Exercise 78 .

Similarly it can be shewn that CFpasses through the middle point of BE .

Thus AD, BE,and CF meet at a point.

92 . Thr ough E draw a s traight line parallel to AB , and let it meet A Cat F. On FC take FG equal to AF. Join GE and produce it to meet ABat H : then GEH shall be the straight line required .

Through F draw a straight line parallel to GH , and let it meet AB atK. In the triangles AFK and EGE the side AF is equal to the side FG byconstruction ; the angle AFK is equal to the angle FGE ,

and the angleFAK is equal to the angle GEE , by I . 29 : therefore FK is equal to GE , byI . 26 . ButFK is equal to EH

,by I. 34 : therefore GE is equal to EH ,

so

that GH is bisected at E .

93. Let ABCD be the given parallelogram , and P the given point on theside AB . On CD take C Q equal to AP j oin A C and PQ intersecting at R .

Through R draw a straight line at right angles to PQ , meeting A D at S,

and CB at T . Then PSQT wi ll be the required rhombus .In the triangles A PR and OQR the sides AP and C Q are equal , by con

struction the angle ARP is equal to the angle CRQ , by I . 15 and theangle RAP is equal to the angle ROQ ,

by I . 29 : therefore PR is equal toQR , and AR i s equal to CR ,

by I. 26 .

In the triangles PRS and QRS the sides PR and QR are equal ; RS 1s

common ; and the angles PRS and QRS are equal being righ t anglestherefore PS 18 equal to QS,

by I. 4 .

In the triangles CRT and ARS the sides AR and CR are equal ; th eangle ARS 13 equal to the angle CRT ,

by I. 15 ; and the angle A SR is equalto the angle CTR , by I . 29 therefore RS is equal to RT, by I . 26.

In the triangles SRP and TRP the sides SR and TR are equal ; the sideRP is common and the angles SRP and TRP are equal being right angles :therefore SP is equal to TP ,

by I . 4 .

In the same manner it may be shewn that TQ is equal to SQ ,and also

equal to TP . Hence PSQT is a rhombus.The construction fails if the straight line through R at right angles to

PQ , instead of meeting AD and CB, meets DC and BA .

T . EX . EUC .

1 8 EXERC ISES IN EUC LID .

94. Let BE intersect AC at G,and DFintersect AC at H . Thr ough

‘G draw a straight line parallel to AD , meeting D E at K.

Then ED is equal and par allel to BF therefore EB is equal and parallelto D E

,so that EGKD i s a parallelogram , and GK is equal to ED ,

and therefore equal to AE . In the triangles AEG and GKH the sides AE and GKare equal ; the angles EA G and KGH are equal , and the angles EGA andKE G are equal , by I . 29 : therefore AG is equal to GH ,

by I . 26 .

Similarly it may be shewn that CH is equal to HG : hence the thr eestraight lines AG , GH ,

H C are a ll equal ; so that AC is trisected ,

I . 35 to 45 .

9 5 . Let 0 be the middle point of DC. Of the two straight lines ADand BC

,suppose AD the less . Through 0 draw a straight line parallel to

AB , meeting AD produced at E ,and meeting BC at F.

Then in the two triangles EOD and FC C the sides D 0 and C C areequal ; the angle EOD is equal to the angle FOC , by I . 15 ; and the

‘ angleOED is equal to the angle CEO,

by I . 29 : therefore the triangles are equalby I . 26 . To each triangle add the figur e AD OFB : thus the figure ABFEis equal to the figure ABCD .

96 . C onstruct a parallelogram by drawing through E a straight lineparallel to AB ; this parallelogram is equal to ABCD by Exercise 95 . Thetriangle AEB is half this parallelogram , by I. 41 : therefore the triangleAEB is half the quadrilateral ABCD .

97 . Let ABCD be a parallelogram ; let 0 be the middle point of thediagonal AC ; through 0 draw any straight line meeting AB a t E and CDat F : then the straight line EOFshall b isect the parallelogram .

In the two triangles A CE and C OFthe sides AO and CO are equal ; theangle A CE is equal to the angle C OE

,by I . 15 ; and the angle OAE is equal

to the angle OCF,by I . 29 : therefore the triangles are equal , by I. 26 . To

each triangle add the figure A CED ; thus the figure AEFD i s equal to th etriangle A CD . But the triangle A OB is half the parallelogr am ABCD ;

therefore the figure A EFD is half the parallelogram ABCD .

98 . Let ABCD be the parallelogram,and P the given point within it .

Bisect AC at O ; j oin P0 and produce it to meet opposite sides of theparallelogram at E and F respectively . Then by Exercise 07, the straightline EPFbisects the parallelogram .

99. Let ABCD be the parallelogram . Bisect A C at O through 0 drawa straight line at right angles to AC

,and let it intersect at E the straight

line drawn through D parallel to AC. Produce E0 through 0 to F, makingOF equal” to OE . Then AFCE will be such a rhombus as is required .

For in the two triangles A OE and C OE the two sides A O,OE are equal

to the two sides CO, OE each to each ; and the right angles A CE , COE are

equal : therefore AE i s equal to CE . Similarly we can shew that AF ise qual to AE ; and also that CF is equal to CE , and to AF. ThereforeAFCE is a rhombus. A lso the triangle AEC is equal to the triangle AD C ,

by I . 37 therefore the rhombus AFCE is equal to the parallelogram ABCD .

100. Let ABC , D EFbe two triangles having the sides AB, BC equal toth e sides D E ,

L'F each to each ; also the angles ABC and D EF together

equal to two right angles : then the triangles shall be equal in area .


rem ainders are equal ; that is the triangle BFC i s equal to the quadrilateralA DFE .

109 . Let ABC be a triangle ; let D be the middle point of CB ,and

E the middle point of CA : ED shall be equal to half of AB .

Through D draw a straight line parallel to CA ,and let it meet A B at

if“; then ED is parallel to AF, by Exercise 106 : therefore AF is equal to ED ,

y I . 34.

In the two triangles CD E , DBF the sides CD , DB are equal by hypothesis ; the angle CD E is equal to the angle DBF, and the angle E CD i sequal to the angle FD R , by I . 29 : therefore ED is equal to ER , by I. 26 .

ifh

l

e

i‘

re

iore AF is equal to ER therefore AF is half of AB ; therefore ED i s

a 0 AB.

110. By Exercise 106 the straight lines EG and EH are both parallelto BD , therefore they are parallel to each other , by I . 30 ; and by Exercise109 they are each equal to half of BD , and therefore they are equal to eachother .

111 . Let D , E , F be the three given points ; through D draw a straightline parallel to EF, through E draw a straight line parallel to FD , andthrough F draw a straight line parallel to D E . Let the fir st and secondstraight line meet at C

,the second and third at A , and the thi rd and first

at B . Then ABC shall be the triangle required .

For by construction BD EF and D CEF are parallelograms , so that BDand DC are each equal to FE ,

by I . 34 : therefore BD is equal to DC , andD i s the m iddle point of BC. Similarly E is the middle point of CA ,

andF is the middle point ofAB .

1 12 . Let ABC be any triangle ; let D be the middle point of BC , andE the middle point of CA : then the triangle BBC shall be one - fourth ofthe triangle ABC .

Join EB. The triangle EBA i s equal to the triangle EBC by I . 38

therefore the triangle EBC is half the triangle ABC . A gain , the triangleD EC is equal to the triangle BED by I . 38 ; therefore the triangle D E C ishalf the triangle BEC ; therefore the triangle DE C is one - fourth th e

triangle ABC .

1 13. EA is equal to ED ,by Exercise 59 ; therefore the angle EAD is

equal to the angle ED A . Similarly the angle FAD is equal to the angleFD A . Therefore the whole angle BAFis equal to the whole angle EDF.

A gain,the triangle EAD is equal to the triangle EBD , and the triangle

FAD is equal to the triangle FCD ,by I . 38 . Therefore AFD E i s equal to

EBD and FCD together . Therefore AFD E i s half the triangle ABC .

1 14. Let ABC , BBC be triangles of equal area on Opposite sides of thesame base BC j oin AD cutting BC

,or BC produced, at F : then shall AF

be equal to FD .

Make the triangle BBC on the same side of D C as DD C , so that BEmay be equal to BA and CE to CA ; then the triangle BCE is equal to thetriangle BOA in all respects

,by I. 8 . Therefore the triangles BCE and

BCD are equal in area . Therefore ED is parallel to BC , by I . 39. Therefore the triangle FCE is equal to the triangle FCD ,

by I . 37 .

In the two triangles CAF and CEFthe side CF is common ; the SidesCA , CE are equal by construction and the angles A CE and ECF are equal ,

EXERC ISES IN EUCLID . 21

since the angles A GB and ECB are equal hence these triangles are equalin all respects

,by I . 4 .

Thus the area of the triangle A CE is equal to the area of the triangleE CF; and therefore the

'

triangles A CE and BCF are equal in area. ThusAF must be equal to D E ; for if these are not equal it can be shewn by the“,

aid of I . 38 , that the areas are not equal.

115 . Let ABCD,BEFO, EGHE be the thr ee parallelograms. Join AF

cutting BC at K ; joi-n BH cutting EFat L : then BLFK shall be half ofBEFC .

AF is parallel to BH,by I . 33. In the triangles ABK, BEL the sides

AB and BE are equal , by hypothesis ; the angles ABK and BEL , and theangles BAK and EBL are equal

,by I . 29 : therefore the triangles ABK and

BEL are equal , and BK is equal to EL ,by I . 26 .

Thus EL is equal to LE,and the triangle ELB is half the parallelogram

LERK,by I. 38 and I . 41 .

In the same way it may be shewn that the triangles ABK and FCK areequal , so that BK is equal to CK,

and the triangle CKFi s half the parallelogram KBLF.

Hence the triangles BEL and ECK are together equal to the parallelogram KELF ; so that KBLFis half BEFO.

116 . The triangle BCG is equal to the triangle BD G , by I . 37. A lso thetriangle ED G is equal to the sum of EEG and BFD

,that is to the sum of

RFC and BEA , by I . 37 . And the triangle BCG is the sum of BEG andCEG . Hence the sum of BFG '

and EFA is equal to the sum of BEG andCFG . Therefore the triangle BEA is equal to the triangle CFG .

1 17 . Suppose AD greater than AB . Join CD ; through B draw astraight line parallel to DC, meeting A C at E j oin ED : then AED will bethe triangle required.

For the triangle EBD i s equal to the triangle EBC , by I . 37 . To eachof these add the triangle ABE ; then the triangle A DE i s equal to thetriangle ABC .

If AD is less than AB the straight line drawn through B parallel to DCwill meet AC p roduc ed ; but the demonstration will not be essentiallychanged .

1 18 . Let D be the given point in BC . Join DA ; through C draw astraight line parallel to DA

,meeting BA produced at E ; join D E : then

D EB will be the triangle required .

For the triangle D EA is equal to the triangle D CA ,by I . 37 . To each

of these add the triangle ADB : then the triangle EDE is equal to thetriangle ABC .

If the given point D is in BC p roduc ed the process will not be essentiallychanged.

119. Let P be the given point in CD ; join PA and PB . Through Cdraw a straight line parallel to PE,

and through D draw a straight lineparallel to PA . Through P draw a straight lin e parallel to AB ,

let it meetthe straight line drawn through C at E , and the straight line drawn throughD at F : then ABEFi s the quadrilateral requir ed .

For the triangle PEE is equal to the triangle P CB; and the trianglePEA is equal to the triangle PD A ,

by I. 37. Therefore the figure ABEFis equal to ABCD .


120. Join PA and PB . Through C draw a straight lineparallel to PR ,

and let it meet AB produced at M ; through D draw a straight line parallelto PA

,and let it meet BA produced at N : then PMN is the triangle

required.

For the triangle PRO is equal to the triangle PBM, and the trianglePAD 13 equal to the triangle PAN, by I. 37 . Therefore the triangle PMNis equal to ABCD .

121 . Let AC, produced if necessary , meet the given straight line at D ;j oin DB ; through C dr aw a straight line parallel to DB , meeting AB , produc ed if necessary, at E then AED is such a triangle as is required .

For the triangle CED is equal to the triangle CEB, by I. 37 : thereforethe triangle AED i s equal to the triangle ABC .

122 . Let ABC be the given triangle, P the given point in the side A C.

Suppose P to be nearer to A than to C. Bisect -BC at D ; j oin AD andPD . Through A draw a straight line parallel to PD

,meeting BC at E .

Join EP : then EP will bisect the triangle ABC .

For the triangle PED is equal to the triangle PAD ,by I . 37 ; to each

of these add the triangle PD C : therefore the triangle PCE is equal to thetriangle A CD . But the triangle A OB is half the triangle ABC by I. 38therefore the triangle PEC is half the triangle ABC .

IfR be nearer to C than to A the side AB must be bisected.

123. Let ABCD be the given quadrilateral , A the given angular point.D raw the di agonals AC and BD ; bisect BD at E ; j oin AE , CE . ThroughE draw a straight line parallel to A C suppose this straight line to befurther from B than A C is , and let it meet DC at G . Join A G : then A Gwill bisect the quadr ilateral .The triangle AEC is equal to the triangle A GO, by I . 37 . To each add

the triangle ABC ; therefore the figure ABCE is equal to the figure ABCG .

But the triangle ABE i s half the triangle ABD,

‘

and the triangle CBEi s half the triangle CBD , by I . 38 : therefore the figure AR CE is half thefigure ABCD Thei eto1 e the figure ABCG is half the figure ABCD .

If the straight line drawn th rough E parallel to A C is new er to B thanAC I s

,it will meet BC instead of DC ; but the demonstration will not be

essentially changed.

124 ; If possible suppose that O is not in the diagonal AC . Let thes traight line through 0 parallel to BC meet AB at E , and DC at F ; andsuppose that A C intersects EFat a point G between 0 and F. ThroughG draw a straight line parallel to AB .

Then the parallelogram GD is equal to the parallelogram GB , by I . 43.

Therefore the parallelogram OD is greater than the parallelogram OB.

But this is impossible , for they are equal by hypothesis . Therefore thepoint 0 cannot fall otherwise than on AC.

I . 46 to 48 .

125 . The angles A CD ‘and‘

BCF are equal , being right angles ; to eachadd the angle A OB; thus the whole ang le BCD is equal to th e wholeangle A CF.


In the two triangles BCD and A CE the two sides BC, CD are equal tothe two sides FC , CA each to each ; and the angle BCD is equal to theangle FCA : therefore FA is equal to BD

,by I . 4 .

1 26 . Let BAC be a triangle, having the angle BAC a cute ; then will thesquare on BC be less than the sum of the squares on BA and A C .

From A draw a straight line at right angles to BA , and cut c fi AD equalto A C ; j oin BD .

Then BD is greate r than BC by I . 24. Now the square on BB is equalto the squares on BA and AD by I . 47 . Hence the square on BC is lessthan the squares on DA and AD,

that i s less than the squares on BA ,

and AC.

127. Let BA C be a triangle , having the angle BA C ob tuse : then wil lthe square on BC be greater than the sum of the squares on BA and A C.

From A draw a straight line at right angles to BA , and cut off AD equalto A C ; j oin BD .

Then BD i s less th an BC , by I . 24 . Now the square on ED is equal tothe squares on BA and AD

,by I . 47 . Hence the square on BC is greater

than the squares on BA and AD,that is greater than the squares on BA

and A C.

128 . Let ABC be a triangle ; and suppose the square on BC less thanthe squares on BA and A C then th e ang le BAC will be an a cute angle .

The angle BA C cannot be a right ang le, for then the square on BCwould be equal to the squares 011 BA and AC, by I . 47 . The angle BA Ccannot be obtuse , for then the square on BC would be greater than thesquares on BA and AC , by

-Exerc ise 127 . Therefore th e angle BAC mustbe an acute angle .

A gain , let ABC be a triangle ; and suppose the square on BC g rea terthan the squares on BA and A C : then the angle BAC will be an ob tuseangle .The angle BAC cannot be a right angle

,for then the square on BC

would be equal to the squares on BA and AC,by I . 47. The angle BA C

cannot be acute,for then the square on BC would be less than the squares

on BA and AC, by Exercise 126 . Therefore the angle BAC must be anobtuse angle .

129 . Let BAC be a triangle,h aving th e angle A a right angle . Let

a straight line meet AB at D,and A C at E : then shall the squares on BE

and CD be equal to the sum of the squares 0 11 BC and DE .

The square on BE is equal to the squares on BA and AE ,and the

square on CD is equal to the squares on CA and AD ,by I . 47 . Therefore

the squares on BE and CD are equal to th e squares on BA,CA

, AE ,AD

th

at i s to the squares on BC, AE ,

- AD , that i s to the squares on BCan D E .

130. D raw through P a straight line parallel to AD ,meeting AB at K

and CD at L . D raw through P a straight line parallel to A D,meeting

£6

” at M and AD at N. Then AK is equal to DL , and KB is equal to LC ,

y I . 34.

The squares on PA and PC are together equal to the squares on AK,

PK, CL , LP , by I. 47 ; that is to th e squares 011 DL, PK,KB

,LP ; that i s


to the squares on DL

,LP

,PA

”

, KB ; that is to the squares'

on PD and PB ,

y I. 47.

131 . Let ABC be a triangle having a right angle at C ; and let thesquare on AC be three times the square on BC. From C draw CD to bisectAB , and CE perpendi cular to AB . Then will the angles A CD

, D OE , ECBbe all equal.Th e square on AB is equal to the squares on A C and BC

,that is to

three times the square on BC and the square on BC, that is to four timesthe square on BC. Hence it may be shewn that AB is twice BC. ButAB is twice DC, by Exercise 59. Thus BC, CD ,

DB are all equal,so that

BCD is an equilateral triangle . Hence the angle BCD is two - thir ds ofa right angle , and A CD is onethird of a right angle, by I . 32 .

A gain,in the two triangles BE C and D EC

,the sides BC and DC are

equal ; therefore the angle CD R is equal to the angle GRD ; and the anglesBE C and DE C are equal being right angles : therefore the angles BCE andD OE are equal , by I . 32. Hence BCE and D OE are each one - third of aright angle ; so that the three angles A CD ,

D CE,BCE are all equal .

132 . Let ABC be a triangle, having a right angle at A ; let E be themiddle point of A C, and F the middle point of AB : then four times thesquares on BE and CF will be equal to five times the square on BC .

For four times the square on BE i s equal to four tim es the square onAB and four times the square on A E , by I . 47 And four times the squareon CF is equal to four times the square on AC and four times the square onAF. Therefore four times the squares on BE and CF are equal to fourtimes the square on AE

,four times the square on AF, and four times the

squares on AB and AC ; that is to four times the square on AE , four tim esthe square on AF, and four times the square on BC ; that is to the squareson A C and AB , and four times the square on BC ; that is to five times thesquare on BC.

133. From D dr aw a perpendicular DM on GB produced . The anglesDBM and MBC are together equal to a right angle ; and so are the anglesCBA and MEC : hence the angles D BM and CBA are equal .In the two triangles DBM

,CBA the sides DB and CB are equal ; the

angles DEM and CBA are equal ; and the right angles DMB and CAB areequal : hence BM is equal to BA

,and DM i s equal to CA

,by I . 26. There

fore GM is equal to twice BA , and the square on GM is equal to four times thesquare on BA . The square on D G is equal to the squares on GM and DM ,

by I . 47 ; that is to four times the square on BA and the square on CA .

Similarly it may be shewn that the square on EFis equal to four times thesquare on CA and the square on BA . Hence the squares on D C and EFareequal to five times the square on BA and five times the square on CA

,that is

to five times the square on BC, by I . 47.

I I . 1 to 11.

134. Let a stra ight line AB be divided into two parts at C, and suppose'

that th e squares on AC and CB are equal to twice the rectangle AC, CB :

then shall AC be equal to CB .

For if AC be not equal to CB suppose A C the gr eater, and construct thediagram of II . 4 ; on CA take CX equal to CB , draw through X a straight


line parallel to AD,meeting HG at Y and DFat Z. Then we have given

that HF and CK are together equal to AG and GE . By our construction wemake XG equal to CK, and YF equal to GE . Thus HZ and CK are together equal to AG ; therefore HZ is equal to A Y. But this is impossiblefor HD is greater than AH. Hence A C and CB cannot be unequal ; that isthey are equal .

135 . By II. 5 the rectangle contained by the parts is always less than thesquare on half the line

,except when the straight line is bisected ; so that

th e rectangle c ontained'

by the parts is grea test when the straight hne 1 sbisected .

136 . Take A C equal to a side of the smaller square ; produce AC to D sothat CD is equal to a side of the larger square : and from CD cut off CBequal to CA . Then by II. 6 the rectangle AD , DB is equal to the differenceof the squares on CD and CB . Thus the required rectangle is found .

137 . By II . 9 the sum of the squares on the two parts is always gr ea terthan double the square on half the straight line

,except when the straight

line is bisected so that the sum of the squares on the parts is lea st when thestraight line is bisected.

138 . Take A C equal to the greater of the two straight lines ; on A Cproduced take CD equal to the less of the two straight lines

,and also take

CB equal to AC. Th en AD is equal to the sum of the two straight lines ,and DB is equal to their difference . And it i s shewn in II . 9 that the squareson AD and DB are together double of the squares on A C and CD.

139 . Let AB be the given straight line to be divided ; KL a side of thegiven square .Make the angle ABE equal to half a right angle . With centre A and

radi us equal to KL describe a circle cutting BE at F. From F draw FDperpendicular to AB : then AB shall be divided at D in the manner requi red .

For it may be shewn as in II . 9 that FD is equal to DB . A lso thesquares on AD and DFare equal to the square on AF, so that the squareson AD and DB a re equal to the square on KL .

A remark may be made like the last sentence of the solution of Exercise 65 .

140. Let AB be the given straight line . Produce AB to C so that A Cmay be equal to the diagonal of the square described on AB and from BAcut off BD equal to BC : then will the square on DA be double the squareon DB .

Since the straight line CD is bisected at B and produced to A , we havethe squares on CA and DA together equal to double the squares on BA andBD , by II. 10. But the square on CA is double the square on BA

,by I . 47 :

therefore the square on DA is double the square on DB .

141 . In the triangles HA C ,FAB the two sides HA

, A C are equalto the two sides FA , AB each to each ; and the right angles HA C ,

FAB areequal : therefore the angle HCA is equal to the angle FBA , that is to theangle HBL . The angle LHB is equal to the angle AHC ,

by I . 15 . Therefore the angle HLB is equal to the angle HA C , by I. 32 . Thus the angleELB i s a right angle .

EXERCISES I N EUC LID .

142 . Since EB is equal to EF, the angle EBF is equal to the angleBEB ; that is the angle OBL is equal to the angle CEL . Therefore theangle FOL i s equal to the angle L OB, by I . 32 and Exercise 141 . Thusthe angle E CO is equal to the angle BOL , and therefore to the angle EOC , byI . 1 5 . Therefore EO is equal to EC by I. 6 ; and therefore also equal to EA .

Thus the angles E C C and E CA are together equal to the two angles EC 0and EA O that is the angle A C C is equal to the two angles A 00 and CAOtherefore the angle A 00 is a right angle by I. 32 .

143. In II. 1 1 it i s shewn that the rectangle AB , BH is equal to th esquare on AH. Therefore the rectangle AH, HB together with the Squareon EH is equal to the square on AH. Thus the rectangle AH, HE i s equalto the difference of the squares on AH and EH , that is to th e rectanglecontained by the sum and the difference of AH and BH . See the note onpage 269 of the Euc lid respecting II . 5 and II. 6 .

II . 12 to 14.

144. Let ABC be a triangle having the sides AB , AC equal . FromBdraw BD perpendicular to CA . By 11 . 13 we know that the square on ACis less than the squares on AB and BC by twice the rectangle A C, CD . Butthe squares on AC and AB are equal , by hypothesis ; therefore the square onBC is equal to twice the rectangle A C

,CD .

145 . See page 293 of the Euc lid .

146 . The squares on CD and A C are equal to twice the squares onAB and CB

,byExercise 145 . A nd AC is equal to AB . Therefore the squa re

on CD is equal to the square on AB together with twice the square on CB .

147 . Let ABCD be th e parallelogram ; j oin AC and BD intersecting at0 ; then A C and BD are bisected at O , by Exercise 78 .

The squares on AB and BC are equal to twi ce the squares on A O and

OB , by Exercise 145 ; and similarly the squares on AD and DC are equalto twice the squares on A O and OD . Therefore the squares on the sidesof the parallelogram are equal to four times the square on A O and fourtimes the square on B0

,that is to the square on A C and the square on BD .

148 . Let ABC be the triangle, and 0 the middle point of AB. ByExercise 145 the squares on A C and BC are equal to twice the square son AO and 00 . Now AO is given , and C C is of constant length : thereforethe sum of the squares on AC and BC is invariable .

149 . Let ABCD be the quadrilateral ; let E ,F

,G , H be the middle .

points of AB , BC , CD ,DA respectively : the squares on A C and BD will be

equal to twice the sum of the squares on EG and FH .

We know that EFGH is a parallelogram,that EH i s half BD , and HG

is half A C : see Exercises 107 and 109 . Therefore the squares on AC andBD are equal to four time s the squares on EH and H G ,

that is to twicethe sum of the squares on the sides of the parallelogram BEGH, that istwice the sum of the squares on th e diagonals EG and FH

,by Exercise 147 .


155 . From D draw D E perpendicular to AB produced , and from E

draw EG perpendicular to AC produced . The angles DBFand CBA aretogether equal to a right angle , by I . 13. The angles A OB and CBA aretogether equal to a right angle , by I. 32. Therefore the angle DEFis equalto the angle A CB.

In the two triangles D RE and BCA the sides DB and BC are equal ;the angle D BF i s equal to the angle ECA

,and the right angles DFB and

BAC are equal : therefore the triangles are equal in allo

respects . Similarlyit may be shewn that the triangles CEG and BCA are equal in all respects .The square on DA is equal to the squares on DB , BA , and twice the

rectangle BA , BF, by II . 12 ; that is the square on DA is equal to thesquares on BC

,BA and twice the rectangle BA , AC : therefore the squares

on DA and A C are equal to the squares on BC,BA

,A C and twice the

rectangle BA,A C. Similarly it may be shewn that the squares on EA and

AB are equal to the squares on BC , BA ,A C and twice the rectangle

BA,A C. Therefore the squares on DA and A C are equal to th e squares on

EA and AB .

1 56 . By 11 . 13 the square on ED is less than the squares on BA andAD by twice the rectangle AB , AE and by the same Proposition the squareon BD is less than the squares on BA and AD by twice the rectangle AC

,

AD ; therefore the rectangle AB , AE is equal to the rectangle AC,AD.

1 57. Let ABC be an equilateral triangle ; suppose AB produced to Dso that the rectangle AD

,DB is equal to th e square on CB : then will the

square on DC be equal to twice the square on CB .

From C draw CE perpendicular to AB ; then the triangles CEA andCEB will be equal in all respects by I . 26 . The square on DC is equal to thesquares on CB, BD and twice the rectangle DB , BE , by II . 12 ; that is to thesquares on CB

,BD and the rectangle DB

,BA that is to the square on

CB and the rectangle AD , DB , by H. 3 ; that is to twice the square on CB,

by supposition.

158 . Let ABC be a triangle , having the angle C a right angle ; fromC draw CD perpendicular to AB : then will the square on CD be equal tothe rectangle AD , DB .

Bisect AB at E . Then one of the two straight lines AD and DB isequal to the sum of AE and ED , and the other is

‘equal to their difference .Hence the rectangle AD , DB is equal to the difference of the squares on AEand ED : see the Euc lid , page 269 . But EC is equal to AE , by Exercise 59 .

Therefore the rectangle AD , DB is equal to the difference of the squares onE C and ED ; that is to the square on CD , by I . 47.

1 59 . U se the same diagram as in Exercise 158 . Then the square onD C is equal to the squares on CD and DB

,by I . 47 ; that is to the

rectangle AD , DB and the square on DB,by Exercise 158 ; that is to

th e rectangle AB,DB

,by lI . 3.

Similarly the square on AC 18 equal to the rec tangleBA , DA .

160. By II . 13 the square on AC together with twice the rectangleAB , BF is equal to the squares on AB , BC and the square on AB togetherwith twice the rectangle A C

,CE is equal to the squares on AC, BC . Hence

the squares on A C,AB , together with twice the rectangle AB , BF and twi ce


the rectangle A C , CE are equal to the square on AB , the square on A C,and twice the square on BC. Therefore twice the rectangles AB , BF andAC

, CE are equal to twice the square on BC ; and therefore the rectanglesAB

,BF and AC

,CE are equal to the square on BC .

161 . Let AB be the straight line which is to be divided . L et L 3! bea side of the given square . Bisect AB at C. From L draw a straight lineLO at right angles to LM with centre M and radius equal to AC describe acircle cutting LO at N. From CB cut off CD equal to LN. Then will therec tangle AD ,

DB be equal to the square on LM .

The rectangle AD,DB is equal to the difference of the squares on A C

and CD,by II . 5 ; that is to the difference of the squares on MN and LN,

by construction that is to th e square on LM , by I. 47 .

II I . 1 to 15 .

162 . Let A be the centre of the circle which i s to be described , B thecentre of the given circle . Join AB ; through B draw a straight line atright angles to AB

,cutting the given circle at L and M then will AL be the

radius of the circle required.

In the two triangles ABL and ABM, the side AB is comm on ; BL isequal to BM ; and the right angles ABL and ABM are equal therefore ALis equal to AM, by I. 4 . Therefore a circle described fr om the centre Awith the radius AL will pass thr ough M,

and so will cut the given circle atthe extremities of a diameter.

163. Each of the straight lines passes through the centre of the circle ,as may be shewn in the manner of III . 1 : thus the straight lines intersectat a fix ed point.

1 64. Let the circles cut each other at B and E . Thr ough B draw anystraight line meeting one circle again at A and the other again at C .

Through E dr aw a straight line parallel to ABC ,meeting the first circle

again at D and the second circle a t F. Then shall A C and D E be equal .Find P the centre of the circle A BED , and Q the centre of the circle

BCFE . From P draw PK perpendicular to AB and produce KP to meetD E at M : then since D E is parallel to AB the angles at M are rightangles. A gain , from Q dr aw QL perpendicular to BC, and produce L Qto meet EFat N then the angles at N are right angles . Thus KLNM is aright - angled parallelogram

,and therefore KL is equal to MN.

KB is half of AB by I I I . 3 , and BL is half of BC : therefore KL is halfof AC. Similarly MN is half of D E. But KL is equal to MN : thereforeAC is e qual to DE.

165 . Suppose that D and F are on the circumference of the circle withcentre A ; and that E and G are on the circumference of the circle withcentre B. From A draw AL perpendicular to FC ,

and AP perpendicular toDC from B draw BM perpendicular to CG and BQ perpendicular to CE .

From B dr aw BX perpendicular to AL, so that BX i s parallel to FG .

A gain from B dr aw a perpendicular BY on PA produced , so that BY is


parallel to DE . Then the angle BA Y wi ll be equal to the angle BA X , since,by supposition , FG and DE are equally inclined to AB .

Now by I . 26 it will follow that BX i s equal to B Y ; also, as in Exercise164, it may be shewn that BX is equal to half of FG, and BY is equal tohalf of D E . Therefore FG is equal to D E .

If the perpendiculars from B instead of meeting AL and PA produced,

m eet LA produced and AP the process i s substantially unchanged .

166. Let A and B denote the centres of the two circles . From theprocess given in the solution of Exercise 165 it follows that the straight linedrawn through the point of intersection of the two circles is always lessthan twice AB except when it is parallel to AB

,and then it is equal to

twice AB . Therefore the greatest possible straight line is obtained bydrawing a straight line through one of the points of intersection of the circlesparallel to the straight line j oining the centres .

167 . Let C denote the centre of the cir cle,A the point in the diameter

,

P and Q the extremities of the chord parallel to this diameter. From Cdr aw CD perpendicular to PQ then PQ is bisected at D,

by III . 3 .

By Exercise 145 the squares on AP and A Q are equal to twice thesquares on AD and PD that is to twice the squares on AC , CD,

and PD ,

by I . 47 that is to twice the squares on AC and CP .

N ow one of the segments of the diameter is the sum of AC and theradius , and the other is th e difference of A C and the radius. Therefore thesum of the squares on the segments is equal to twice the square on AC andtwice the square on the radius that is to twice the squares on AC and CP ;that is to‘the squares on AP and A Q . See Exercise 138 .

168 . Let 0 be the middle point of AB then the sum of the squares onAP and BP is equal to twice the square on OP together with twice thesquare on AO . Hence we require OP to be the least possible . Join O withthe centre of the circle , and let the joining line cut the circumference at Qthen Q is the required point : for OQ is less than any other straight linedr awn from O to the circumference

,by III. 8 .

169. Let ABC be a circle,and let F be its centre ; let D RE be another

circle,and let G be its centre ; let the second circle fall within the first, and

let them touch at B . Let AC and DE be parallel diameters, A and D beingtowards the same parts . Then B

,D

,and A will be in a straight line.

Join BD and AD . If they are not in a straight line let BD producedmeet AC at H. Join FG ; then FG produced will pass through B , byI I I . 1 1 . The angle GDB i s equal to the angle FHB,

by I . 29 . Thereforethe angle FHB is equa l to the angle FBH . Therefore EH i s equal to FB.

Therefore EH is equal to FA which is absurd. Therefore B,D , and A

cannot lie otherwise than in a straight line.If the circles touch externa lly the process will be sim ilar ; but then

the extremities of the diameters must be taken towards opposite parts.

1 70. Let C be the centre of the larger circle , ‘

A the centre of the other.Let FAE be a chord of the larger circle at right angles to A C ; and let Hbe one of the points where it Cuts the smaller circle . Let BHD be anotherchord of the larger circle

,and let it be at right angles to FAE . Of the two

EH and EH let EH be th e less ; and of the two RH and DH let DH be theless . Then will EH be equal to EH

,and EH equal to DH.

EXERCISES IN EUCL ID . 31

From C dr aw CK perpendicular to,

BD ; then A CEH will be a square .The chordsFE and BD are equally distant from C , and are therefore equal ,by III . 14 . Th erefore AE and BK, the halves of these chords, are equal.But AH is equal to HK therefore EH is equal to EH and therefore EHis equal to DH .

171 . Let A be the given point, C the centre of the circle . Join CA andthrough A draw a ' chord BAD at right angles to A C : this shall be theshortest chord through A .

For draw any other chord EAF through A ; and from C draw theperpendicular C G on EAF. Then since C GA is a right angle CA G i s lessthan a right angle

,by I . 32 . Therefore C G is less than CA

,by I . 19 .

Therefore EAF is nearer to the centre than BAD ; and therefore EAF isgreater than BAD

,by III. 15 .

172. From 0 draw the radius OB parallel to PN,so that OB and PN

are on opposi te sides of the diameter on whi ch ON li es. Join PB .

The angle NPB is equal to the angle OBB,by I . 29 ; the angle OBP is

equal to the angle OPB by I . 5 : therefore the angle NPB is equal to theangle OPB. Thus the angle OPN i s bisected by the straight line PB ; sothat so long as P is on the same side of the fixed diameter the straight linebisecting the angle always passes through the fixed point B .

Produce B0 to meet the circumference again at B’. Then if P is on

the other side of the fixed diameter,the straight line bisecting the angle

will pass through B’.

173. Take 0 the centre of the circle DBCE ; join OD and OE . TakeP the centre of the circ le on which A and B lie, and Q the centre of thecircle on which A and C lie .The angle PAB is equal to the angle PBA

,and therefore equal to the

angle GED ,and therefore equal to the angle ODB, by I. 5 and I . 15 .

Therefore PA is parallel to DO by I . 27 . Similarly QA is parallel to OE .

But PA and QA form one straight line by III. 12 . Therefore OD and OEform one straight line

,which is a di ameter of the circle DBCE , and is

parallel to PQ .

174. Let ABCD be the quadrilateral figure. Let the circles describedon AB and BC as diameters intersect at E . Then the straight line whichbisects EB at right angles will pass through the centres of the circles ,by I I I . 1

,so that it will bisect AB and BC. Therefore thi s straight l ine

will be parallel to AE and to E C,by Exercise 106 . Hence AE and E C are

in one straight line ; and the common chord EB is at right angles to A C.

Similarly the common chord of the circles described on CD and DA asdiameters is also at right angles to A C ; and therefore the two commonchords are parallel .In like manner the common ch ord of the circles described on AD and

AB as diameters is parallel to the common chord of the circles described onCB and CD as diameters .

175 . Let C be the centre of the given circle,A the given point in the

gi ven straight line . On thi s straight line take AB equal to the radius ofthe g iven circle ; j oin BC ; at C draw the radius CD , making the angleBCD equal to the angle CBA

,and on the same side of CB. Produce D C

to meet the given straight line at 0 . Then 0 will be the required centre.


For the angle OCE is equal to the angle OBO, by construction ,there

fore OC is equal to OB , by I. 6 . And BA is equal to CD , by construction ,

therefore OD Is equal to GA ; and the circle described from the centre 0with the radius OA will pass through D , and will touch the given circleat D .

I I I . 16 to 19 .

176 . In th e diagram of III. 17 let FD be produced to meet the outercircle at H ; j oin EH cutting the inner circle at K ; j oin AK : then AKw illtouch the circle BCD .

This may be shewn precisely as in III. 17. A lso DH is equal to DE byI I I . 3 ; therefore AK will be equal to AB .

1 77 . Let C be the centre of the given circle. From C draw a straightline perpendicular to the given straight line meeting it at A , and intersecting the circumference at B . Through B draw a straight line at right anglesto CB , this touches the given circle by III. 16 C orollary ; and it i s parallelto the given straight line by I. 28.

178 . Let C be the centre of the given circle. D raw the radius CAparallel to the given straight line. Through A draw a straight line at rightangles to CA ; this touches the given circle by III. 16 Corollary ; and it isperpendicular to the given straight line by I. 29.

179 . Let C be the centre of the circle, A the end of the diameter .D raw A D a t right angles to CA , and make it of the given length . JoinCD cutting the circum ference at B ; from B draw a straight line at rightangles to CB and let it meet CA produced at E . Then EB will be therequired tangent .For in the two triangles CA D , CBE the angle C is common , the right

angles CAD and CBE are equal , and the sides CA and CB are equal ;hence A D i s equal to BE , by I. 26. And BE touches the circle at B , byIII. 16 Corollary.

180. Let 0 be the centre of the two circles. Let ABC be a chord - of

the outer circle,touching the inner circle at B ; let D EF be another chord

of the outer circle, touching the inner circle a t E : then will A C be equalto DE.

Join OB and OE . The angles at B and E are right angles , by III. 18 .

Thus A C and DE are equally distant from 0 : therefore A C is equa l to D E,

by III. 14 .

181 . Let 0 be the centre of the given circle, P the given point. Inthe circle place a chord AB equal to the given straight line . From 0 drawa perpendicular OC on AB . With centre 0 and radius 00 describe a circle,and from P draw a straight line touching this circle at E and meetingthe given circle at the points D and F. Then PDEF shall be the straightl ine required .

For by Exercise 180 the straight lines AB and D E are equal .

182 . Let MON be the diameter ; let tangents be drawn at M and N ;let AB be the portion of another tangent which is intercepted between thetangents at M and N then A OB shall be a right angle.Let AB touch the circle at E . Join A C and BC.


In the triangles CAE and CAM the sides AE and AM are equal , byExercise 176 ; the sides EC and MC are equal ; and the right angles AECand AM0 are equal : therefore the angles A CE and A OM are equal , by I. 4 .

Sim ilar ly the angles BCE and BCN are equal. Therefore the angle A OBi s equa l to the two angles A OM and BCN . Therefore the angle A OB is aright ang le , by I. 13.

183. Let AB be the given straight line, C the centre of the given circle .D raw a straight line DE parallel to AB

,and at a distance from it equal to

the radius of the required circle which is given. With centre C describea circle having its radius equal to the sum of the r adii of the given circleand the required circle ; let this circle cut DE at the points F and G : theneither F or G may be taken as the centre of the required circle.For the distance of F from AB is equal to the radius o f the required circle ,

and therefore the circle described from F as centre with the given radiuswill touch AB . And the distance of F from C is equal to the sum of theradii of the given circle and the required circle ; and therefore the circledescribed from F as centre will touch the given circle . Similarly G maybe taken as the centre of the required circle.If the line AB lies without the given circle

, D E must be drawn on thesame side of AB as C is .

184. Let 0 be the'

c entre of the given circle. Let s'

circle describedfrom C as centre touch the given straight line at A , and touch externallythe given circle at B . From 0 draw a straight l ine parallel to. CA to meetthe circumference o f. the g iven circle a t D, so that .C D and CA may beon opposite sides of 00 . Then D will be a fixed point ; and D,

B,A will

lie in one straight line,For OD and C A are parallel , b y construction ; a nd the angles at A are

right angles, by III. 18 therefore if D O be pro duc ed to meet the givenstraight line the angles at the point of intersection w ill be right angles.Thus OD is a fixed straight line

,

“and D i s a fixed point .Join AB

,BD . The angle BOD is equal to the angle BCA by I. 29.

Therefore the angles ODB and OBD together are equal to the angles CABand CBA by I. 32 . But the angle OBD .is equal to th e angle ODB, and theangle CBA is equal to the a ngle CAB,

by I. 5 . Therefore the angle OBB

is equal to the angle CBA . ThereforeDB 3 119 3 4 are in the same straightme.

185, See the Euc lid, page 295

186 . Let ABC , D EFbe the circles. I t i s required t o draw a straightline touching the circle ABC , and so that the part of it intercepted by D EFmay be equal to a given straight line .In th e circle DEFplace the straight line DE equal to the given straight

line . Find G the centre of this circle , and wi th G as a centre describe acircle touching D E . D raw a straight line AFE to touch the latter circleand to touch ABC , cutting the circle D EF at F and H : this shall be thestraight line required.

For since D E and EH are equally distant from the centre of the circleD EF

, they are equal : therefore EH is of the required length.

1872Let A and B be the centres of the two circles . In the circle

having Its centre at A place a chord PQ of the length which is to ber . Ex. E0 0. 3

34 EXERCISES. I N EUC LID .

intercepted by this circle from the required straight line . In the circlehaving its centre at B place a chord RS of the length whi ch is to be interc epted by this circle from the required straight line . With A andB as centresdescribe circles touching these chords respectively. D raw a straight line totouch the two circles thus described : this will be the straight line requir ed.

For let this straight line meet the given circle which has A for centreat the points K and L , and the given circle whi ch has B for centre at th epoints M and N. Then KL and PQ are equally distant from A , andtherefore they are equal by III. 14. Similarly MN and RS are equal .

188 . Let ABCD be a quadrilateral, the sides of which touch a circlethen will AB and CD together be equal to BC and D A together.Let AB touch the cir cle at P

,let BC touch it at Q, let CD touch it at R ,

and let D A touch it at S. Then AP is equal to A S, BP to BQ , C Q to CR ,

and DR to D S, by Exercise 176 . Therefore the sum of AP , PB , CR , RD i sequal to the sum of AS, BQ , CQ , D S; that is AB and CD together areequal to BC and D A together.

189 . Let ABCD be a parallelogr am which is described about a circle.Then AB and CD together are equal to BC and D A together , by the preceding Exercise . But AB is equal to CD , and BC is equal to DA , by I . 34.

Hence AB, B C , CD, D A are all equal , so that the parallelogram is arhombus.

190. If possible suppose that DE does not touch the c ircle but cuts it.D raw DFto touch the circle at F and produce D E to meet AE producedat G.

Then D E i s equal to DB , and GF i s equal to GC , by Exercise 176.

Therefore D G i s equal to the sum of DB and GC . But, by hyp othesis ,D E is equal to the sum of DB and EC ; add EG to both ; thus DE and

EG are equal to the sum of DB , E C , and EG ; that is to the sum of DBand C C . Therefore D E and EG are equal to D G ; that is two sides of atriangle are equal to the third , which is impossible by I. 20. Therefore :D E does not cut the circle . In the same way it may be shewn that DEdoes not fall without the circle . Therefore DE touches the circle.

- 191 . Take the diagram of Exercise 188 . Let 0 be the centre of thecircle. D raw OA , OB , 00, OD , OP, OQ , OR , OS.

In the triangles OAS, OAP the side 0A is common ; the sides OS andOP are equal ; the sides A S and AP are equal by Exercise 1 76 . Thereforethe angles A OSand AOP are equal , by I. 8 . Similarly the angles BOP and BOQare equal ; also the angles COQ and C OR are equal ; and the angles D OBand D OS are equal. Therefore the angles A OP

,BOP, C OR , D OR are

together equal to the angles A OS, BOQ , C OQ , D OS ; therefore each set offour angles are together equal to two right angles

,by I. 15 . C orollary 2 .

192 . Let CA and CB be the two radii which are at right angles . Leta straight line touch the circle , and m eet CA produced at P,

and CB produced at Q . From P draw PM to touch the circle, and from Q draw QN totouch the circle ; then PM will be parallel to QN .

It may be shewn as in Exercise 191 that the angle MPQ is double theangle OPQ ,

and that the angle N QP is double the angle C QP . Thereforethe two angles MPQ and N QP are double the angles OPQ and C QP . Butthe angles OPQ and C QP are together equal to a right angle, by I . 32

36 EX ERC ISES I N

198 . Let ABC D be the quadrilateral , having BC parallel toAD .

" LetK be the centre of a circle whi ch touches AB at E, BC atF, CD at G, andD A at H. Through K draw a straight line LKM paral lel to AD , andterminated by AB and DC. Then LM will be equal to one- fourth of

“theperimeter of ABCD .

It may be shewn that AB is bisected at L , and D C bisected at Msee Exercise 36.

The angle EBK i s equal to the angleFBK, by Exercise 176 . The angleFBK is equal to the angle LKB, by I . 29 . Therefore the angle LBK i sequal to the angle LKB. Therefore LX is equal to LB, by I. 6 . ThusLK is half AB . Similarly MK is hal f DC. Thus LM i s half the sum ofAB and DC ; that is LM is one- fourth of the perimeter of ABCD , byExercise 188 .

199. Let A be the fixed point in the fixed straight line . Let a circlewhich touches the fixed straight line at A cut the fixed parallel straight lineat B and C. D raw a tangent to this circle at B , and let it meet the fixedstraight line through A at F ; and draw from A the perpendicular AE onthis tangent : also draw from A the perpendicular AD on BC.

The angle FBA is equal to the angle FAB , by Exercise 176 ; and istherefore equal to the angle ABD , by I. 29 . Hence in the two trianglesABE and ABD the side AB is common ; the angle ABE is equal to theangle ABD ; and the right angles AEB and ADB are equal : therefore ABis equal to AD

,by I . 26 . Hence the straight line FB will touch at E

the circle described from A as a centre with a radius equal to the fixedlength AD.

Sim ilarly the tangent at C to the circle which passes through A , B, Cwill touch the same fixed circle.

200. Let A and B be the two given points . Let C be a point on theconvex circumference of the given circle such ’ that the tangent to the circleat C i s equally inclined to AC and CB . The sum of A C and BC shall beless than the sum of the straight lines drawn from A and B to any otherpoint on the circumference.For take any other point on the circumference , as P ; join PA and PB .

Let PA cut at E the tangent at C, and Join EB. Then the‘ sum of PE andPB is greater than EB

,by I. 20. Therefore the sum of PA and PB is

greater than the sum of EA and EB. But the sum of EA and EB is

greater than the sum of CA and CB , by page 306 of the Euc lid . Thereforethe sum of PA and PB is greater than the sum of CA and CB .

201 . Join CD . The angle CD A is equal to the angle CAD ,by I. 5 .

The angles CDB and BDE are together equal to a right angle, by III. 18 .

The angles CBA and CAB are together equal to a right angle , by I . 32 .

Therefore the angle BDE is equal to the angle CBA . A lso the angle DBEi s equal to the angle CBA , by I . 15 . Therefore the angles BDE and DBBare equal ; and the triangle DBE is isosceles, by I. 6.

202 . Let 0 be the centre of the circle . Produce 00 to a point F suchthat CF is equal to 00 : then by compari ng the triangles POO and PCF itm ay be shewn that PF is equal to P0 ,

and also OFis equal to OP , so thatBOFi s an equilateral triangle . Therefore the angle POC is two - thirds of aright angle. Therefore the angle EPA is o ne - third of a right angle , by

ExERC IsEs IN’

EUCLID . 37

I .

"

32. Therefore the angle PEA is two - thirds of a right angle,by I . 32.

Therefore the angle CED is two - thirds of a right angle , by I. 15 .

The angle COA is equal to the angles OBC and OCB, by I. 32 , that i s totwice the angle OBC , by I. 5 : therefore the angle OBC i s one - thi rd ofa right angle. The angle BAD is a right angle . Therefore the angle ADBis twothirds of a righ t angle. Thus each of the angles CD E and CED isequal to twothirds of a right angle ; and therefore the angle E CD is alsotwo- thirds of a right angle , by I . 32. Thus the triangle DEC is equiangular , and therefore equilateral.

I II . 20 to 22.

203. The four angles ABD,BD C , D CA , CAB are equal to four right

angles. The angle BAC i s fixed, and the angle BD C is constant, by III. 21 :ther efore the sum of the angles ABD and A OB i s constant.

204. The angle RAB is half the sum of the angles PAB and QAB ; andthe angle RBA is half the sum of the angles PBA and QBA . Thus theangle R together with half the sum of the angles PAB , QAB , PBA , QBA isequal to two right angles. A gain

,half the sum of the angles Q , QAB , QBA

is equal to a right angle ; and half the sum of the angles P ,PAB , PEA is

equal to a right angle . Thus half the sum . of the angles P and Q togetherwith half the sum of the angles PAB , QAB , PBA , QBA is equal to the angleR together with half the sum of the angles PAB , QAB , PBA , QBA . Ther efore the angle R is equal to. half the sum of the angles P and Q, and istherefore constant by III . 21 .

205 . In the quadrilateral Q CPD the angles QCP , CPD ,PDQ are all

constant by II I . 21 . The four angles of a quadrilateral are together equalto four right angles . Therefore the angle CQD is constant.

206 . The angles at Q and R are constant by III . 21 ; therefore thethird angle of the triangle formed on QR as a base by QA and RB producedis constant , by I. 32 .

207; The angles OD O and AD C are together equal to two right anglesby I. 13 ; but the angle OD C i s equal to the angle ABC therefore the anglesABC and A D C are together equal to two right angles. Therefore a circlec an be described round ABCD : see the Euc li d , page 276 .

208 . Let ABCD be the quadrilateral inscribed in a circle . Let AB produced through B , and D C produced through C, meet at O.

The angles BAD and BCD are together equal to two right angles,by III .

22 . The angles BCD and BCG are together equal to two right angles byI. 13. Therefore the angles BAD and BCD are together equal to the twoangles BCD and B00 . Therefore the angle B00 is equal to the angleD A O. Similarly the angle OBO is equal to the angle AD O.

209 . Let ABCD be a parallelogram inscribed in a circle. By II I . 22

the angles A and C are together equal to two right angles ; by I. 34 theangles A and C are equal : therefore each of the angles A and C i s a rightangle .

Similarly each of the angles B and D is a right angle.

EXERC ISES IN EUCLID .

210. Let ABC be a “triangle inscribed in a circle. Let D, E , F he

points in the segments exterior to the triangle cut off by BC, CA , ABrespectively. The angles CAB and CDB

'

a re together equal to two rightangles, by HI . 22 ; so also are the angles CBA and CEA ,

and the anglesA GB and AFB . Therefore the angles in the exterior segments togetherwith the angles of the triangle are together equal to six right angles ;therefore the angles in the exterior segments are together equal to fourright angles , by I. 32 .

211 . Take the diagram of III . 22. Let ABCD be a quadrilateralinscribed in a circle. The angle in the segment exterior to AB and theangle A GB are together equal to two right angles , by III . 22. So also arethe. angle in the segment exterior to BC and the angle CAB. Also theangle in the segment exterior to CD and the angle CBD . A lso theangle in the segment exterior to DA and the angle DBA . Therefore theangles in the four exterior segments together with the angles A CB, CAB,

CBD,DBA are equal to eight right angles . But the angles A CB,

CAB,

CBD , DBA together make up the angles of the triangle ABC , and aretherefore equal to two right angles. Hence the angles in the four exteriorsegments are equal to six right angles.

212. Let A OB be a triangle inscribed in a circle such that the angleA GB is one - third of two right angles . In the segment cut off by AB on theopposite side of C take any point D on th e circumference. Then the anglesA OB and ADB are together equal to two right angles , by II I . 22 . But theangle A CB is one - third of two right angles , by construction. Therefore theangle ADB is twothirds of two right angles . Thus the angle in the segment ADB is double the angle in the segment A CB.

213 . Let A CB be a triangle inscribed in a circle such that the angleA CB is one - sixth of two right angles . I n the segment cut off by AB o n

the Opposite side of 0 take any point D on the circumference . Then theangles A CB and ADB are together equal to two right angles , by I I I . 22.

But the angle'

A CB is one sixth of two right angles by construction . Therefore the angle ADB is five- sixths of two right angles . Thus the angle inthe segment ADB is five times the angle in the segment ACB.

214. Let ABCD be a quadrilateral . Let a side AB be produced to E,

and suppose that the angle EBC is equal to the angle AD C .

The angles ABC and EBO are equal to two right angles by I . 13. Butthe angle EBC

»

is equal to the angle AD C ,by supposition. Therefore the

angles ABC and ADC are together equal to two right angles . Thereforea circle can be circumscribed round ABCD : see the Euc lid , page 276 . Thenthe angles ADB and A CB will be equal by III. 21 . In like manner anyo ther side of the quadr ilateral will subtend equal angles at the oppositeangles of the quadrilateral .

215 . Let ABCDEF be a hexagon inscribed in a circle ; suppose thatAB is parallel to D E ,

and B C parallel to BE : then will CD be paralleltoJoin AD . The angle ABC i s equal to the angle FED by Exercise 35 .

The anglesFED and FAD are together equal to two right angles by I II . 22 ;

so also are the angles ABC and AD C . Therefore th e ang les FED and

EXERCISES I N.EUCLID . 39

FAD are together equal to the angles ABC and AD C . Therefore the angleFAD is equal to the angle AD C . Therefore FA is parallel to DC, by I . 27.

216. Let the strai ght lines which respectively bisect the angles APCand A QC meet at H the angle PHQ will be a right angle .The angle PC Q and the angle BAD are together equal to two right

angles , by I . 15 and III. 22. Therefore the angles PQC and QPC are togetherequal to the angle at A . Now the angle C QH and half the angles at Aand B are equal to a right angle by I. 32 ; so a lso the angle CPH togetherwith half the angles at A and D are equal to a right angle . Thus theangles C QH , GPH ,

together with the angle at A , and half the angles at Band D are equal to two right angles ; therefore the angles C QH , CPH , andA are together equal to a right angle

,by I II . 22. Therefore, by what is

shewn above,the angles HQP and HPQ a re together equal to a right angle .

Therefore the angle PHQ is a right angle , by I. 32.

217. Let ABCD be a quadrilateral inscribed in a circle. Let a straightline EFGH be drawn meeting the circumference at E and H,

the side ADat F, and the side BC at G ; and suppose the angle AFG to be equa l to thea ngle BGE : then will EH also make equal angles with AB and DC.

Sinc e the angles AFG and BGE are equal the arcs ED and ABE aretogether e qual to the arcs HC and EAB : see the Euc lid , page 294. Therefore the differenc e between the arcs H C and ED is equal to the differenceb etween the arcs EH and EA . Therefore HE makes equal angles with ABand D C : see the Euclid , page 294.

218 . Let EFGH be a quadrilateral which can be inscribed in a circle,so that the angles E and G are together equal to two right angles. Supposealso that a circle can be inscribed in this quadrilateral touching the sidesEF, FG , GH ,

HE at A , B ,C,D respectively. Then will the straight lines

BD and AC be at right angles to each other.Let 0 be the centre of the circle ABCD . The angle between A C and BD

is measured by half the sum of the arcs DA and BC ; see the Euc lid , page294, and is therefore equal to half the sum of the angles DOA and B00 .

Now the angles ODE and OAE are right angles ; therefore DOA and D EAare together equal to two right angles

,by I. 32 . But the angles D EA and

BGC are together equal to two right angles,by supposition . Therefore the

angle DOA is equal to the angle BGC . Similarly the angle BOC i s equalto the angle AED . Therefore the angle between AC and BB is equal tohalf the sum of the angles AED and BGC , that i s to a right angle.

III. 23 to 30.

219 . Let A C and BG be chords of two equal arcs in a circle ; let ABand C G join the extremities of these chords towards the same parts : thenAB and C G Shall be parallel.The angle ABC is equal to the angle BC G by III. 27 : therefore C G i s

par allel to AB by I. 27.

220. Let AB and CG be parallel chords in a circle ; let them be joinedtowards the same parts by AC and BG,

and towards opposite parts by A Gand BC : then AC will be equal to BG, and AG to BC.

40 EXERCISES IN EUCLID.

Because AB is parallel to C G the angle ABC is equal to the angle BC G’,

by I. 29 : therefore the arc AC is equal to the arc BG ,by III. 26 : therefore

the straight line AC is equal to the straight line BG , by I II . 29 .

A gain , it ha s been shewn that the arc AC is equal to the arc BG ; toeach add the arc CG . therefore the arc A CG is equal to the arc BGC :

therefore the straight line A G is equal to the straight line BC, by III. 29 .

221 . Suppose C and D on opposite sides of A , and C and E on oppositesides of B. Then the angle D AE i s the sum of the angles A GB andAEB, which are invariable by III. 21 , and is therefore invariable. Thereforethe arc DE is invariable, by III. 26 .

But C and D may be on the same side of A , and C and E on the sameside of B . Thus the angleFOD is equal to the angle A CB, and is thereforeinvariable

, by III. 21 . A lso the angle AEB is invariable,by III. 21 .

Therefore the angle EAD . which is the difference of FOD and AER,by

I . 32 , is also invariable . Therefore the arc DE is invariable, by III. 26 .

222 . Let the straight linewhich bisects the angles m eet the arc CB at

a point P : thenwill EP be equal to BB.

The angles ECP and EBB are together equal to two right angles, by I II .

22 ; the angles,

ECB and_

D OP are equal to two right angles, by I . 13

therefore the angle EBP i s equal to the angle D CF,that is to half the

angle D CB. Again, the angle B P is equal to the angle BCP , by III. 21 ,that

'

I s to half the angle D CB. bus the angles EBP and BEP are equal ;thereforeEP and BP are equal, by I. 6.

223. Let ABCD be a quadrilateral figure inscribed In a circle ; produceCD toany point E ; let the straight lines which bisect the angles ABC andADE meet at F : then F shall b e

‘

onthe circumference of the circle.Let AD and BF intersect at H. The angle AD E i s equal to the angle

ABC ,by III. 22 : therefore the angle HDF is equal to the angle HEA .

The angle DEF is equal to the angle BHA,by I. 15 . Therefore the angle

DEH i s equal to the angle BAH,by I . 32 . Therefore F is on the circum

ference of the circle. See the Euc lid, page 276 .

If BF meet AD produced at H, the proof will be similar , using th e c onverse of III. 22 instead of that of III . 21 .

224. Let C be the centre of the given circle . From D as centre withradius equal to CD describe a. circle cutting AB produced at F. Join FDand produce it to meet the given cir cle again at E . Then will the arc AEbe three times the a re BD .

For the angle A CE i s equal to the two angles CEFand OFF,

'

by I . 32therefore the angle A CE is equal to the two angles CD E and GED , by I. 5therefore th e angle A CE is equal to the angle D OFand twice the angleDFC ,

by I . 32 ; therefore the angle A CE is equal to three times the angleD CF, by I . 5 . Hence the arc AE is three times the arc BD , by III . 26.

225 . The points P and Q will always lie on a fixed segment of a circlede scribed on AB : see the Euc lid , page 276. A lso a s the angle A QB isgiven , and the angle C is given , the angle QBP i s invariable. Hence thestraight line PQ is always of the same length .

226 . Let AB be the common chord of the two equal circles . ThroughA draw a straight line meeting one circle at C, and the other at D ; drawBC and BD ; these straight lines shall be equal.


For the angle A CB is equal to the'

angle A DB, by II I . 27 and II I . 28

therefore BC is equal to BD by I. 6 .

227. In the two triangles BOP, ROQ the side OB is common ; theangle BOP is equal to the angle ROQ , by supposition ; and the right anglesBPO and BQO are equal : therefore BP is equal . to BQ , b y I . 26 .

Now AB is equal to BC , by III. 26 and II I . 29 . Therefore the square onAB is equal to the square on BC. Therefore the squares on AP and PBare equal to the squares on C Q and QB . But PB has been shewn equal toQB . Therefore the square on AP is equal to the square on CQ . ThereforeAP is equal to C Q .

228 . Suppose AB to lie between AL and AM, and AM nearer to th ecentre than AL . From B dr aw BP perpendicular to AM and BQ perpendicular to AL produced. Then , as in Exercise 227 , we have PM equal toQL . Therefore the sum of AL and AM is equal to twice AP ; and this is afixed quantity, because the side AB is fixed, and so are the two angles PABand APB .

Next let AB not lie between AL and AM, and suppose AM nearer to thecentre than AL . From B dr aw BP perpendi cular to AM and BQ perpendicular to LA produced. Now LB and MB are equal by Exercise 222 ; andthus we find as in Exercise 227 that L Q and PM are equal . Hence thedifference of AM and AL will be found to be equa l to twice AP, and this asbefore is a fixed quantity.

229 . LetFE and 0A be“ pr oduced to meet at H. Bisect OFat K,and

b in EX .JThe triangles EFK and E CK are equal in all respects by I. 8 ; so that

EKO i s a right angle : therefore KE and OH are parallel by I . 28 . Therefore the angle FEK is equal to the angle EH 0 , and the angle KEO is equalto the angle EOH, by I. 29. Therefo re the angle EH 0 i s equal to theangle EOHThe angle GOB Is equal to the tvvo angles OGH and OHG , by I . 32 ;

therefore the angle GOB is equal to the two angles CEG and OHG by I. 5 ;therefore the angle GOB is equal to the angle EOH and twice the angleOHG ; therefore the angle GOB i s et al to three times the angle EOH .

Hence the arc BG i s three times the arc AE, by III. 26 .

230. Let AB be the given base, and ABC one of the triangles. Th e

point C will be s ituated on the arc of a c ertain fixed segment of a circledescribed on AB as a base . see the Euc lid

,page 276 . Let the angle at C be

bisected by a straight line,and let this straight line he produced to cut at D

the circumference of the cir cle of which’ the segment i s part. Then sincethe angles A CD and BCD are equal the arcs AD and BD are equal ; so thatD is the mi ddle point of the arc ADB. Hence D is a fixed point

,and the

straight lines which b isect the vertical angles all pass thr ough this fixedpoint.

231 . Let the 1mm c ircles touch at A . Let K be the centr e of the smallercircle

,and L the centre of the larger circle . Let BD C be a chord of the

larger circle, touching the smaller circle at D . Then BD and D C willsubtend equal angles a t A .

Produce AD to cut the larg er circle at F. Join KD and LF. Then A'

,

K, L are in one straight line,by 111 . 11 . The angle KAD is equal to the


angle KD A , and the angle LAD is equal to the angle LEA ,by I. 5 : there

fore the angle KDA is equal to the angle LEA : therefore LFis parallel toKD , by I. 28 . But ED is at right angles to BC , by III . 18 therefore LFis atr ight angles to BC , by I. Therefore LFbisects BC , by II I. 3. Therefore the arc BF is equal to the arc CF. Therefore the angle BAF is equalto the angle CAF, by II I . 27.

I II . 31 .

232. Let AB be the given hypotenuse, ~ and ABC one of the triangleshaving the angle at C a r ight angle. Let D be the middle point of AB .

Then D C is equal to DA by Exercise 59 , so that C is on the circumferenceof a circle having D a s centre , and DA as radius, and therefore having ABas diameter.

233. Let ABC be a triangle having the sides A C and BC equal . LetD be the middle point of AB . Then t he triangles AD C and BD C are equalin all respects , by I. 8 ; so that the angles AD C and ED O are right angles.Hence the circle described on A C as di ameter will pass through D ,

byExercise 232 ; and also the circle described on BC as diameter will passthrough D . Thus the circles intersect at D .

234. Let ABCD be a rectangle inscribed in a circle. Since the angleAD C i s a right angle A C is a diameter of the circle. Let 0 be the centreof the circle. The rectangle is double of the triangle AD C by I. 34. The

triangle D A C has a fixed base, namely the diameter of the circle ; but itsheight is always less than OD except when D OA is a right angle. Thereforethe greatest rectangle that can be inscribed in a circle is one which hasits diameters at right angles to each other ; th is by I . 4 has its sides allequal

,and is therefore a square.

235 . Suppose C and E on opposite sides ofAB , and C and A on oppositesides of FE . CD is equal to half AB,

by Exercise 59 . Therefore a circledescribed from the centre D with radius CD will pass through B , E , A , F.

The angle ECB i s half the angle EDE , by III. 20 that is half a rightangle

,and is therefore half the angle A CB, by I II . 31 . Thus EC bisects

the angle A CB.

Similarly the angle FCA is half the angle FD A , that is half a rightangle. ThusFC bisects the angle between CA and BC produced.

236 . Let 0 be the centre of the circle. Let E and C be on the sameside of AB . EFis parallel to BC therefore the angle OEB i s equal to theangle EBC ,

by I . 29 . But the angle OEB i s equal to the angle OBE , by I. 5 .

Therefore the angle OBE is equal to the angle EBC . Thus EB bisects theangle OBC .

Similarly if CB be produced to D it may be shewn that the angle OBDi s bisected by BF.

237. A c ircle described on A C as diameter will pass through D and E ,

by Exercise 232 . Then A CE and ADE will be angles in the same segment ,and therefore equal by 111 . 21 .

238 . Th e angles ABC and ABD are both right angles , by I I I . 31 .

Therefore BC and BD are in one straight line by I. 14. That is the straightline CD passes through B .


and therefore equal to the angles APM and PAM ; and therefore equalto a right angle , by Therefore QR touches the circle AQM at Q .

Similarly it may be shewn that QR touches the circle BEM at R.

245 . A circle described on BC as diameterwill pass . through F and Dalso a cir cle described on FD as diameter will pass through E and G ; seethe Euc lid , page 276 . By the aid of the latter circle we have the angleFEG equal to the angle FD G : by the aid of the former circle we have theangleFBC equal to the angle FD G . Therefore the angleFEG is equal tothe angle EBO. Therefore EG is parallel to BC ,

by I .‘

28 .

246 . Take the case in which the chords BC'

and'

A C are p roduc edthrough C to meet the other circumference. Let HC be a diameter of thecircle ABC , and let it be produced to cut ED at K.

The angle CED is equal to the angle CAB,by 111. 21 ; a nd there

fore equal to the angle CHB, by III. T h e angle ECK is equal to theangle HCB , by I. 15 . Therefore the angle EKG is equal to the angle HBC ,

by I. 32 . Therefore the angle EKC is a right angle,by III. 31 .

The figure becomes modified in other cases,but the demonstration

remains substantially unchanged.

247 . Let ABC be a triangle having a r ight angle at G. Let squaresbe described on BC, CA , AB outside the triangle ; and let D , E , F re

spec tively be the intersections of the di agonals In these squares.Then the angle AFB is a right angle

,and therefore a circle would go

round ACBF : see in the Euc lid the note on III. 22 . The angleFCA willbe equal to the angle EBA , by III . 21 . Therefore the angle FCA i s halfa right angle . A l so the angle A CE is half a right angle . Therefore theangle ECE is a right angle. So a lso is ECB. Then use

248 . D raw a straight line at right angles to CA at the point A ; let thi smeet the cir cumference of the given cir cle at B . then B will be the pointrequired .

For a circle described on CB as diameter will g o through A , and willtouch the given cir cle at B . Let P be any other point on the circumferenceof the given circle. Join CP cutting at Q the circle j ust described ; andjoin A Q .

Then the angle CBA is equal to the angle CQA , by I I I . 21 . The angleC QA i s greater than the angle CPA by Therefore the angle CBA isgreater than the angle CPA .

249. LetFG produced meet AB at H. The angles ED G and FEG areright angles

,by III. 31 : therefore a circle would go round ED GE, by page

276 of the Euc li d . Therefore the angle DFG is equal to the angle D EGand therefore equal to the angle DBA , by III . 21 . Hence in the trianglesAFE and ABD the angle at A is common ; and the angle AFE i s equalto the angle ABD ; therefore the angle AHF is equal to the angle A DB,

and is therefore a right angle,by III. 31 .

250. Let two equal circles touch externally at B . Let AB be a diameterof one circle

,and BC be a diameter of the other. D raw BD , a chord

'

of thefirst circle ; and BE at right angles to

'

BD a chord of the second circle .The straight line D E shall be parallel to A C and equal to the distancebetween the centres of the given circles.


Join AD and CE . The angle ADB is a right angle, by I II . 31 ; theangle DBE is a right angle by supposition , therefore the angles ADB andDBE are equal . therefore AD is parallel to BE , by I . 27. Therefore theangle BAD is equal to the angle GBE,

by I. 29 . The angle ADB is equalto the angleBEC , by III. 31 . Al so AB is equal to BC . Therefore BEis equal to AD

,by I. 26 . And it was shewn that BE is parallel to AD .

Therefore DE is equal and parallel to AB , by I. 33 .

251 . Let A CBD be the rh ombus, and AB the shorter diagonal. On

AB as diameter describe a circle cutting A C at H, BC at K, AD at L , BDat M. Join AK and BH ,

" intersecting at E . Join AM and BL intersectinga t

'

F. Then AEBFwill be a rhombus, having i ts angles equal to those ofD A CB.

Since A C is equal to BC the angle CAB is equal to the angle CBA . Inthe triangles BAH and ABK the angles HAB and KBA are equal, as wehave just shewn ; the angles AHB and AKE are equal, being right anglesby IR . 31 : therefore the angles ABE and B

'

AK are equal,by I. 32 . There

fore EA is equal to EB,by I . 5 .

A lso AC is parallel to BD , by Exercise 76 ; therefore the angle HABis equal to the angle MBA ; and the right angle AHB is equal to the righta ngle EMA ; therefore the angle ABH is equal to the angle BAM. Therefore BE is parallel to AF. Similarly AE is parallel to BF. Thus thefigure AEBFis a paral lelogram . And since EA has been shewn equal toEB the parallelogram is a rhombus .

A gain,the angle AER is equal to the angle HEK, by I. 15 . The angles

HEK and HCK are together equal to two right angles , because the anglesEHO and EKU are right angles. The angles HOK and CAD are equal totwo right angles

,by Therefore the angles H CK and CAD are to

gether equal to the angles HOK and HEK. Therefore the angle CAD isequal to the angle HEK, that is

‘to the angle AER . This shews that theangles of one rhombus are equal to the angles of the other.

252 . Let AB , CD be two chords of a circle ; let them meet, producedif necessary, at E at right angles. The squares on AE , EB, CE , ED will betogether equal to the square on the di am eter.Let 0 be the centre of the circle. D raw OG perpendicular to AB , and

OH perpendicular to CD . Then AB is bisected at G,and CD is bisected

at H,by II I . 3. The squares on AE and EB are together double of the

squares on AG and EG and the squares on CE and ED are together doubleof the squares on CH and HE . see the Euc lid , page 269. A lso EG i sequal to H0 , and EH is equal to GO

,by I. 34 . Therefore the squares on

AE,EB

, CE , ED are equal to double the squares on AG , OG , CH, HO ;that i s to double the squares on OA and OC , by I . 47 ; that is to four timesthe square on OA , that is to the square on the diameter of the circle.

I II . 32 to 34.

253. Produce BC to meet the circumference again at E ; and producePD to meet the cir cumference again at F.

Then PD is equal to FD by III . 3 . The triangles BPD and BED areequal in all respects, by I . 4 ; so that the angle BPD is equal to the angleDFD and therefore equal to

‘

the angle BEP ,by III. 21. The angle BPA

46 EXERC ISES IN EUCLID'

.

i s equal to the angle BEP ,by 111: 32 . Therefore the angle BPA is equal to

the angle BPD .

254. Let two circles touch each other at the point B . Let a straightline ABC m eet one circle at the points A and B

,and the other circle at the

p oints B and C. The segments cut ofi b y AB will be respectively similar tothose cut off by BC .

D raw a straight line EBFthrough B , at right angles to the“

straight linewhi ch j oin s the centres of the circles ; then this straight line touches bothcircles by III. 16 Cor . The angles in the segments cut offby AB are equal toth e angles ABE and ABFrespectively, by III. 32 ; so also are the angles inthe segments cut off by BC . Hence the segments cut offby AB are respectively similar to those cut ofi by BC, by the definition of similar segments .

255 . The angle DAP is equal to the angle PQA , by III. 32 ; and therefore equal to the angle QAB, by I. 29 . The angle DPA is equal to theang le QBA , by I . 13 and III . 22. Therefore the angle PDA is equal to theangle BQA

,by I. 32 .

256 . Suppose G to fall between A and H. The angle GBD i s equal tothe angle GAB

,by IH . 32 ; the angle GAB is equal to the angle HDB, by

III. 21 : therefore the angle GED i s equal to the angle HDB. ThereforeHD i s parallel to BG , by I. 27.

The figure becomes modified in other cases but the demonstration remainssubstantially unchanged.

257 . Produce CA to any point E, and DA to any pointF. The angleCAF is equal to the angle ABC , and the angle D AE i s equal to the angleABD , by II I . 32 . But the angle CAF is equal to the angle D AE , by I . 1 5 :therefore the angle ABC i s equal to the angle ABD . Therefore AB , produc ed if necessary, bisects the angle CBD .

258 . D raw PE the tangent at P, so that PE and PA are on the sameside of PB . The angle EPA is equal to the angle PBA , by III. 32. Theangle PBA is equal to the angle PCD ,

by I . 13 and III . 22 . Therefore the

fingl

le EPA is equal to the angle POD . Therefore EP is parallel to CD ,

y 27.

259 . Let A be any point in the circum ference of a circle. From Adr aw the chord AB

,and the tangent AC. From D

,the middle point of the

arc AB , dr aw DE perpendicular to the chord AB , and DE perpendicular tothe tangent A C . Then D E will be equal to DF.

The angle DAF is equal to the angle DBA , by III. 32. The angle DBAis equal to the angle D AB, by III. 27 . Therefore the angle DAF is equal tothe angle D AE . Then in the two triangles DAF and D AE the side AD iscommon ; the angle DAF is equal to the angle D AE ; and the right angleD EA i s equal to the right ang le DEA : therefore D E is equal to DE , byI. 26.

260. Th e angle ABN i s equal to the angle A QM, by III . 32 . The rightangle ANP i s equal to the right angle AMQ . Therefore the angle PAN isequal to the angle QAM, by I. 32. Therefore the angle NAM is equal tothe angle PAQ .

A gain,since ANP and AMP are ri ght angles a circle would go round

ANPM ; see the Euc lid , page 276. Therefore the angle ANM is equal to


the angle APQ, by III . 21. Therefore the angle AMN is equal to the angleA QP , by I. 32 .

261 . The angle RPS is a right angle, by III . 31 . The angle QPS isequal to the angle PAB, by 111 . 32 and I . 15 . But the angle PAB is equalto the angle OSE,

by I . 32 ; that is equal to the angle PSQ . Thereforethe angle QPS is equal to the angle QSP .

A gain,since RPS i s a right angle the angles PSR and PBS are together

equal to the angle RPS, by I. 32 . And it has been shewn that the angleQPS i s equal to the angle QSP ; therefore the angle QPR is equal to theangle QRP . Therefore by I. 6 we have QS and QR each equal to QP ; sothat QS is equal to QR .

262 . Let AB be the given base, and D the point in the base on whichthe perpendicular fall s. On AB describe a segment of a circle containingan angle equal to the given vertical angle . From D draw a straight lineat right angles to AB

,and let it cut the arc of the segment at C. Then

ABC is the triangle required. For this triangle has the given base, andthe g iven vertical angle ; and the perpendicular from the vertex on thegiven base meets the base at the assigned point .

263. Let AB be the given base . On AB describe a segment of a circlecontaining an angle equal to the given vertical angle . From A draw astraight line AD at right angles to AB

,and equal in length to the g iven

altitude . Thr ough D draw a straight line parallel to AB , and let it cut thearc of the segment at C and E . Then the triangle A CB will satisfy all theprescribed conditions ; and so also will the triangle AEB.

264. Let AB be the given base. On AB describe a segment of a circlecontaining an angle equal to the given vertical angle . From the middlepoint of AB as centre with a

- radius equal to the given length describe acircle , and let it cut the arc of the segment at C and E . Then the triangleA CB will satisfy all the prescribed conditions

,and so also will the triangle

AEB.

265 . Let AB be the given base. On AB describe a segment of a c irclecontaining an angle equal to the given vertical angle. From the middlepoint of AB draw a straight line at right angles to AB

,and let it meet the

arc of the segment at C. Then the triangle ABC is the greatest trianglethat can be constructed under the prescribed condi tions .For the straigh t line D C passes through the centre of the circle of whi ch

the segment forms part , by I II . 1 . Hence a straight line through C parallelto AB will be a tangent to the circle, by III . 16 C or . If any other pointexcept C on the arc of the segment be taken as the vertex of the triangle thevertex will be nearer to AB than C ; so that the triangle will have a lessheig ht than A CB, and therefore will be less than A CB.

266. We suppose that B is nearer to A than C is . We may regardOB as th e base of the triangle ; then the height of the triangle will alwaysbe less than 00 except when the angle COB is a right angle

,and then the

height of the triangle is OC . Hence the triangle OBC is greatest when theangle COB is a right angle. Then as the angles OBC and OCB are equaleach of them will be half a right angle.


Therefore on CA describe a segment of a cir cle containing an angleequal to hal f a right angle, and take the point at whi ch the arc of thi s segment cuts the given cir cle for the point C.

267. Suppose A and B the fixed points. Let C be a point in which thetwo straight lines meet. Then all the points of intersection which are on‘

the same side of AB as C' i s will l ie on the arc of a segment of a circle

described on AB with an angle equal to A CB : see the Euc lid , page 276 .

All the straight lines whi ch bisec t these angles pass through a fixed point,

by Exercise 230.

Similarly a second fixed point is obtained by considering the straightlines bisecting the angles formed at points whi ch lie on the side of .ABopposi te to that we have hitherto considered.

268 . Let AB be the side wh ich is opposite to the g iven angle. On ABdescribe a segment containing an angle equal to the given angle

,and also a

segment containing an angle equal to hal f the given angle. From A ascentre with radi us equal to the sum of the other two sides describe a circle

,

and let D be one of the points where it intersects the second descr ibedsegment . Join AD and let it out the first described segment at C. Thenthe triangle A CB will satisfy the prescribed conditions.The angle A CB is equal to the sum of the angles CDB and CBD , by

‘ I. 32 ; also the angle A CB is double the angle CDB by construction :therefore the angle CDB i s equal to th e angle CBD : the refore CD is equalto CB

,by I . 6 .

Hence CA and CB together are equal to CA and CD together ; that'

is toAD ; that is to the given quantity. And AB and the angle A CB also havethe prescribed values.

III . 35 to

269. Let two circles cut one another at the points A and B, From anypoint T in AB produced suppose a tangent TP drawn to one circle , and atangent TQ to the other : then will TP be equal to TQ .

For by I I I . 36 the r ectangle TA , TB is equal to the square on TP , and

also to the square on TQ ; therefore the square on TP is equa l to the squareon TQ ; therefore TP is equal to TQ .

270. Let PQ denote the common tangent ; and let AB produced meetPQ at T . Then TP is equal to TQ , by Exercise 269 ; therefore PQ isbisected at T.

271. Since the. angles AD C and AEC are equal a circle will go roundAFDC. Therefore the rectangle BC , BD is equal to the rectangle BA, BE ,

by III . 36, C orollary .

272 . Let AB be the common chord of the two circles. Through anypoint C in AB draw D CE a chord of one circle , and ECG a chord of theother . The rectangle AC, CB is equal to the rectangle DC, CE , and alsoequal to the rectangle FC , CG ,

by III . 35 ; therefore the rectangle D C , CEi s equal to the rectangle EU,

CG . Therefore a circle will go through D , F,E , : see the Euc lid , page 277.

reclau

see0.


273. Let BC be the given straight line , and C the fixed point in it. L et

A be the given centre. Join AC ; and describe a circle on AC as diameter.In it place a straight line CD equal to a side of the g iven square . Withcentre A and radius equal to AD describe a circle . thi s will b e the circlerequired.

For let this circle cut BC at E and F. The angle A D C is a right angle ,by III . 31

,therefore CD touches the circle DEF, by III . 16 Cor . ; therefore

the rectangle CE ,CF is equal to the square on CD by III . 36.

274. Let K be the middle point ofBC . The square on GE i s equal tothe rectangle GB

,GC , by I I I . 36 . Thus the square on GE i s equal to the

difference of the squares on GK and BK ; see the Euc lid , page 269 . Therefore the square on GK exceeds the square on GE by the square on BK.

Therefore four times the square on GK exceeds four tim es the square on GEby four times the square on BK. Therefore four times the square on GKexceeds the square on AE by the square onBC : see Exercise 270. And itmay be shewn that K is the middle point of GH ; so that four times thesquare on GK is the square on GH . Thus the square on GH exceeds thesquare on AE by the square on D C .

To shew that K is the middle point of GH we may proceed thus. FirstAE and DFwill be equal ; for each of them will be a side of a right - angledtriangle in which the hypotenuse is the dis tanc e p f th e centres of the cir cles,and a side is equal to the difference of the radii of the circles : see theEuc lid , page 295 . Next the square on GK i s equal to the squares on GEand BK ; and the square on HK is equal to the squares on HF and CK :

therefore the square on GK is equal to the . square on EX therefore GK isequal to HK.

275 . Let two of the circles intersect at A and B. Let T be the fixedpoint such that the tangent TP to one circle is equal to the tangent TQ tothe other. Then A , B ,

and T shall be on a straight line .For if not , j oin TA and le t TA , produced if necessary, cut one circle at

L and the other at M. Then by III. 36 the square on TP i s equal to therectangle TA

,TL ; and the square on TQ i s equal to the rectangle TA ,

TM . Therefore the rectangle TA , TL is equal to the rectangle TA , TM ;

which is impossible. Therefore TA can not cut the circles at any otherpoint than B .

276 . D escribe a circle on BC as diameter ; and let the straight line DEwhen produced cut this circle at N and O .

The rectangle N E,E 0 is equal to the difference of the squares on D N

and D E ; see the Euc lid , page 269. But by the square on D N isequal to the rectangle BD , D C and the rectangle N E , E0 is equal to therectangle AE , EC . Therefore th e rectangle A E , EC i s equal to the difference of the rectangle BD , D C and the square on D E . Therefore the squareon DE is equal to the difference of the rectangles BD

,D C and AE

,E C .

A gain , the rectangleFN ,FO I s equal to the difference of the squares onDE and D N ; see the Euc lid , page 269 . So that the square on DE is equalto the square on D N together with the rectangleFN , FO. But the squareon D N is equal to the rectangle D C , DB,

by III . 35 ; and the rectangleFN ,F0 is equal to the rectangle FA ,FB, by III. 36 , C orollary . Therefore

the square on DE is equal to the sum of the rectangles BD,D C and

T . EX . EUC .


277 . Let AB be the given diameter. In the circle place a straight lineBP equal to a side of the given square . Join AP and produce it to meetat T the tangent to the circle at B . Then T is the required point .For the rectangle TP , TA i s equal to the square on TB, by III. 36 ;

therefore the rectangle TP,PA together with the square on TB is equal to

the square on TB, by I I“

. 3 . Therefore the rectangle TP , PA is equal to thedifference of the squares on TB and TP ; that is to the square on BP , ,

byI II . 31 and I. 47.

IV . 1 to 4.

278 . Join AB. The straight lin es through A and B to touch the circlemake with AB angles which are together less than two right angles

,for each

of them is less than one right angle. Therefore,by A xiom 12

,these straight

lines will meet .

279 . The angles ABC and A OB are together less than two right angles ;much m ore therefore the angles made with BC by the straight lines whichbisect ABC and A CB are together less than two right angles . Therefore ,by Axi om 12 , these straight lines will meet.

280. By I . 47 the square on DA is equal to the squares on D G andGA

,and also equal to the squares on DE and EA ; therefore the squares on

D G and GA are equal to the squares on DE and EA . But the square onD G is equal to the square on D E ; therefore the square on GA is equal tothe square on EA ; therefore GA is equa l to EA .

Therefore the angle DAG is equal to the angle D AE , by I. 8 , so that theangle EA G is bisected by AD .

281 . Let 0 be the centre of the circle inscribed in the triangle ABC.

The straight line OGA bisects the angle D AE ,by Exercise 280. Hence

the angles A OD and A GE are equal , by I . 32 . The angle AD G is equal tothe angle GED , by III . 32 ; and is therefore half the angle D OG , by III. 20 ;that is half the angle GOE . Also . the angle GD E is half the angle GOE ,

by I II . 20. Therefore D G bisects the angle AD E . Similarly EG bisectsthe angle AED . Therefore G is the centre of the circle inscribed in thetriangle ADE , by IV . 4.

282 . Let it be required to describe a cir cle which touches the Side BC ofa tri angle

,and also touches the sides AB and AC produced.

Produce AB to any point K, and AC to any point L, Bisect the anglesKBC , L CB by the straight lines DB, D C meeting at the point D . Then Dwill be the centre of the required circle . The demonstration will be likethat of IV. 4.

In like manner suppose it required to describe a circle which touches theside CA , and also touches the sides BC and BA produced .

4 Produce BC toany point M, and BA to any point N. Bisect the angles A OM, CAN by thestraight li nes CE

,AE meeting at E . Then E will be the centre of the

required circle.We have now to shew that D , C ,

E lie on one straight line.The angles A CB, BCL ,

L CM , MOA are together equal to four rightangles

,by I. 15 , C orollary 1 . The angle BCD is half the angle BCL , and

the angle E CA is half the angle MCA,by construction ; also the angle

A CB is half the sum of the two angles A CB, L CM, by I. 15 . Hence the

5 2 EXERCISES IN EUC LID .

B and C respectively : it is required to describe a circle which shall touch thetwo straight lines and the given circle .Take 0 the centre of the given circle. Join OA cutting the c ircumference

at D . D raw a straight line through D at right angles to OD,and let it meet

AB at E,and AC at F. Bisect the angle AED by a straight line EG meeting

A D at G. Then a circle described from the centre G with the radius GD willsatisfy the prescribed conditions .For from the triangles A OB and A OG we can shew that AG bisects the

angle BA C . Hence the circle described from G as centre with radius GDwill touch the sides of the triangle AEF, by IV. 4 ; and it will also touch thecircle BD C .

If A O produced cut the given circle at K then another solution can b eobtained by using K instead of D .

288 . Take the diagram of IV. 3. Then LA is equal to LC,by Exercise

1 76 therefore the angle LAC is equal to the angle L CA therefore each -of

them is less than a right angle,by I . 17 . The angle LAC is equal to the

angle ABC , by III. 32 therefore the angle ABC is less than a right angle.Similarly BCA and CAB are each less than a right angle.

289 . Let ABCD be a quadrilateral , in which the sum of the sides ABand CD is equal to the sum of the sides BC and AD ; and each angle is lessthan two right angles then a circle can be inscribed in the quadrilateral.Bisect the angles ABC and BCD by the straight lines BO and CO meeting

at 0 . From 0 draw OE perpendicular to BC. Then a circle described withO as centre, and OE for radius will touch AB , BC, and CD . This may beshewn by the process of IV. 4 . It then remains to shew that it will alsotouch AD .

Let BA and CD be produced to meet at S. Suppose i f possible that thecircle described in the manner prescribed does not touc h AD but cuts it .Suppose a straight line LM dr awn parallel to AD , cutting SA at L and SDat M and touching the circle. Then by Exercise 188 the sum of LM and BCis equal to the sum of LB and MC. But

,the sum of AD and BC is equal to

the sum of AB and DC, by supposition. Therefore LM and BC are equalto the sum of LA , AD , MD and BC ; that i s LM i s equal to the three sidesAL

,AD

,MD of the quadrilateral A LMD . But this is impossible : see I . 20.

Similarly we may shew that the circle described in the manner prescribed cannot fall completely within the quadrilateral ABCD .

And in like manner we may treat the case in which ABCD is a parallelogram , so that neither pair of opposite sides will meet when produced.

290. D raw SPT touching both circles at P, cutting HK at S and LM atT . Let HL cut the straight line which joins the centres of the circles at B ;and let KM cut this straight line at C . They will cut it at right angles .

L T and MT are each equal to PT , so that T i s the middle point of LM .

Hence by Exercise 89 BL and CM are together equal to twice PT ,that is to

LM . In this way we find that HL and KM are together equal to -HK andLM and then , by Exercise 289 a circle can be inscribed in HKML .

To s hew that HL will cut BC at right angles, see the construction andfigure on p . 295 of the Euc lid . A CB is a right angled triangle In which A Ci s the difference of the radii of the two circles . When the common tangentis drawn b elow AB similarly situated to DE, the Corresponding right angled


triangle will h ave its sides equal to those of A CB by I. 47 . Hence the radiifrom A to the two points of c onta c t. on the circle of which A is the centre

. will make equal angles with the line AB. Then by I. 26 we have the re

qui red result.

291 . L et ABC be a triangle . Let D be the centre of the circle whichtouches BC

,and also AB and AC produced. Let E be the centre of the

circle which touches CA , and also BC and BA produced. Let F be the centreof the c ircle which touches AB , and also CA and CB produced . ThenAD

,BE , CF will intersect at the centre of the circle inscribed in the

triangle ABC .

Let 0 denote this centre. Then , as in Exercise 280, we can shew that0A bisects the angle BA C . In like manner we can shew that DA bisectsthe angle BA C . Thus the straight line OA coincides in direction with DAso that 0 falls on DA . Similarly 0 also falls on EB, and on EC .

292. Let ABC represent the triangle ; AB and A C being the sides givenin position . D escribe a cir cle which touches AB produced at D , and A C produc ed at F, and touches BC at E . Then , as in Exercise 286 , we can shewthat AD and AF are each equal to half the perimeter of the triangle ABCand so D and F are fixed points. Thus the circle touches the fixed straightlines at fixed points, and so it is a fixed cir cle.

293. Take the diagram of IV. 3. Join MK and KN. We shall shewthat the angle MKN is equal to the sum of a right angle together withhalf the angle MLN . The angles KMA and KME are equal , so that eachof them is half the angle LMN . Therefore the angle MKB, half theangle LMN

,and a right angle, are equal to two right angles by I . 32 .

Similarly the angle NKE,half the angle LNM , and a right angle are equal

to two right angles. Therefore by addition , the angle MKN , half the angleLMN , and half the angle LNM are equal to two right angles ; that isare equal to a right angle together with half the angles at L ,

'

M , N byI . 32 . Therefore the angle MKN is equal to a right angle together with halfthe angle at L . We shall now apply this to the present Exercise . LetPQ be the given base . D raw a straight line parallel to P Q and at adistance from it equal to the given radius : then the centre of the inscribedcircle will be on this straight line. A gain describe on PQ a segment of acircle containing an angle equal to the sum of a right angle and half thegiven vertical angle : then from what is shewn above the centre of the inscribed circle will be on this segment. Hence either of the points of intersection oi this segment and the straight line drawn parallel to PQ maybe taken as the centre of the inscribed circle : the circle may be drawn ,and then tangents to the cir cle from P and Q, produced to meet , will formthe other two sides of the required triangle.

IV. 5 to 9 .

294. From F drawFG perpendicular to BC . SinceFB i s equal to FCthe angle FBC is equal to the angle FCB also the right angleFGB is equalto the right angleFGC therefore BG is equal to CG

,by I . 26 .

.

295 . The tangent at A to the circle described round AD E makes withAD an angle equal to the angle AED , by III . 32. The tangent at A to the

54 E XERC ISES IN EUCLID.

circle described round ABC makes with ADB an angle equal to the angleA UB, by III. 32 . But the angle AED i s equal to the angle A CB,

by I. 29 .

Therefore the tangent at A to the circle described round ABC coincides withthe tangent at A to the circle described round A D E .

296. Let ABC b e a triangle . Let O be the centre of the circle describedabout the triangle, and also the centre of the cir cle inscribed in the trianglethe triangle will be equilateral .Because 0 is the centre of the circle described about the triangle OA ,

OB ,00 are all equal . Because 0 i s the centre of the circle inscribed in the

triangle OA , OB , and C C bisect the angles at A ,B

,and C respectively

,by

IV. 4. Since 0A i s equal to OB the angle OAB is equal to the angle OBAthat is half the angle BAC is equal to half the angle ABC therefore theangle BA C is equal to the angle ABC ; therefore A C is equal to BC, by I. 6 .

Similarly BA is equal to CA . Thus the triangle ABC is equilateral .

297. Let ABC be a tr iangle ; let P be the centre of the inscri bed circle,and Q the centre of the circumscribed triangle : and suppose that P , Q , andA .are on one straight line : then AB will be equal to AC.

From Q dr aw QM perpendicular to AB , and QN perpendi cular to AC. In

the two triangles A QM and A QN the angle QAM is equal to the angle QAN ;the right angles AMQ and ANQ are equal ; and the side A Q i s common :therefore AM is equal to AN by I. 26. But AM is half AB , and AN is halfAC, by III. 3 : therefore AB is equal to AC.

298 . .Let LM denote the common chord . The rectangle PL,

PM is

equal to the square on PA,by II I . 36 the rectangle PL ,

PM is equal to therectangle PB , PC , by II I . 36 Corollary : therefore the rectangle PB , PC isequal to the square on PA . Therefore PA touches the circle which passesthrough A , B, and C. Therefore a s this circle and the circ le AML havethe common tangent PA they touch each other.

299 . Let EP be the tangent at E to the circle described round the triangle ECD , and suppose P and C to be on the same side of ED then EP '

will be parallel to AB .

The angle PEC is equal to the angle ED G, by I II . 32 . The angle ED Gi s equal to the angle ABC , by I II . 22 and I. 13. Therefore the angle PEC isequal to the angle ABC . Therefore EP is parallel to AB, by I . 27 .

300. Let A and B be the two given points ; j oin AB, and produce it tom eet the given straight line at C . On the given straight line take D suchthat the square on CD is equal to the rectangle CA ; CB, II . 14. Th en thecircle which passes through A ,

B , and D is the circle required .

For Isinc e the - rectangle .CA , C B is equal to the square on CD the straightline CD touches the circle which p asses through A , B , and D , by III. 37 .

See the Euc lid, p . 296 .

301 . Let A and B be the given points ; join AB , and produce it to m eetthe given straight line at C. Bisect AB at E from E draw a straight lineat r ight . angles to AB and in this straight line take any point F suchthat AF

‘

i s greater than hal f the given chord. From F a s centre , withradius equal to FA

,describe a circle ; this will also pass through B . Through

C dr aw a straight line CKL by Exercise 1 81 such that the length KL intercepted by this c ircle shall be equal to the given chord. O n the given s traight


line take CM q I al to CK. Then the circle described round A , B, and Mwill satisfy the prescribed conditions.For let this circle cut the given straight line at N. Then the rectangle

CM, CN is equal to the rectang le CA , CB , by III. 36 C orollary ; that is to

the rectangle CK, CL. But GM is equal to CK ; therefore CN is equal toCL . Therefore MN is equal to KL ; that is to the given chord.

302 . Let AB be the g iven straight line in which the centre is to lie ; letAC and BC be the other two given straight lines. Bisect the angle A CE bya straight line meeting AB at D. From D draw a straight line perpendicularto AC and meeting it atE ; from E on the straight line EC cut off EM equalto half the g iven chord . Then the circle described from the centre D

,with

the radius DM , i s the required circle .

For let this circle cut CA again at N ; then MN is bisected at E ,by I II . 3

therefore MN is equal to the given chord .

A gain, from D draw a straight line perpendicular to BC meeting it at F.

Then from the triangles CD E and CDF we see that DF is equal to D E .

Therefore the chord of the circle cut from CB will be equal to that cut fromCA

,by III. 14.

303 . Let ABC and KLM be two triangles, having the bases BC and LNequal

,and the angles BAC and LKM equal . Let F be the centre of the

circle circumscribing ABC,and G the centre of the circle circumscribing

KLM : then FB wi ll be equal to GL .

For , since the angle at A is equal to the angle at K,the angle BFC I s

equal to the angle L CM , by III . 20 ; therefore the angles FD C and FCBtogether are equal to the angles GLM and CML together ; but the anglesFBC and FCB are equal, and so also are the angles GLM and GML ,

by I . 5 .

Therefore the angle FBC i s equal to the angle C LM ,and the angle FCB is

equal to the angle CML . But BC is equal to LM,by supposition ; therefore

FB is equal to GL , by I . 26 .

304. Let A be the point from which the tangent is to be drawn ; let Band C be the two given points through which the circle is to pass . Join AB ;and in thi s straight line, produced if necessary, take the point D such thatthe rectangle AB

,AD is equal to the square on the given tangent

,by I. 45

C orollary . Then D is also a point through which the circle must pass,by

III . 36. Thus we have only to describe a circle passing through the th ree.

points B , C , and D ; and this can be done by IV. 5 .

305 . The angle APB . is half the angle A CB,by III . 20 that i s half a

right angle ; that is equal to BA C . Therefore AB touches the circle . described round ANP : see III. 32 .

306 . Suppose that AB and CD are joined towards apposi te parts by ADand BC. Through E draw a straight line GEH touching the circle AER .

Then the angle AEG is equal to the angle ABE , by III . 32 . But the angleAEG is equal to the angle D EH , by and the angle ABE is equal tothe angle ECD , by I. 29 ; therefore the angle DEH i s equal to the angleECD . Therefore GEH touches the circle ECD see III . 32 . Hence the twocircles have a common tangent at E , and therefore touch each other.

So also the proposition may be demonstrated when AB and CD are joinedtowards the sam e parts by A C and BD

,and A C and BD are pI odII c ed to

meet. See Exercise 295 .

5 6 EXERC ISES IN EUCLID .

307. By II I . .14'

the centre of the required circle m uSt be equally distantfrom the three sides of the triangle and hence it easily follows that it mustcoincide with the centre of the circle inscribed in the triangle.

308 . Join B0 . The angle BAF is equal to the angle CAF, by IV. 4.

Therefore the arc BF is equal to the arc CF, and the straight line BF to thestraight line CF

,by III . 26 and III. 29. A gain , the angle OBF is equal to

the angles OBC andFBC ; that is to the angles OBA and FAC ; that is tothe angles OBA and OAB ; that is to the angle BOF,

'by I. 32 . ThereforeBF is equal to OF, -by I. 6.

309 . Let ABCD be the quadrilateral figure ; let BC and AD be produc ed to meet at P , and let BA and CD be produced to meet at Q .

The angles D RP and D CP are together equal to two right angles , byIII . 22 ; and So al so are

‘

the“

angles D RQ and D AQ . Therefore the anglesD RP

,D RQ, D CP , and DA Q are together equal to four right angles . But

the angles D CR and DAQ are together equal to two right angles , by I . 13and III . 22 . Therefore the angles DEP and D RQ are together equal to tworight angles. Therefore RP and R

’

Q are iri one straight line , by I . 14.

310. D escribe a circle round A OB: Then the straight line which bisectsthe angle A CB will bisect the a re cut off by AB , by III . 26 . A gain thestraight line drawn from the mi ddle point of AB at right angles to AB alsobisects the a re cut off by AB see III. 30. Hence the point D is on the arecut ofi by AB ; Therefore the angles AGB and ADB are together equal totwo right angles; by I II ; 22 .

311 . D raw the tangent to the circle at C and let it meet AB at 0 . Thenthe angle OCE i s equal td the angle CD E , by III . 32 ; that is equal to theangle CBA , by III . Therefore 00 is equal to OB, by I . 6 ; and O is thecentre of the semicircle. Thus the tangent at C to the circle is at rightangles to the tangent at C to the semicircle .Similarly the tangent to the circle at D passes through 0 , and the circ le

cuts the sem icircle at right angles at D .

312 . Let P, Q, R , S be the centres of the circles described round thetriangles A CB, BOG , COD , DOA respectively. N ow P and S are both onthe straight line which passes through the middle point of OA and is at rightangles to CA , by IV. 5 . Thus PS is at right angles to AC. Similarly QR isat right angles to AC. Therefore PS is parallel to QR,

by I. 28 .

Similarly PQ i s parallel to SR .

313. The angle CED is equal to the angle ECD , by I. 5 . But the angleEGD i s equal to the angles ECB and BCD that is to the angles ECB andBA C ,

by III . 32 . And the angle CED is equal to the angles ECA and BAC,by I. 32 . Therefore the angles E CB and BAC are equal ' to the anglesE CA and BA C therefore the angle ECB is equal to the angle ECA .

314. By the process given we see that F is the centre of the circle described round the triangle ABC , so that AF is equal to BF and equal to CF.

Now in the triangle BEC the side BC is of given length the angle BFC istwice the angle BAC, by III. 20, and is therefore a given angle ; the anglesFBC and FCB are equal, and therefore each of them is given, by I. 32.

Hence the triangleFBC is completely known , and soED is of constant length.


315 . The angle ABD is equal to the angle A CE, by I. 5 ; and the angleA CB is equal to the angle A EB,

by III . 2 1 : therefore the angle ABD isequal to the angle BED . Therefore AB touches the circle described roundBED : see III. 32 .

316 . Let E be the point of contact of the circles,TE the common

tangent.Then the angle TED is equal to the angle DEE

,by I II . 32, and the angle

TEC i s equal to the angle D AE ; therefore the angle D EC is equal to theangle AEB, by I . 32.

317. Let C denote the centre of the circle. Then HKmust be as nearto C as possible, by I I I . 15 ; so that the angle HCK must be as great as possible : therefore the angle HPK must be as great a s possible , by III. 20.

Thus the problem is reduced to that which is given on page 308 of theEuc lid.

318 . The centres of both circles will lie on the straight line which bisectsthe angle formed by the two given straight lines . In the given circle place achord at right angles to this straight line containing the centres, and cuttingoffa segment containing an angle .equal to the given angle. Let PQ denotethis chord . We have finally to describe a circle passing through P and Qand touching a given straight line ; for thi s see page 296 of the Euc lid .

319 . Suppose the angles A CE and ABC to be acute . Let OA intersectEF at G . The angle A OB is equal to twice the angle A CB, by III . 20 :therefore the angles BAO and ABO are together equal to the excess of tworight angles over twice the angle A CB; but the angles BAO and ABO areequal , by I . 5 : therefore the angle BA O is equal to the excess of a rightangle over the angle A CB.

A gain Since BFC and BEC are right angles a circle will go round BFEC :see page 276 of the Euc lid. Therefore the angles AFE and A CB are equal,by III . 22 and I . 13.

Thus the angles AFG and FAG are together equal to a right angle ; andtherefore the angle A GE is a right angle .The process requires but a slight modification if either of the angles

A CB and ABC is obtuse .In like manner it can be shewn that OB is perpendi cular to FD and 00

to DE .

320. Let ABCD be the square, P any point on the circumference. SinceABC is a right angle A C i s a diameter of the circle

,and ABC i s a right

angle : see 111 . 31 . Therefore the squares on AP and CP are together equalto the square on A C . Sim ilarly the squares on BP and DP are togetherequal to the square onBD . Therefore the squares on AP

,BP , CP , DP are

together equal to the squares on'

A C and BD , that is to twice the squareon AC.

321 . Suppose ABCD to be a rectangle described about a circle . ByExercise 188 the sum of AB and CD is equal to the sum of BC and DA ;that is twice AB is equal to twice BC : therefore AB is equal to BC, and therectangle is a square.

322 . Let ABCD be a rectangle . Join A C and BD , intersecting at 0 .

By Exercise 78 the diagonals A C and BD are bisected at 0 ; also OB is equal


to OC , by Exercise 59 . Thus OA ,OB , OC , OD are all equal ; and the circle

described from the centre 0 with the radius OA will pass through B,C,and D,

and thus will be desc r ib cd a b out the rectangle .

323. Let 0 be the centre of the circle ; let A OB and COD be twodiameters.

_

D raw tangents at'

A,B , C, D thus forming a quadrilateral

figure having for sides NAK, KD L ,LBM , MON : this quadrilateral figure

will be a rhombus .The straight lines MB and MC are equal , by Exercise 176 ; therefore the

angle MOB is equal to the ‘

angle MOC ,by I . 8 , so that MO bisects the angle

’

BOC ; similarly KO bisects the angle A OD : therefore the angle MOB isequal to the angle KOD .

In the two triangles MOB and KOD the angle MOB is equal to the angleKOD ; th e angles MBO and KD O are equal being rig ht angles ; and OB isequal to OD ; therefore BM is equal to DK. A lso LB is equal to LD

,by

Exercise 176 . Therefore LM is equal to LK.

Sim ilarly it may be shewn that LM is equal to MN, and MN to NK ; sothat the figure NKLM is a rhombus.

IV. 10.

324. The angle A CD is equal to the two anglesBD C and CBD , by I . 32,

But it is shewn in IV. 10 that the angle BD C is e ual to the angle BAD ,

and that the angle CBD i s twice the angle BAD ; erefore the angle A CDis equal to three times the angle BA D .

325 . The triangle BCD is sh ewn in the c ourse of IV. 10 to have each ofthe angles BCD and CBD double of the angle CDB. A lso the triangle A CD_

has each of the angles CAD and CDA one ! third of the angle A CD see

Exercise 324.

326. Suppose F the point at which the circles intersect again . ThenAF is equal to AD . A lso the angle AFD is equal to the ‘angle A DB, byIII . 32 ; the angle A DF is equal to the angle AFD , by I. 5 : therefore theangle FAD is equal to the angle BAD ,

by I . 32 . Thus the angle ADE istwice the angle DAF. Bisect the angle A DE by the straight line D G meeting the circumferenc e of the small circle at G . Then the five angles AD G,

GDF, FAD,D A C , ADC are all equal ; and therefore CD is the side of a

regular pentagon inscribed in the small circle .

327. Let KL be the given base . Make the angles MKL and MLX eachequal to the angle CAD of IV. 10. Then . the angles MKL and MLK are

equal to the angles CAD and ADC ; therefore th e angle XML is equal to theangle A CD : therefore the angle KML is three times the angleMXL or MLK.

See Exercise 324.

328 . This is shewn in the course of the solution of Exercise 326

329 . The angle BAG is twice the angle BAD ,by Exercise 328 ; but

'

theangle ABD is twic e the angle BA D , by IV. 10 : therefore the angle BAG isequal to the angle ABG, and each of them is twice the angle AGB .

330. Let CA produc ed meet the larger circle at G , and let DC producedme et the larger circle at H then the triangle GCH will be of the same kindas the triangle ABD . .


of AF and EC is equal to the sum of BF and CF, that is to the sum of BFand AB .

336. Let ABCD E be the regular pentagon . Join A C , AD ; throughC draw a straight line CF paral lel to AB , meeting AD at F.

A circle may be described round the pentagon by IV. 14. Then AD isparallel to BC, by Exercise 219 ; thus ABCF is a parallelogram ; and thetriangle ABC is equal to the triangle AFC, by I . 34. The triangle AED i sal so equal to the triangle ABC . Thus the regular pentagon exceeds threetimes the triangle ABC , namely by the triangle FCD . Therefore the triangleABC is less than a third of the pentagon .

A gain AC is equal to AD ; AB and BC are together greater than A C , byI . 20; therefore AB and BC are together greater than AD . But BC is equalto AF

,by I . 34 ; therefore AB is greater than FD ; therefore AF is greater

than FD . Therefore the tri angle CFD is less than the triangle CAF. Thusthe regular pentagon falls short of four times the tri angle ABC

,namely by

the excess of the triangle A CFover the triangle CFD . Therefore the triangleABC is greater than a fourth of the pentagon.

337. Let ABC be the equilateral triangle, O the centre of the circledescribed round it . From 0 draw OD a perpendicular to BC and produce itto meet the circumference at E . Then BE and EC are sides of the hexagon .

For from the triangles BOD and COD we can shew that ‘

OD bisects theangleBOC . Therefore the angle BOD is equal to the angle BAC , by II I . 20.

Thus the angle BOE i s equal to the angle of an equilateral triangle . Butthe angles OBE and OEB are equal ; therefore each of these is the angle ofan equilateral triangle. Therefore BE i s equal to B0 .

In this way we can shew that each Side of the hexagon is equal to theradius of the given circle ; and each angle of the hexagon is equal to twicethe angle of an equilateral triangle : thus the hexagon is a regula r h exagon .

A lso the triangles OBC and EEC are equal ; and in this way it -.follows

that the regular hexagon is double the given equilateral triangle .

338 . Let A 1 , A 2, A be a regular quindecagon inscribed in theg iven cir cle . D raw the straight lines A 1A 3, A 3A 8 , A SA”

cutting off arcswhich are to one another in the proportion of 2

,5,8 . The angles which

stand on these arcs will also be in thi s proportion ; and therefore A 1A 3A 8wi ll be the triangle required.

339. We can shew by the method used for Exercise 334 that a regularhexagon is obtained ; we proceed to shew that the area is onethird of thearea of the original hexagon. Let ABCD EF be the given regular hexagon ;let 0 be the centre of the circle described about the hexagon . 0 willalso be the centre of the circle described about the derived hexagon . Let EBand A C intersect at G ; let AC and BD intersect at H.

The angles AEG and BAG are equal so are the angles BCH and HBC .

Then it may be shewn that all the angles of the triangle GBH are equal .Thus AG , GH , H C are all equal . A lso F0 is parallel to AB by Exercise219 ; therefore the triangle ABC is equal to the triangle A BO .

A gain , the angle GOH will be onesixth of four right angles , so tha tGOH will be an equilateral triangle ; therefore it will be equal to the triangleGBH , that i s to one - third of the triangle ABC , that is to one - third of thetriangle OAB . In this way we can shew that the area of the derived hexagonis onethird of the area of the origina lh exagon.


‘

61

340; Let A ,B, C , D ,

be consecutive angular points of the equilateralfi gure inscribed in a circle : this figure will be equiangular . We will shewthat the angles ABC and BCD are equal.For

,the angle ABC stands on an a re which consists of the whole circum

ference except the two smaller arcs cut off by AB and BC respectively ; andthe angle BCD stands on an a re which consists of the whole circumferenceexcept the two smaller arcs cut off by BC and CD respectively . Now thesmaller a re cut off by AB is equal to the smaller arc c II t off by CD , byIII . 28 . Therefore the arc ABC is equal to the arc BCD . Hence the a re onwhi ch the angle ABC stands is equal to the a r e on which the angle BCDstands. Therefore the angle ABC is equal to the angle BCD , by 111 . 27.

VI. 1 , 2.

341 . In the diagram of IV. 10 take E on AD such that AE may be equalto A C : then it may be shewn that the triangle CDB is equal to the triangleCAE . Therefore the triangle GED i s to the triangle A OB as the triangleA CE is to the triangle A CD ; that is as AB is to AD , by VI . 1 ; that is as A Cis to AB . A gain , the triangle A CD is to the triangle ABD as A C is to AB ,

by VI. 1 . Therefore the triangle CBD is to the triangle ACD as the triangleA CD is to the triangle ABD . Therefore the triangle A OD is a mean proportional between the triangles CBD and ABD .

342 . The triangle AFE is equal to the triangle ED E , by I. 34 ; thetriangle FD R i s equal to the triangle FDC, by I . 37 therefore the triangleAFE is equal to the triangle FDC. Hence the triangle BED i s to thetriangle AFE as BD is to D C , by VI. 1 . A ga in, the triangle AFE i s to thetriangle ED C as AE i s to E C , by VI . 1 ; that I s , as BD is to D C , by VI. 2 .

Therefore the triangle BED is to the triangle AFE as the triangle AFE is tothe triangle ED C ; so that the triangle AFE i s a mean proportional betweenthe triangles BFD and ED C .

343. From a point 0,within an equilateral tri angle ABC ,

let perpendicular s OP

,OQ , OR be dr awn on the sides BC, CA , AB respectively. A lso

dr aw CD from C perpendicular to AB . Then in the same manner as inVI. 1 . Cor . it may be shewn that triangles on equal bases are to one anotheras their altitudes ; thus the triangle OBC i s to the triangle ABC as OP I s toCD , also the triangle OCA is to the triangle ABC as OQ is to CD ,

and thetriangle OAB is to the triangle ABC as OR is to CD . Hence the sum of thetriangles OBC , OCA , OAB Is to the triangle A BC as the sum of OP, OQ , ORis to CD . But the sum of the triangles OBC , OCA ,

OAB is equal to thetriangle ABC . Therefore the sum of OP, OQ , OR is equal to CD .

344. Let ABC be a triangle . From A draw AD perpendicular to BC,and from D on DA take DK equal to one third of DA . From B draw BEperpendicular to AC, and from E on EB take EL equal to one thi rd

c EB.

Through K dr aw a straight line parallel to BC ; and through L drawastraight line parallel to A C : let these straight lines meet at 0 : th is shall bethe point requir ed .

For the triang les ABC and OBC have the same base, but the height ofORC is one - third of the height of ABC ; therefore the tr iangle OBC is one

62 EXERC ISES IN EUCLID ,

third of the triangle ABC , as inVI .

‘

1 . In like manner the triangle OCA is

one - third of the triangle ABC . Therefore th e triangle OAB is also one - thirdof the triangle ABC .

345 . CF is to FB as AF is to EB, and also D G is to GB as AE i s toEB

,by VI. 2 . Therefore CF is to FB as D G

,is to GB . Therefore FG is .

par allel to CD, by VI. 2 .

346 . Let ABC be a triangle. From any point K in the base AB drawKL parallel to AC meeting CB at L ,

and KM parallel to BC meetingCA at M.

The diagonals CK and LM intersect at 0 the middle point of CK, byExercise 78 . Thr ough O draw a straight line POQ parallel to AB , meetingAC at P, and BC at Q . Then CP is to PA as CO is to OK ; but C O is equalto OK ; therefore CP is equal to PA . Thus P is the middle point of CA .

Similarly Q is th e middle point of CB . Thus 0 is on the fixed s traightline which joins the middle points of AC and BC.

347 . Let the straight line from D parallel to BCmeet AC atE . ProduceAD to meet BC at F. Then AD is equal to DF,

by I. 26 . An d AD is toD E as AE is to EC , by VI . 2. Therefore AE is equal to EC .

348 . The triangle BED i s equal to the triangle CED , by I. 37 thereforethe triangle DFB is equal to the t riangle EFC .

The triangle ADE is to the triangle BDF as AD is to DB ,by VI. 1 ;

that is a s AE is to E C , by VI . 2 ; that is as the triangle AEF is to thetriangle EFC , by VI. 1 that is as the triangle AEFi s to the triangle BDF.

Thus the triangle ADE is to the triangleBDFa s the triangle AEFis to thetriangle BD E. Therefore the triangle ADFis equal to the triangle AEF.

349 . Let AF produced meet BC at H. The triangle BFH is to the triangle BFA as FH is to FA ,

by VI. that Is as the triangle CHF is to thetriangle CFA . But from what is shewn in the last Exercise we find that thetriangle AFB is equal to the triangle AFC . Therefore the triangle BEHis to the triangle BFA as the triangle CFH is to the triangle BFA . Therefore the triangle BFH i s equal to the triangle CFH. Therefore BH i s equalfto CH .

350. Let ABCD be a quadrilateral figure, having the sides AB ,and DC

parallel. Let a‘straight line parallel to these side‘s meet AD atE

,and BC a t

'

F. Then DE will be to EA as CFi s to FB.

Of the two sides AB and DC suppose AB the greater,From C draw a

straight line C GH parallel to DA meeting EF at G,and AB at H. Then

CE and GA are parallelograms ; and therefore CG is equal to D E ,and GH

is equal to EA .

Now CG is to GH as CFis to FB, by VI. 2 ; therefore D E is to EA asCF is to FB.

351 . Let P be the given point. Bisect PA at Q . From Q draw a straightline parallel to AC, meeting BC at R . Join PR and produce it to meet A G,

produced if necessary , at S.

Then PR is to RS as PQ is to QA , by VI. 2. But PQ is equal to QA ;therefore PR is equal to RS.


VI. 3, A .

352 . AF is to FC as AD is to DC. by VI . 3 ; and so also AE is to EBas AD is to DB . But DC is equal to DB . Therefore AF is to FC as AE isto EB. Therefore FF is parallel to BC

,by VI. 2 .

The are DEC i s bisected at B : see III. 30. Therefore the angleD GO is b isec ted

g

by'

GE , by IH . 27 . Therefore D G is to GC as D E i s to E C .

“

Similarly D E is toFC as D E is to EC . Therefore D G is to G0 as D E is toFC . Therefore also D G is to DE as C O is to FC ,

by V . 16 .

354. Let AB be the given straight line . From A as centre with anyradius describe a circle . From B as centre with a radius equal to doublethe former describe another cir cle , cutting the form er at C. Join AC andBC, and bisect the angle A CB by a straight line meeting AB at D .

Then A C is to CB as AD is to DB,by VI . 3. But BC is twice A C ;

therefore DB is twice AD . Bisect DB at E . Then AD,DE

,EB are all

equal ; so that AB is trisected.

355 . The angle CPD is bisected by PA therefore CA is to AD as CP isto DP, by VI. 3 . The angle APB is a right angle , by I II . 31 ; therefore PBbisects the angle between CP produced and DP. Therefore CB is to DB asCP is to PD

,by VI. A . Therefore CA is to AD as CB is to DB ; therefore

CA i s to CB as AD is to DB , by V. 16 .

356 . On AB as diam eter‘

desc r ibe a circle . Bisect the arc AB at C.

Join CD and produce it to meet the circumference again at E . From Edraw a straight line at right angles to D E , and let it meet AB produced atP. Then P will be the required point .For

,since the arc AC is equal to the arc BC, the angle AFB is bisected

by ED , by III. 27 . Therefore AB is to EB as AD is to DB , by VI . 3. AndSince the angle D EP is a right angle EP bisects the angle between AE produc ed and BE . Therefore AE i s to EB as AP is to BP , by VI. A . Therefore AD is to DB as AP is to BB.

357 . AB is equal to AE , by supposition ; therefore the angle AED i sequal to the angle A BC ; also the angle EAD is equal to

‘ the angle BACth erefore ED is equal to BC, by I . 26 . Now AC bisects the angle BAD , andAE is at right angles to AC ; therefore BC is to C D as BA is to AD , by VI. 3 ;and BE is to ED a s BA is to AD

,by VI. A . Therefore BE is to ED a s BC

is to CD . But ED is equal to BC . Therefore BE is to BC as BC is to CD ;so that BC is a m ean proportional between BE and CD .

358 . BD is to D C as BA is to A C ; therefore the difference of BD andD C is to their sum as the d ifi erenc e of BA and AC is to their sum : see theEuc lid , p . 310. Now the difference of BD and D C is twice DO, and their sumis twice OB . A lso twice DO is to twice BO as D O is to B0 , by V. 15 . Therefore D O is to B0 as the difference of BA and A C is to their sum .

359 . Suppose E is on BC produced through C. By VI . 3 and VI. A wehave BD to DC as BE is to EC . Therefore the difference of BD and D Cis to D C as the difference of BE and E C is to E C . That is twice OD i s toDC as twice 00 is to CE . Therefore OD is to D C as 00 is to CE . Th ere:fore OD is to the sum of OD and DC as 00 is to the sum of OC and CE ;

that is OD is to 00 as 00 is to OE . But 00 is equal to OR ; therefore OD is

64 EXERCISES I N EUC I’

.ID .3

to OB as OB is to OE ; so that OB is a mean proportional between ODand OE.

360. Let ABC be a triangle, let D, E ,F be points in BC , CA ,

AB,re;

spec tively such that DE and D E make equal angles with BC, ED and EFmake equal angles with CA

,FE and FD make equal angles with AB . Then

AD, BE ,

CF shall be at right angles respectively to BC,CA , AB .

Let AD and FE meet a t ‘G. Since the angles GEA and D EC are equalAE bisects the angle between D E produced and GE ; therefore D E is toEG as DA is to GA . Similarly DF i s to FG as DA is to GA . ThereforeD E is to EC as DF i s to FG ; therefore D E is to DFas EG i s to FG .

Therefore D G bisects the angle FD E , VI. 3 . Thus the angle GD E i s equalto the angle GD E ,

and the angleFDB is equal to the angle ED O thereforethe angle GDE is equal to the angle GD C , so that each of them is a rightangle .

Similarly it may be shewn that BE is at right angles to A C, and CF atright angles to AB .

VI. 4 to 6 .

361 . Let ABC and DEFbe triangles on equal bases AB and DE ,and

between the same parallels ABDE and CF. Let a straight line be drawnparallel to ABD E , meeting A C at K, BC at L , DF at M, EF at N. Thenthe triangle CKL shall be equal to the triangleFMN .

KL is to AB as CL is to CB,and MN is to DE as EM is to FD

,by VI . 4.

But CL is to CB as FM is to FD , by Exercise 350. Therefore KL is to ABas MN is to D E . Therefore KL is to MN a s AB is to D E . But AB is equalto D E , by supposition . Therefore KL is equal to MN. Therefore thetriangle CKL is equal to the triangleFMN , by I . 38 .

362 . CE is to AB asFE is toFB, and ED is to AB as GD is to GB,by

VI. 4. But CE is equal to ED , by supposition. Therefore FE is to F3 asGD is to GB . ThereforeFG i s parallel to ED by VI. 2 .

363. D raw any straight line thr ough C. D raw BN and AM perpen

di culars on thi s straight line . Then BN i s to AM as CB is to CA,by VI. 4 ;

that is BN has to AM a constant ratio.

364. Let . A and B be two fixed points , and suppose a straight line topass between them and to cut AB at 0. D raw AM and EN perpendicularson MCN . Th enA M i s to AC as BN i s to BC , by VI. 4 . Therefore AM is

to EN as AC is to BC . Thus if the ratio of AM to BN is given the ratio ofA C to BC is also given ; and therefore C is a fixed point.

365 . Let A , B , C be three given points. Suppose a straight line to pass.

through a point D between A and C and also through a point E betweenB and C. D raw AF, CG , BH perpendiculars on this straight line. ThenAF is to CG as AD is to DC, by VI. 4. Therefore Since the ratio of AF toCG is known the point D is known . Similarly the point E is known. Thusthe required straight line i s obtained by joining D E .

366. Let A '

and B be the points from which the perpendiculars are to bedr awn , C the -point through wh ich the straight line is to be drawn .


Join A C and produce it to D , making AC to CD in the given ratio. JoinBD ,

and through C draw a straight line E CFperpendicular to BD : thenECF wi ll be the required straight line .For draw AM perpendicular to E CF

,and let EFand BD intersect at N.

Then CM is to CN as A C is to D C , by VI. 4 : thus CM has to CM theassigned ratio.

367 . The angle BED is equal to the angle EFC , by I. 15 ; the angleDBF is equal to the angle FEC , by I . 29 . Therefore the triangle BFD i sequiangular to the triangle EFC . Therefore BD is to CE as BF is to FE .

But ED is equal to BA , and CA is equal to CE therefore BA is to A C asBF is to FE . Therefore AF is parallel to CE , by VI. 2.

368 . From P draw PY perpendicular to AB,and from Q draw QZ

perpendicular to CD . Then the triangle PMY is equiangular to the triangleQNZ therefore PM is to QN as PY is to QZ, Thus the ratio of PM toQN is constant .A gain

,let NM and QP produced if necessary meet at R . Then the tri

angle PMR is equiangular to the triangle QNR ; therefore RP is to RQ asPM is to QN so that the ratio of RP to RQ is constant therefore R is afixed point .

369. Let ABCD be a quadr ilateral figure , in which AB is parallel toCD and equal to twice CD. Join AC and BD intersecting at 0, Then OCwill be one - third of AC.

The angle DOC is equal to the angle BOA , by I . 15 ; the angle GCD i sequal to the angle OAB, by I . 29 ; therefore the triangle D OC i s equiangular to the triangle BOA . Therefore A O is to CO as AB is to DC, byVI . 4. But AB is twice DC ; therefore A O i s twice CO . Bisect AO at Pth en AP, P0 , 00 are all equal and 00 is one- th ir d of AC.

370. The angles CA T and CBT are right angles ; therefore a circle willgo round CA TB : see page 276 of the Euc lid. Therefore the angle CABis equal to the angle C TB,

by III . 21 ; therefore the angle ABN is equal tothe angle C TB. Al so the right angles ANB and CBT are equal . Thereforethe angle NAB is equal to the angle BOT, b y I . 32 . Thus the triangleBAN is equiangular to the triangle ECT ; and therefore BT is to BC asBN is to NA , by VI . 4.

371 . Through E draw OE parallel to AB,meeting BC at 0 . Then AB

is to A C as OE Is to E C ,by VI . 4 ; that is as OE is to BD , by Supposition ;

that Is as EFIs to DE,by VI. 4.

372 . Let P be the centre of the circle which passes through A , C, andany point D in BC ; let Q be the centre of the circle which passes throughA

,B

,and D.

The angle APC is equal to twice the excess of two righ t angles overA D C , by III. 22 , 20, that is the angle APC is twice A DB ; also the angleA QB is twice A DB, by III . 20 ; therefore the angle A QB is equal to theangle APC . Thus the isosceles triangle A QB i s equiangular to the isoscelestr iangle APC , and therefore PA is to QA as AC is to AB .

373. Let ABC be a triangle. Suppose that the perpendicular from therequired point on BC 18 to be to the perpendicular on CA in the ratio of Xto Y.

T . EX . EUC .

66 EXERCISES IN EUCLID ;

D rawa straight line parallel to‘BC, a t the distance X froni it ; and draw

a straight line parall el to CA at a distance Y from it let these straight linesmeet at D

.Then the perpendicular from D on BC is equal to X , and the

perpendicular‘

from D onCA is equal to Y.

Join CD ; take any point P on CD ,and draw PM perpendi cul ar to BC

and PN perpendicular to CA . Then PM is to X as CP is to CD , by VI . 4 ;

and also .PN i s to Y as CP is to CD therefore PM is to PN as X is to Y.

A gain , suppose that the perpendicular from the required point on' CA is

to be to the perpendicular on AB as Y is to Z . Then as before we find a pointE such that the perpendi cular from it on AC is Y, and the perpendicularfrom it on AB is Z . Join AE then, as before , we can shew that the perpendi cular s from any point in AE on CA and AB are in the ratio of Y to Z .

Let CD and AE be produc ed to meet at 0 . Then from what has beenshewn it follows that the perpendiculars from O on BC, CA , AB are pr0portional to X ,

Y, Z respectively.

374. Let A EB be one triangle , and AFC the other. Suppose that AEand AF are homologous . D raw EM perpendicular to AB, and EN p erpen

dicular to AC and produce EM and FN to meet at P. Let D be the angleof the rectangle opposite to A .

Then the triangle ABM is equiangular to the triangle AFN and thereforeAM is to AN a s AB is to AF,

that is as AB is to AC . Therefore AM is toMP as AB ~ is to BD ; therefore the triangle AMP is equiangular to thetriangle ABD ,

by VI . 6 therefore P is on the straight line AD .

If EB and AF are homologous it will be found that the perpendi cularsmeet on BC .

375 . L et GE produced through E , and CA produced through A ,-meet

at the point P. Then CP i s to KP as CG is to KE , by VI . 4. Let EH'

produced through H, and CA produced through A , meet at the point Q .

Then CQ is to KQ as CF is to KH ,byVI . .4 . But the triangle C GK is equi

angular to the triangle KEA therefore CG is to GK as KE is to EA,that is

CG is to CF as KE i s to KH ; therefore CG is to KE as CF is to KH .

Therefore CP is to KP as C Q is to KQ : therefore the poInts P and Qcoincide ; so that GE

,EH , and CA , produced, meet at a point .

376. Let PQ and AC , produced if necessary, mect‘at L . Then LP is to

L Q as AP is to C Q , by VI. 4 . Let PQ"

and BD,produced if necessary

,meet

at M. Then MP is to MQ as PB is to QD . Now AP is to PB as C Q is to QD ,

by supposition . Therefore LP i s to L Q as MP is to MQ . Therefore Lcoincides with M.

If instead of having given that AP’

is to PB as C Q is to D Q we have APIS to PB as D Q is to Q C , we can shew that PQ , AD , BC meet at a point.

377. Let the straight line parallel to AB cut AC at M,and BC at N.

From M and N draw straight lines parallel to BD meet-ing AB at P and Qrespectively. Then PA will be equal to QB .

From C dr aw a straight line parallel to AB meeting BD at ‘

E . Thensince the straight line AD is bisected at C the straight l ine BD i s bisected at "

E ,byThe triang le MAP is equiangular to the triangle D OE so that MP .

is to PA as D E is to EC , by VI . 4 . A gain,the triangle NBQ is equiangular


384. From A draw AE perpendi cular to BD , from B draw BFperpendicular to A C, from C draw C G perpendicular to BD , and from D drawD H perpendicular to AC. Let AC and BD intersect at O .

The triangle OAE i s equiangular to the triangle OBF, and therefore 0A isto OE a s OB is to OF ; therefore the triangle OFE is equiangular to thetriangle OAB , b y VI. 6 ; therefore the angle OEFis equal to the angle OABand the angle OFE to the angle OBA . Similarly the angle OEH is equal tothe angle OCB, that is to the angle OAD . Therefore the angle FEH isequal to the angle BA D . And the angle EFH has been shewn equal to theangle ABD . Therefore the angle EHE i s equal to the angle BD A . Therefore the triangle FEH i s equiangular to the triangle BA D ; and thereforesimilar to it

,by VI . 4 .

Sim ilarly the triangle FGH is similar to the tri angle BCD . Hence it willfollow that EFGH i s similar to ABCD .

385 . Let A be the centre of one circle,and C that of the other. Let

them intersect at the given point B . Then , by supposition , BA and BC arefixed directions . Let a straight line touch the former cir cle at E

,and the

latter at F ; and let EFproduced meet AC produced at D . Then D is thepoint at which two tangents to both circles will intersect.Then DA is to D C as AE is to CF, by VI. 4 ; that is as AB is to BC.

Therefore DB bisects the angle between AB produced and BC, by VI. A :

thus D is on a fixed straight line .Now a s BA and BC are fixed direc tions other cases may occur in which

instead of A we have some point on AB p roduced through B, or in whichwe have instead of C Some point on CB p roduc ed through B . Thus weobtain a second fixed straight line , namely that which bisects the angle ABC .

VI. 7 to 18 .

386. Let the circles touch each other at the point B ; let one circle touchthe straight line at C, and let the other circle touch the straight line at D .

D raw CA a diameter of the former circle , and DE a diameter of the latter.D raw the straight line BF touching the circles at B ,

and meeting CD at F:

Thus FB, FC ,FD are all equal , by Exercise 176. Therefore a circle

described with F as centre will go through B , C, and D therefore CBD willbe a right angle, by III. 31 . Also EBD i s a right angle , by II I . 31 . ThusGBE i s a straight line , by I .

—l 4 . Similarly DBA is a straight line .N ow the angle BCD i s equal to the angle CAB, and the angle BD C i s

equal to the angle D EB,by 111 . 32 . Therefore the triangle A CD i s equi

angular to the triangle CD E therefore ED i s to D C as D C i s to CA .

387 . Let EDK be the given arc , and EGK the remaining part of thecircum ference. Bisect the arc EGK at F. D ivide EK at H so that EH maybe to BK in the given ratio. Join FH and produce it to meet EDKa t Lthen L will be the point required .

For since the arc EFi s equal to the.

arc FK the angle ELF is equal tothe angle ELK, by III . 27 . Therefore EL i s to LK as EH is to HK, byVI. 3 : so that EL i s to LK in the given ratio .

388.Let ABC be the triangle ; draw CE p arallel to AB , and make CE toCB as CB is to BA . Join BE cutting A C at D . D raw D E parallel to AB,

meeting BC at F.

EXERCISES I N EUCLID . 69

Then D E is toFB a s CE i s to CB, by VI . 4 that is as CB is to BA , byconstruction that is as CF is to FD

,by VI. 4 . Thus DFis to FB as CF is

to FD ; therefore FB‘i s to FD as FD is to CF, by V. B. Thus FD i s a.

mean proportional betweenFB andFC .

389 . The angles BDA and ADC are equal being right angles ; also EDis to DA as DA is to DC, by supposition . Therefore the triangle BDA isequi angular to the triangle A D C , by VI . 6 . Thus the angle BAD is equal tothe angle A CD ,

and the angle ABD is equal to the angle CA D . Thereforethe angle BAC is equal to the two angles ABC and A GB ; therefore the angleBAC is a right angle .

390. Since BD is to BA as BA is to BC the triangle BDA is equiangularto the triangle BA C , by VI. 6 ; therefore the angle BAC is equal to the angleBD A ; that is the angle BAC is a right angle .

391 . CA is to CP as CP is to CB , by supposition ; therefore the triangleA GP is equiangular to the triangle P CB, by VI. 6 ; ther efore the angle CPAis equal to the angle CBP .

392 . Let the centre of the circle in any position be at a point C ; and letthe circle touch the straight line OA at the point B .

PQ is a third proportional to OP and P C so that OP is to P C as PC isto PQ ; therefore the angle OCQ is a right angle , by Exercise 389 . ProduceQC to meet OA at N. Then

,

in the triangles OCQ , OCN the angle COQ isequal to the angle C ON because the triangles C OP ,

C OB are equal , byExercise 176 the right angle OC Q i s equal to the right angle OCN and00 i s comm on therefore Q C is equal to CN ,

by I. 26 .

From Q draw QM perpendicular to OA . Then the triangle QNM is equiangular to the triangle CNR ; therefore QM is to CB as QN is to CN , byVI. 4. But QN h a s been shewn to be twice CN therefore QM is twice CB .

Thus Q is always on a straight line which is parallel to GA and at a di stancefrom it equal to twice the g iven radius.

393. Let A S and BT be the parallel straight lines , AB being a diameterof the circle ; let C be the centre of the circle. Then SCT i s a right angle

,by

Exercise 182 . Therefore CP is a mean proportional between SP and PT ,by

VI. 8 , C orollary . Therefore the rectangle SP,PT is equal to the square on

CP,byVI . 17 ; thus the rectangle SP ,

PT i s constant.

394 . Suppose D the point in the side AB of the triangle ABC , and letD E be parallel to BC. Then the triangle A DE must be equal to the triangleDBC ; therefore AD is to DB as BC is to D E , by VI . 15 . ButBC is to D Eas AB is to AD , by VI . 4. Therefore A D is to DB as AB is to AD ; therefore the rectangle AB, DB must be equal to the square on AD . Thus ABmust be divided a t D in the manner of II . 11 .

395 . The triangles ECA andBCD are equiangular ; therefore E C is to CAas CB is to CD , by VI. 4 ; therefore the triangle ECD is equal to the triangleA CB, byVI . 15 .

396 . The triangle ABE is equiangular to the triangle CBF thereforeAB is to BE as CB is to BF, by VI. 4 ; therefore AB is to CB as BE is toBF therefore the triangle ABFis equal to the tr iangle CBE , by VI. 15 .


397. Let ABCD be a quadrilateral figure inscri bed in a circle ; let”

A C

and BD intersect at O .

The angle A01), is equal to the angle BOC ,by I. 15 ; the angle DAO is

equal to the angle CEO , by and the angle ADO is equal to the angle8 00 ,

by III. 21 : therefore the triangle A OD is equiangular to the triangleBOC . .Therefore DO is to AO as C O i s to B0 , by VI. 4 . Th erefore

‘

the

rectangle DO,OB is equal to the rectangle CO , OA ,

by VI. 16 .

Similarly it may be shewn that the triangle COD i s equiangular to thetriangle BOA .

398 . Let EFand CD meet at M. Then G0 is to EM as C O is to CM,

and LO is to FM as C O is to CM, by VI . 4. Therefore G0 is to EM as LOis to FM. Therefore G0 is to LO as EM is to FM. Similarly KO is to H Oa s EM is to FM. Therefore G0 is to LO as KO is to H0 . Therefore therectangle GO , H O is equal to the rectangle L O,

KO,by VI . 16 .

399. The angles DFC , A CD ,BD C and half of!A CB make up two right

angles,by I . 32 . The angles D GC , A CD , BD C and half of ADB also make

up two right angles. Therefore th e angles DFC and half of A CB are togetherequal to the angles D GO and half of A DB. Therefo re the angle DFC is

equal to the angle D GC ,by II I . 21 . Therefore a circle will go round D GFC ,

by page 276 of the Euc lid . Therefore the rectangle EG, ED is equal to therectangle EF, E C , by III . 36 , C orollary. Therefore EFis to EG as ED is toE C , by VI . 16 .

400. Let ABC be the triangle . From A draw a straight line -

A D ,meeting

BC at D ; and also draw from A a straight line AE , meeting the circumferenceof the circumscribing cir cle at E , such that the angle A CE is equal to theangle A DB.

Then the angle ABD is equal to th e angle AEC , by the angleADB 18 equal to the angle A CE ,

by cdnstruc tion ; therefore the angle BADis equal to the angle EAC ,

by Thus the triangle BAD is equiangularto the triangle EA C and therefore AB is to AD as AE i s to AC , by VI. .4 ;and therefore the rectangle AB , A C i s equal to the rectangle AD,

AE .

401 . AC is to CE as CD is to CB ; and the angle A CE is equal to theangle DCB therefore the triangle A CE i s equiangular to the triangle D CB,

by VI . 6 , so that the angle CEA is equal to the angle OBB. Therefore acircle will go

;

round CEFA ; and the point E will bisect the arc AEB,

because the angle A CE i s equal to the angle BCE . If AB and the angleA CB are given , this circle will be a fixed circle, and E will be a fixed pointon the circumference. See p . 276 of the Euc lid .

402 . Let DFGE be the square ; F being on the side AC, and G on theside BC. Then the triangle A DE Is equiangular to the triangle GEB ; therefore EB i s toEG as DFI s to DA . Therefore the rectangle AD

,BE i s equal

to the rectangle EG DF, by V .I 16 , that Is to the rectangle EG ED ; that

is to the square DFGE .

403 . T he triangle AFE is equiangular to the triangle CFB ; therefore EFi s to FB as FA is to FC , by VI. 4 . In like manner from the triangles GFCand BFA we haveFB to FG as FA to FC . Therefore EFIs to F!) as FB isto FG ; therefore the rectangle EF, EC is equal to the square on FB, by

'

VI


404. In the triangle ABC suppose that AB is equal to AC. From Adraw a straight line meeting BC at D,

and produce AD to meet at E the circum ferenc e of the circle described round ABC .

The angle AFB is equal to the angle A CB, by III . 21; the angleA CB i sequal to the angle ABC ,

by : therefor e the angle AFB i s equal to theangle ABD . Thus the triangle

{

AEB is equiangular!to th e triangle ABD , by

Therefore DA is to AB as AB i s to AE,by VI. 4 ; therefore the

rectangle DA , AE is equal to the square on AB, by VI. 17.

405 . Let T be one of the points of contact of the given tangents , andP one of th e points of intersection of the two circles . Then the square onET is equal to the rectangle EA

,EH , by VI . 8 , C orollary ; and ET is equal

to EP ; therefore the square on EP is equal to the rectangle EA , EH ; therefore EP touches the cir cle HPA , by 111 . 37.

VI. 19 to D .

406 . In the diagram of IV 10 suppose a straight line drawn from Cparallel to BD , meeting AD at F. Th en FB bisects the angle ABD : seeExercise 63. The triangle A CFwill be to the figure BCFD as BD is to BA .

The triangle A CFis equiangular to the triangle ABD ; and therefore thetriangle A CE is to .the triangle ABD as the square on A C is to the square onAB , by VI . 19 that is as the rectangle AB

,BC is to the square ou AB , by

IV . 10 that i s as BC is to AB . Therefore the triangle A CE is to the figureBCFD as BC is to AC, see V . E . But AC is equal to BD ; therefore thetriangle A CE is to the figure BCFD as BC is to BD ; that is as BD is to BA,for the triangle BCD is equiangular to the triangle D A -D .

407. Let EB be a side of the regular polygon , K the centre of the‘ circl e.

Let CB be half the side of the circumscribed figure of half the number ofsides ; C, E , K being In one straight line. From E draw EL perpendicularto

d

BK ; then EL“

15; half the side of the Inscribed figure of half the number ofSI esLet X , Y, Z denote the area s of the three figures respectively in descending

order of m agnitude. Then X is to Z as . the ,triangle CBK is to the triangle

ELK ; that is in the duplicate ratio of CB to EL ,by VI . 19 . A lso X is to Y

as the triangle CEK is to the triangle EBK ; that is as CK is to EX , byVI. 1 ; that is a s CB is to EL , byVI . 4 . Thus X is to Z in the duplicateratio of X to Y. Therefore Y is a mean proportional between X and Z .

408 . Join EG cutting AF at P,and HK cutting FC at Q . The triangles

A EF and FH C are equiangular ; therefore AE is to AF as EH is to FC .

But AP is half of AF, and FQ is half of FC ; therefore AE is to A P as FMis to FQ . Therefore the triangle AEP I s equiangular to the triangle PH Q,by VI . 6 , so that the angle APE is equal to the angle FQH . ThereforeEy

'

P 1s parallel to HQ , by I . 28 .

409 . Let ABC be the triangle. From C draw C H perpendicular to AB,“

produced if necessary ; and complete the rectangle AFCH. D escribe arectangle AED G similar to AHCF,

and equal to the triangle ABC,so that

glm a

ybe on A H and G on AF

,by VI. 25 . Then D will fall on A C, by

26


Then th e triangle AED is half the rectangle AED O , and is thereforeequal to half the triangle ABC . Thus the straight line ED satisfies theassigned condi tions .

410. Let ABC , DEF be the two isosceles triangles which are to oneanother in the duplicate ratio of their bases BC, EF. Then ABC and DEFshall be similar triangles.For if the ang le ABC be not equal to the angle DEF, one of them must

be the greater. Let ABC be the greater , and make the angle C -BG equal tothe angleFED . Similarly make the angle BCG equal to the angle EFD .

Then GBO and ABC being isosceles triangles the point G will fall withinthe angle BA C , so that the triangle ABC is greater than the triangle GBC .

A lso D EFand GBC are similar triangles .N ow, by supposition , the triangle D EF is to the triangle ABC in the

duplicate ratio of EF to BC ; and the triangle D EF is, by VI. 19 , to thetriangle GBC in the same ratio ; therefore the triangle D EF is to thetriangle ABC as the triangle D EFi s to the triangle GBC . Therefore thetriangle A BC is equal to the triangle GBO, which is absurd. Therefore thetriangle ABC is similar to the triangle D EF.

411 . The rectangle contained by the two segments i s known , for it isequal to that of the segments of any chord of the circle through the point.A lso the ratio of the sides of the rectangle is known. Hence the rectanglecan be constructed by VI . 25 .

412 . The rectangle contained by the line and one segment is known , forit i s equal to the square on the tangent . Al so one side of the rectangle isto be double the other. Hence the rectangle can be constructed by VI . 25 .

413. The straight line CD is divided similarly to AB . The straightline EB is divided s imilarly to AB : see VI. 2 . A lso it is shewn in II. 1 1that the rectangle CF,

FA is equal to the square on AC ; therefore CF isdivided at A in the required manner. And

‘

KG is divided at H similarly tothe way in which CF is divided at A .

414. Let BC denote the given base ; on BC describe a segment of acircle containing an angle equal to the given vertical angle. Then thediameter of this circle is known.

By VI. C the perpendicular from the vertex on the base of the triangle i sknown . Hence we must dr aw a straight line parallel to BC at a distancefrom it equal to thi s known perpendicular ; either of the intersections of thisstraight line with the arc of the described segment may be taken for the required vertex of the triangle.

415 . Let ABC be an equilateral triangle ; suppose a circle to be describedround the triangle , and let P be any point on the circumference of this circle.D raw PA , PB , PC.

Suppose P to be between A and C . Then APCB is a quadrilateral ihscribed ih a circle ; and therefore the rectangle PB , AC is equal to the sumof the rectangles PA , BC and PC, AB,

by VI. D . But AC, BC, and AB areall equal

,by supposition. Therefore PB is equal to the sum of PA and PC.

416 . Since the angles ABD and A CD are right angles a cir cle would goround ABD C , by page 276 of the Euc lid ; therefore the rectangle AD, BC is

EXERCISES IN EUCLID .

equal to the sum of the rectangles AB , CD and AC, BD ,by VI. D . But AB

is equal to A C, therefore the angle ABC is equal to the angle A CB,by L a

therefore the angle D -BC is equal to the angle D CB ; therefore DB 15 equalto DC

,by I. 6 . Thus each of the rectangles AB

,CD and AC , ED is equal

to the rectangle AB , DB. Therefore the rectangle AD,BC is equal to twice

the rectangle AB , DB.

417. Suppose that the straight line through A falls without the triangleABC , and that the perpendi cular CE is less than FG.

The angle AFC is a right angle by I . 8 ; therefore a circle would go roundAECF, by page 276 of the Euc lid . Therefore the rectangle A C , EFis equalto the sum of the rectangles AE , FC and AF, EC. From F draw EH per

pendicula r to EC produced ; then E C is the excess of EH over HC, that isequal to the excess of EC over HC. Therefore the rectangle AC, FF is equalto the excess of the rectangles AE , FC and AF, FG over the rectangle AF,

HC. Now the triangle AFG is equiangular to the triangle HFC ,for AF and

EH are respectively perpendicular to FC and FG ; so that FA is to AG asEC is to HC, by VI. 4 ; and therefore the rectangle FA , HC is equal to therec tangleFC , AG.

Therefore the rectangle A C , EFis equal to th e excess of the rectanglesAE

,FC and AF, FG over the rectangleFC , AG, that is equal to the rectangles

AF, FG andFC , EG .

The demonstration will remain substantially the same for other forms ofthe diagram.

XI . 1 to 12 .

418 . Let PA , PB be two equal straight lines dr awn from a point P to aplane . Let PN be perpendicular to the plane .The angles PNA and PNB are right angles ; therefore the square on PA

is equal to the sum of the squares on PN and NA , and the square on PB 13

equal to the sum of the squares on PN and NB . But PA is equal to PB ;therefore NA is equal to NB . Therefore the angle PAN is equal to the anglePEN , by I. 8 .

419 . Suppose AB , A C to be two straight lines in one plane equally 1hc lined to another plane ; and let the planes intersect in BC. From A dr awAD perpendicular to the second plane . Then the angle ABD is equal to theangle A CD by supposition ; the right angle ADB is equal to the right angleA D C ; and AD is common to the two triangles ADB and A D C . ThereforeAB is equal to AC , by I. 26 ; and therefore the angle ABC is equal to theangle A GB, by I. 5 .

If the point A is in the line of intersection of the two planes this m ethoddoes not apply. Then take AB equal to A C ; from B and C draw BF, CGperpendicular to the line of intersection of the planes , and from F and Gdr aw FD , GE in the second plane perpendicular to the line of intersectionof the planes ; and from B and C draw BD ,

CE perpendicular respectivelyto the straight lines FD and GE . Then BD , CE will be perpendicular to.the second plane by the construction in XI . 11 ; and the angles DFD , andOGE will each measure the inclination of the two given planes, by XI . D ef. 6 .

Join AD,AE . Then AB is equal to AC ; the angle BAD is equal to the

angle CAE , by supposition ; and the right angle BDA is equal to the rightangle CEA therefore BD i s equal to CE , by I . 26.

74 EXERC ISES IN EUC LID .

The angle BFD is equal to the angle C GE,for each measures the in

c lination of one plane to the other ; therefore BF is equal to CG, by - I . 26.

A lso AB is equal to AC ; therefore AF is equal to AG see I . 47 Thereforethe angle BAF is equal to the angle CAG, by I. 8 .

420. Let D CE be the Straight line in the'

plane. Through B drawInthe plane a straight lineFBG at right angles to BC. Then GBFi s at rightangles to both BC and BA

,and 1 s therefore at right angles to the plane ABC ,

by XI . 4. Now D E and GFare both In one plane,and both at right angles

to BC , therefore DE is parallel to GF,by Therefore D E is at right

angles to the plane A CB, by XI . 8 . Therefore D E is at right angles to A C ;that Is AC Is at right angles to D E .

421 . In the plane ABC draw GCFparallel to AB ; produce CD to meetAB at K ; and from K draw KL parallel to D E .

Then DE i s perpendicular to the plane ABC by supposition ; thereforeLK 1s perpendi cular to this plane

,by XI . 8 ; therefore BKL Is a right angle .

And EKO is a right angle, see the Euc lid, p . 313. T herefore BK is perpendicular to the plane containing LX , DE , and KC. Therefore G CFi s perpondicular to this plane, by XI . Therefore C C is at right angles to CE .

422. Let A and B be the two g iven'

points . From A draw A C perpend icular to the plane , and produce AC to D making CD equal to CA . JoinDB intersecting the given plane at P . Then A P and BP are together lessthan any two other straight lines which can be drawn from A and B to meetin the plane.For take any point Q in the plane ; j oin A Q and D Q . Then in the two

triangles ACQ and D C Q the sides A C and DC are equal ; the side CQ iscommon ; and the right angle A CQ is equal to the right angle D CQ : therefore A Q is equal to D Q ,

by I. 4 . In the same manner it can be shewn thatAP is equal to DP. Therefore the sum of AQ andBQ is equal to the sum ofD Q and BQ ; and is therefore greater than DB,

by But DB is thesum of DP and PB ,

and 1s therefore equal to the sum of AP and BP . Thusthesum of AP and BP 1s less than the sum of A Q andBQ .

423. Let OA , OB ,00 be the three equal straight lines , meeting at 0 .

From 0 draw OP perpendicular to the plane ABC —j oin PA , PB,PC.

The angles OPA , OPB, GPG are right angles. Hence by I . 47 it can b eshewn that PA

,PB , PC are all equal ; so that P is the centre of the circle

described round the triangle ABC .

‘

424. Let the three straight lines meet at 0 .

'

Take on these straigh1

lines equal lengths OA,OB , OC . From 0 draw OP perpendicular to the

plane ABC . Then OP is the required straight line .For PA

,PB

,PC are all equal

,by Exercise 423 ; therefore the angle:

POA ,POB

,POC are all equal , by I. 8 .

425 . Since EC ,DFare perpendicular to the same plane they are parallel

by XI. 6 : therefore the points E ,C, D , F are in one plane .

Let CF,produced if necessary

,meet AB at G. D raw the straight lint

GH parallel to EC or DF. Then GH 18 at right angles to the plane CABby XI . 8 ; and therefore the angle AGH 1 8 a right angle. Similarly a stra ighline GK drawn parallel to ED will lie In the p lane EGD ,

and will be at righangles to the plane D AB ,

therefore the angle A GH i s a right angle. Thu

EXERC ISES IN EUCLID:

perpendi cular to the plane A CD , by XI. 1 1 ; and so also is BL. Thus theperpendicular from B on the face A CD is three times the perpendicular fromF on the face . And the perpendicular from B on the face A CD is equal tothe perpendicular from A on the fac e BCD by reason of the symmetry of theregular tetrahedron .

431 . Since the angles at the vertex a re right angles each of the three facesmeeting at the vertex is at right angles to the other two . Let BAC be theequila teral base, and D the vertex. Then since BA is equal to BC

,and BD

is common to the triangles BDA and ED O, we have AD equal to CD,by

I. 47 . Similarly AD is equal to BD .

Now let P be any point of the base. From P draw Pm perpendicular tothe plane AD C , and Pn perpendicular to the plane A DB. Then the planemPn will by XI. 18 be perpendicular to both the planes AD C , ADB,

andtherefore also to AD their line of intersection . A lso the plane mPn isparallel to ED O by XI . 14 ; let it cut AC in c , AB in b , AD in d . Then bei s parallel to BD , and c a is parallel to CD by XI. 16. A lso (ID is equal tothe perpendicular from P on the plane ED O.

The triangle Pmc i s equiangular to the triangle ED O ; therefore Pm isequal to m e , by VI. 4. Thus the sum of Pm and Pn is equal to the sum ofm e and m a ; that is equal to a c ; that is equal to dA , since a c is parallel toDC. Therefore the sum of Pm , Pu, and dB i s equal to AD , and is thereforeconstant .

432. Let OA,

‘ OB,00 be the three straight lines which meet at 0 .

Through O draw any straight line OQ meeting the plane ABC at a point Qwithin the triangle ABC .

Then by the aid of I. 20we can shew that the sum of QA , QB,and QC is

less than the sum of AB , BC , and CA , but greater than half the sum of AB ,

BC, and CA : see Exercises 22 and 441 . N ow in precisely the same manner '

the present Exercise may be established , using XI. 20 instead of I. 20 ; andinstead of a straight line, as QA , the corresponding an

'gle QOA .

433. Let one plane cut the three straight lines at A , B , C respectivelylet another plane, parallel to thi s, cut the straight lines a t K, L , M,

respec

tively let a third plane cut the straight lines at P, Q , R,respectively

,where

P, Q , R are not in a straight line. And suppose AKP , BLQ ,

OMB to be cutin the same ratio by the planes. Then the plane PQR will be parallel to theother two planes.For if not draw a plane through PQ parallel to the planes ABC and

KLM ; and let this cut the straight line CMR at S.

Then CR is to RM as BQ is to QL ,by supposition ; and CS is to SM as

BQ i s to QL , by XI . 17 : therefore CR is to RM as CS is to SM ; which isimpossible . The condition that P, Q , R should not b e on one straight lineis necessary ; for otherwise an indefinite number of planes could pass thr oughP : Q : R '

434. Let AB and CD be the two straight lines. A t the point A draw AKparallel to CD ; and at the point C draw CL parallel to AB. Thus theplanes BAK, LCD are the planes required.

For these planes are parallel , by XI . 15 ; and the former plane passesthrough the straight line AB , and the latter through the straight line CD.

andC,


435 . Let the planes ABCD , KLMN be parallel. Let a plane cut ABCDin OP, and KLMN in RS. Let another plane cut ABCD in OQ , and KLMNin RT . Then the angleFOQ will be equal to the angle SRT.

For RS is parallel to OP, and RT i s parallel to OQ , by XI. 16 ; thereforethe angle POQ is equal to the angle SRT , by XI. 10.

436 . The plane ABC is at right angles to both planes , by XI . 18 ; andtherefore to their common intersection , by XI . 19 . Therefore any straightline in the plane ABC is perpendicular to their common intersection ; andBC is such a straight line ; therefore BC is perpendicular to their commonintersection .

437. Let AB and BC be two consecutive sides of a polygon obtained bycutting a prism by a plane ; let a b and be be the corresponding sides of thepolygon obtained by cutting the prism by a plane parallel to the former.Then A o. and Bb are parallel by the definition of a prism ; and AB and a bare parallel , by XI. 16 ; therefore ABbd is a parallelogram : therefore AB isequal to a b

,by I . 34. Similarly BC is equal to bc ; and so on . A l so the

angle ABC is equal to the angle a bc by XI . 10.

438 . Let AB and BC be two consecutive sides of a polygon obtained bycutting a pyramid by a plane ; let a b and bc be the corresponding sides ofthe polygon obtained by cutting the pyramid by a plane parallel to theformer . Let 0 be the vertex of the pyramid .

Then AB is parallel to a b , and BC is parallel to be, b y XI . 16 ; thereforeAB is to a b as OB is to Ob , by

'VI . 4 . Similarly BC is to bc as OB i s to Ob .

Therefore AB is to a b as BC is to bc . A lso the angle ABC is equal to theangle a bc , by XI . 10. Since these results hold for any corresponding pair ofconsecutive sides the polygons are similar.

439 . The angle ABC is equal to the angle a bc , by XI. 10. A lso AB is toa b as PB is to P b, byVI .

'4 ; that is as p b is to pB,by supposition ; that is

as be is to BC, by VI. 4. Therefore the triangle ABC is equal to the trianglea bc , by VI . 15 .

440. Let OC be the line of intersection , where O is in EF, or-EFpro

duc ed . The plane'

A OE contains AB and is therefore perpendicular to theplane A OC , by XI . 18 ; similarly the plane A OB contains AE and is therefore perpendicular to the plane EOC . Therefore OC

'

I S perpendicular to theplane A OE , by XI . 19 ; therefore 00 i s perpendi cular to the straight lineEFm that plane.

I. 1 to 48 .

441 . The sum of BP and CP is less than the sum of BA and CA , byI. 21 ; similarly the sum of CP and AP is less than the sum of CB and AB;and the sum of AP and BP is less than the sum of A C and BC. Thus twicethe sum of AP,

BP,and CP i s less than twice the sum of AC, BA ,

and CB ;and therefore the sum of AP

,BP , and CP i s less than the sum of A C, BA ,

and CB .

442 . Th e angle APR is equal to the angle BQS, by I . 29 . The ahgleAPR is equal to the angle ARP , and the angle EQS i s equal to the angleESQ , by I . 5 ; therefore the angle ARP is equal to the angle ESQ . Therefore AR is parallel to BS, by I. 28 .

EXERCI SES“

IN EUCLID .

’

£ 443. Le't ABCD be a parallelogram,and P ab y poInt within it.

’

The

triangles PAB and PD C will be together half the parallelogram .

Through P draw a straight lineD

parallel to AB and DC, meeting AD atK,

and BC at '

M . Th e triangle PD C 1s half the parallelogram KD CM,by I. 41 ,

and also the triangle PAB 1s half the parallelogram KAEM. Therefore thetriangles PD C and PAB together are equal to half the

_sum of the parallelograms KD CM and KABM ; that Is to half the parallelogram ABCD .

444. Let ABCD be a quadrilateral,such that .the diagonal A C bisects it

Then the triangle ADC is equal to the triangle ABC therefore BD is bisectedby AC

,by Exercise 1 14 .

445 . Let ABCD be the quadrila teral figure . The triangle ABC is equalto the triangle ABD ; for each is half the quadr ilateral by supposition

”

.

Therefore DC is parallel to AB,by I . 39 .

Similarly BC is parallel to AD.

446 . Produce CA to M,and BA to N, m aking CM equal to EN .

In the triangles MCB and NBC the side CB is common ; the side MC isequal to the side NB , by con struction ; and the angle MCB i s equal to theangle NBC

,by I . 5 . Therefore the side MB is equal to the side N C , and the

angle BMC is equal to the angle ONE .

In th e triangles BMA and CNA we have the side MB equal to the sideNC

,and the angle EMA equal to the angle CN A , as just shewn ; also the

side MA equal to the side NA ; therefore the angle MAB i s equal to the angleN A C , by

447. From the given straight line cut off BC equal to th e g iven length .

Join A C ; draw from A a straight line AP meeting CB at P,and making the

angle CA P equal to the angle A CP . Suppose that P falls between B and C.

Then AP is equal to CP ,by I . 6 ; and

is equal to CB .

If P does not fall between B and C the problem is impossible . But ththe diference between AP and PB may be made equal to the given length .

448 .

'

In the fir st case of I . 26 suppose the triangle D EFappli edthe triangle ABC

,so that EFmay be on BC

,and the triangle

the same side of BC as the triangle ABC is . Then sinc

i s equal to the angle A CB, and the angle D EFis equal tthe triangle D EFwill coincide with the triangle ABC

’

,

equal to it .In the second case of I . 26 suppose the triangle D

triangle ABC , so that D E may bside of AB as the triangle A BC .

coincides with the angle DFE and Is equal to it ;to the angle A OB, by supposition , therefore theangle A CB. But

2thi s is impossible

,by I .

otherwise than on A C . Therefor e the triangletriangle ABC, and is equal to it ,

449 . Let ABC be a . triangle having the sidesstra ight line EDFmeet A Bat E

,meet BC at _

D ,

EXERCISES IN EUCLID“; 79

F; and let FF be bisected at D : then AE : and AF together will be equal toAB and AC together.Fro 'm F drawFG parallel to AB , meeting BC produced at G . In the

triangles EDB and FD C the sides ED and0

FD are equal,by supposition ;

the angles EDB and ED G are equal , by I. 15 ; and the angles EBD andEGD ar e equal, by I. 29 : therefor e EC is equal to EB,

by°Also the angle 'FCG i s equa l

'

to the angle A CB, by I . 15 ; and is thereforeequal to the angle EBD , by Thus the angle FGC is equal to th e

angle F ;C G and therefore EC is equal to FC , by I . 6 . Thus EC is equalto EB ; therefore the sum ofAB and AC I S equal to the sum of AE and AF.

450. Suppose the Straight line DME to meet AB at D,and to meet A C

produced at E ; and let AD be equal to AE then will BD be equal to CE .

Through C draw CF parallel to AB, meeting D E at F. In the trianglesD MB and MFC the sides BM. and CM are equal , by supposition the angles

.

DMB and FMC are equal , by and the angles DEM and MCF areequal, by I. 29 : therefore DB is equal to FC ,

by. Th e angle CFE is equal to the

5 angleA D E

,by I . 29 ; the angle A DE

i s equal to the angle A ED , by I . therefore the angle CEE is0

equal tothe angle CEF.

'T herefore CF is equal to CE , by I . 6. Thus CE is equal

to'

BD .

451 . Let AB be one of the di agonals, and the triangle ABC half of one ,

of the parallelograms . Let 0 be the middle point of AB and join OC .

Then 00 i s half of the other diagonal, by Exercise 78 . Thus AB and OCare given ; and it is manifest that the triangle ABC has its greatest possiblevalue when 0 0 is at right angles to AB ; and then AC is equal to BC,by I . 4 . Similarly any oth er

e

two adjacent sides must be equal , so thatthe parallelogram will be a rhombus.

452. If AD be not equal to BD and CD it must be either greater orless than them . If possible suppose AD greater . Then the angles ABDand A CD are together greater than BAD and CAD together

,by I . 18 ; th e1 e

fore the three angles of the triangle ABC are greater0

than two right angles ;but this I s impossible , by I. 32 . Therefore AD cannot be greater thanBD .

Similarly it may be shewn that AD cannot be less than BD .

453. Let AB and CD be two equal straight lines which intersect at rightangles . The quadrilateral A CBD will be equal to half the square on AB or.

Through C and D draw straight lines parallel to AB,and thr ough A and“

B draw straight lines p arallel to CD thus a parallelogram is formed ; ' and i th as all its sides equal , and all its angles right angles , so that it is a squareAl so each triangle ABC

,ABD is half the corresponding rectangle of which

AB is a side. Therefore the sum of the two triangles is half the sum of thetwo rectangles , that is half the square on AB .

454. Let ABC be the given triangle, D the given point within it. D rawAD and produce it to meet BC at E . On DA take DFequal to D E ; andthrough F draw a straight line parallel to BC , meeting AB at G,

and AC atH. Produce GD to meet BC at K, and produce HD to meet BC at L : then,will GHKL be a parallelogram .

In the triangles C DFand KDE the sides DFand D E are equal, by construction ; the angles GDE and KDE are equal, ° by I . 15 ; and the angles


EGD and EKD are equal , by I. 29. Therefore CF is equal to KE , by I . 26.

In like manner by comparing the triangles HDF and LDE we find thatHE is equal to LE . Thus GH is equal to LK ; and therefore GL is equaland parallel to HK, by I. 33 .

Instead of j oining the point A with D we might j oin B or C with D .

Thus three solutions occ ur ; but we must have AD greater than D E in orderthat the first solution may be possible ; and similar conditions hold withrespect to the other two solutions.

455 . Let AB be one of the given sides . On AB describe a triangle ABDhaving its area equal to the given area . Through D dr aw a straight line D Eparallel to AB . With centre A and radius equal to the other given sidedescribe a circle cutting DE at C. Then A CB will be the triangle required .

For the sides AB and AC have the prescribed lengths ; and the area ofABC is equal to the area of ABD

,by I. 37, and therefore has the prescribed

value.

456. Let AB be the given base . A t the point B make the angle ABDequal to half the difference of the angles at the base. With centre A andradius equal to the given difference of the sides describe a circle meeting BDat D and E , and let E be the nearer of the two points to B . Produce AEto any point F ; at the point B make the angle EEG equal to the angleFEB ; and let BG meet EFat C. Then ABC will be the triangle required .

For EC is equal to BC, by I . 5 ; therefore the difference of AC and BC isequal to AE . The angle ABC is the sum of the angles EBC and ABE ; theangle CAB is the difference of the angles CEB and ABE, by I . 32 . Thusthe angle ABC exceeds the angle CAB by twice the angle ABE , that is by theprescribed excess. And AB is the given base.

457. On AB take AF equal to the given straight line. Bisect the angleBAC by the straight line AE . From F draw a straight line at right anglesto AB

,meeting AE at G. From G draw a straight line meeting AB at P,

m aking the angle AGP equal to the angle GAP. From P draw PQ perpendic ula r to AC.

The angle AGP is equal to the angle GAP by construction ; therefore APis equal to GP, by I. 5 . The angle GPFis equal to the sum of the anglesA GP and GAP, by I. 32 ; that

’

is to twice the angle GAP ; that is to theangle PAQ . The angles PFG and A QP are equa l being right angles . Therefore PF is equal to A Q , by I . 26 . Therefore the sum of A Q and AP, is equalto AF ; that is to the prescribed sum .

458 . Let BC be the base of a triangle , D the middle point of the baseand let the angle BAC of the triangle be a right angle ; then AD is equal toBD : see Exercise 59 or Exercise 452 .

Next let BEC be the triangle,and suppose BEC an acute angle . From C

draw CA perpendicular to BE ; then CA falls within the triangle CBE .

The angle D AE is greater than a right angle , and the angle D EA is less thana right angle ; therefore D E i s greater than DA ,

by I . 19. ButDA is equalto BD , by the fi rst case therefore D E is greater than BD .

Finally let BFC be the triangle,and suppose RFC an obtuse angle.

From C draw CA perpendicular to BF produced ; then CA falls without thetriangle OEF. The angle AFD is greater than the angle FBD , by I . 32 ; andtherefore grea ter than the angle BA D , by I. 5 ; therefore DFis less than DB.


459 . Let ABCD be a square . Take on AB a point E , on BC the pointF, on CD the point G , and on DA the point H,

such that AE,BF, CG, and

DH are all equal : then will EFGH be a square .For we have EB, FC ,

GD , and HA all equal . Thus the triangles HAE ,

EBE, FOG , and GDH are all equal, by I . 4 : therefore the figure EFGH isequilateral. It is also rectangular . For the angle HEA is equal to theangle EFE ; therefore the angles HEA and FEB are together equal to theangles EFB and FEB togeth er ; that is to a right angle, by I. 32 . Therefore the angle HEFis a right angle by I. Similarly the other angles ofthe figure EFCH are right angles .

460. Let AB be the given base,P the point thr ough which a side is

to pass. Join AP and produce it to E ,and

'

cut off a part AD equal to theg iven difi erenc e of the sides . Join BD . A t the point B make the angleD BF equal to the angle BDE let D E intersect BF at C. Then ABC isthe required triangle.For the angle BD C is equal to the angle DBC , by construction ; therefore

BC is equal to DC, by I . 6 . But AC exceeds CD by AD ; therefore A Cexceeds BC by AD . Thus AC exceeds BC by the prescribed length ; andAC passes through the given point P.

461 . From AB cut off AE equal to A C , and join ED . Thus thetriangles AED and A CD are equal in all respects by I. 4 . The angle BEDis greater than the angle AD E ; by I. 16 ; therefore the angle BED is greaterthan the angle A D C . The angle ADC is greater than the angle ABD

,by

I . 16 ; therefore the angle BED is greater than the angle EBD ; thereforeBD is greater than DE

,by I. 19 . But D E i s equal to DC therefore BD i s

greater than DC.

462 . Let ABC be a triangle,having the angle BA C triple the angle

ABC .

Make the angle BAE equal to the angle ABC , and let AE m eet BCat D . Then the angle DAC is double the angle ABC . And the angle ADCis equal to the angles ABD and BA D ,

by I . 32 ; therefore the angle ADCis equal to twice the angle ABD ; therefore the angle ADC is equal to theangle DAC. Thus BAD and CAD are isosceles tr iangles.

463. Let ABC be a triangle,having the angle BA C equal to double the

angle ABC .

From C as centre, with radius equal to BC describe a circle meeting BA,

produced at D . Then BCD is an isosceles triangle ; and therefore the angleADC is equal to the angle ABC . The angle BAC is equal to ' the two anglesADC and A CD

,by I . 32 ; and it is also equal to twice the angle ABC , that

is to twice the angle AD C . Therefore the angle ADC is equal to the angleA GB ; therefore AD is equal to AC, by I. 6. Thus BCD and CAD areisosceles triangles .

464. Let ABC be the triangle having AB equal to A C . Let D be themiddle point of AB . Produce AB to E so that EB is equal to BA . ThenCE will b e equal to twice CD.

T . EX . EUC


Produce CD toFmaking DE equal to CD ; and j oin BF. Then the twotriangles ADC and BDF are equal in all respects by I. 4 ; so that FB isequal to CA

,and the angle DBFequal to the angle DAC.

In the two triangles OEFand OBE the side CB is common the side BFis equal to AC

,that is to AB

,that is to BE the angle FBC is equal to the

angles FBA and ABC , that is to the angles DAC and A CE, that i s to theangle EBC , by I. 32 ; thereforeF0 i s equal to EC , by I . 4 . ButF0 is equalto twice D C , by construction ; therefore EC is equal to twice DC.

465 . Suppose D and E'

of‘

the'

preceding Exercise to denote the fixedpoints ; then every point C is a point on the required locus, so that the locusis a cir cle having its centre at A and its radius equal to AB .

466. Let H be opposite C in the parallelogram D OFE , and let K beopposite C in the parallelogram GCEK. D raw AH

,HB, BK, and KA .

Then HD i s equal and parallel'

to BG , and DA is equal and parallel toGK ; thus the angle HD A i s equal to the angle BGK,

and HA is equal andparallel to BK. Similarly HB is equal and parallel to AK. ThereforeAHBK i s a parallelogram ; and its diagonals bisect each other by Exercise78 . But C is the middle point of AB

,and therefore also of HK.

467. Through F draw a straight line parallel to BC , meeting AB at G ,and j oin EG . From BA cut offBH equal to DF, and join EH . Then thetriangle BHE is half the rectangle BE

,EH and is therefore equal to half

the rectangle BE,DF. We have then to shew th at the triangles AFF and

BEH are together equal to half the rectangle ABCD .

The triangle BEH is equal to the triangle EGA , by I . 38 ; therefore thetriangles AEF and BEH together are equal to the figure EGAF; andtherefore equal to the two triangles ECF, and GFA ; that is to half therectangle EGEG together with half the rectangle GFD A that is to .half therectangle ABCD .

1

468. Take the case in which D is without the triangle ABC , and BDb etween BA and BC ; also suppose F to be on BA produced and E on CAproduced.

Then ED is equal to EC , as they are radii of the same circle ; ands imilarly FD is equal to FB. Then ED and AF together are equal to ECa nd AF together ; that is to EA , AF, and AC together . A gain FD andEA together are equal to FB and EA together ; that is to EA , FA , and ABtogether, that is to EA , AF, and AC together. Therefore ED and AFtogether are equal to FD and EA together .Similarly the other cases which arise from modifications of the diagram

m ay be treated. For instance , if D be within the triangle ABC ,the point

F on AB,and the point E on AC , we shall find that EA and ED together

are equal to FA and FD together.In all cases the sides of the quadrilateral taken in order are AE , ED , D E,

FA ; and if one of the sides BA , CA ,requi res to be produced , the other also

will have to be produced.

469 . From AB cut ofi AE equal to the required length . From E drawEH perpendicular to AC. Bisect the angle HEB by the straight line EMm eeting A C at M ; from M draw MP st right angles to A C, meeting AB atP ; then P will be the required point .

J


'

473 . Let a straight line parallel to’

the given straight line meet CA at Land CB at M. A t C make the angle L CK equal to ML O ,

and let CK cutAB at D . Through D draw a straight line parallel to LM meeting CA atE and CB at F. Then will EFbe bisected at D .

For the angle D E C is equal to the angle ML C ,by I . 29 ; and the angle

D CE i s equal to the angle ML C by construction ; therefore the angle D CEi s equal to the angle D EC . The angles D OE and D OFa re together equalto a right ang le ; the angles D E C and DFG are together equal to a rightangle, by I . 32 ; therefore the angle D CF is equal to the angle DFG.

Therefore D E and D E are each equal to DC,by I . 6 ; therefore D E is

equal to D E, so that FE is bisected at D .

474. Bisect BD at E . Then BE is equal to BC. The angle ABC isequal to the angles BEC and BCE

,by I . 32 . Therefore the angle ABC is

twice the angle BEC. Therefore the angle BEC is half the angle ABC ,and

is th erefore e qua l to twice the angle BAC, by supposition . And the angleEBC is equal to the angles BAC and BCA

,by I . 32. Therefore the angle

EBO is equal to twice the angle BA C . Therefore the angle EEC is equalto the angle BEC ; therefore E C is equal to BC, by I. 6 . Thus EBC isequi lateral .

A gain , EC i s equal to ED ; therefore the angle BEC i s equal to twicethe angle BBC . Therefore the angle BD C i s equal to the angle BA C .

Therefore the angle A CD i s equal to the angle ABC , by I . 32 . Then thetriangle ABC is equiangular to the triangle A C D .

475 . Take the di agram of I . 43. L et BD and HG intersect at L .

Through L draw a straight line parallel to AD , meeting AB at M and DCat N.

The complements AL and LC are equal , by I. 43 ; to each add KN.

Then the parallelograms AK and MFtogether are equal to the parallelogramKC. Therefore the difference of KC and AK is equal to MF.

Now the parallelogram MK is twice the triangle BLK,and the parallelo~

gram LFis twice the triangle D LX , by I. 41 : therefore the parallelogramMFis twice the triangle BKD . Therefore the difference of the parallelograms KC and AK is equal to twice the triangle BKD .

476. Take BC equal to the given side ; draw CD at right angles toBC, and make it equal to th e difference between the hypotenuse and .theother side . A t the point B in the straight line BD mak e the angle D BEequal to the angle BD C ,

and let BE meet DC produced at A : then ABC isthe triangle required .

For the angle ABD is equal to the angle A D B, by construction ; there.fore AD is equal to AB,

by I. 6 ; but AD exceeds A C by CD ; therefore ABexceeds AC by CD . Thus the hypotenuse AB exceeds th e

’

side AC by theprescribed length . AISO the side CB has the prescribed length .

The process requires the angle CBD to be less than the angle CDB; itwill be found that this leads to the c ondition that the hypotenuse must beless than the sum of the sides , which i s of course necessary by I. 20.

477. The triangleEBC is half the triangle ABC ,by I . 38 ; the triangle

BED is half the triangle EBC , and is therefore one - fourth of the triangleABC . Bisect A G at H ; then the triangle ABH is equal to the tr ianglGBH ,

and the triang le AHE is equal to the triangle GHE : therefore th


triangle BHE is half the triangle BAE , and is therefore equal to the triangleBD E . Therefore GH is equal to GD , by Exercise 114. Thus AG is doubleof GD .

478 . Let BC meet AE at F, The angle D A C is equal to the angleD CA ; see Exercise 59 . Suppose F to be between D and C. The angleEFC is equal to the. ang les EDFand DEF, by I . 32 . A lso the angle EFCis equal to the angles DCA and FA C

,by I. 32 ; that is to the angles DAC

and FAC that i s to twice the angle FAC together with the angle D AF;that is to the angles BAC and D AF. Thus the sum of the angles ED E andD EF is equal to the sum of the angles BAC and D AF. But the rightangles EDFand BAC are equal ; therefore the angle DAF is equal to theangle D EF; therefore DA is equal to D E ,

by I. 6.

479 . The triangle A CE is half the rhombus, and the triangle ABC ishalf the square ; therefore these triangles are equal ; therefore FEB is astraight line parallel to AC

,by I . 39 . D raw BH and EX perpendicular to

A C. Then BH is equal to AH by I . 6 ; and therefore equa l to half A C ;and therefore equal to half AE . Therefore EK is half of AE , by I. 34.And the angle AKE is a right angle. Therefore if EK be produced to apoint L such that KL is equal to KE , the triangle ALE is equilateral.Thus AEK is an angle of an equilateral triangle , and therefore it is twothirds of a right angle .Therefore the angle KAE is onethird of a right angle , by I , 32 . A lso the

angle BAC is half a right angle ; therefore the angle BAE is one - sixth ofa right angle. The angles CAF and BAE are equal by Exercise 11 ; therefore each of them is onesixth of a right angle. Thus the angles BAE ,

EAF,

and FAC are all equal .

480. From G draw GM perpendicular to AB , and GN perpendicularto AC. Then the angle NGM is a right angle . The angle EGD is also a

right angle , Therefore the angle EGN is equal to the angle D GM .

In the two triangle s EGN and D GM the side EG is equal to the side D G ;the angles EGN and D GM have been shewn to be equal ; and the right anglesEN G and DMG are equal : therefore GM is equal to GN by I. 26 . ThusAMGN is a square, and the diagonal AG bisects the angle BAC so that thelocus of G is the straight line which bisects the angle BA0.

481 . We Shall first shew that a rectangle is greater than any parallelogram on the same base with the same perimeter.Let ABCD be a parallelogram

,and ABEE a rectangle on the same base

AB,and having the same perimeter ; then AF is equal to AD and there

fore the perpendicular from D on AB is less than AF. Let DC producedcut AF at G and BE at H. Then the parallelogram ABCD is equal toABHG, by I . 35 ; and is therefore less than the rectangle ABEE.

Next we shall shew that a square is greater than any rectangle havingthe same perimeter.Let ABCD be a rectangle , AB being longer than AD . Let AEFG be

gsqgare having the same perimeter

,E being on AB

,and G on AD pro

uceSince the perimeters are equal EB is equal to GD

,and BC is less

than GF; therefore the rectangle E C is less than the rectangle D E ; therefore the rectangle AC is less than the square EG .

8 6 EXERCISES IN EUCLID :

482. L et ABCD be the square in which the square of given area is to beinscribed . Join AC ; a nd bisect it at E . From E as centre with a radiusequal to half the diagonal of the square of given area describe a circle cuttingA B at F and produce FE to meet CD at G . Through E draw a straightline at right angles to FG ,

meeting BC at H, and AD at K. ThenFHGKwill be the square required .

For FE is equal to GE and HE i s equal to KE , by Exercise 36 . Thenfrom the triangles FEH and GEH we shew that EH i s equal to GH andfrom the triangles HEG and KEG we shew that HG is equal to KG. Inthis way we shew that the figure EH GK is equilateral.The angle AEB i s a right angle, and so also is the angleFEH

‘

; thereforethe angle A EFis equal to the angle BEH . A lso the angle EAFis equal tothe angle EBH

,each being half a right angle . And EB is equal “to EA .

Therefore EF is equal to EH ; therefore the angle BEH is equal to theangle EHF; therefore each of them is half a right angle . In this way weshew that the figureFHGK is rectangular.

483 . The triangle BD C is twice the triangle A D C ,and the triangle BDE

i s twice the triangle A D E ; see I . 38 . Therefore the triangle BFC i s twicethe triangle FA C . Similarly the triangle BFC is twice the triangle FAB.

Therefore the triangleBFC i s equal to the sum of the triangles FAC and FAB ;so that the triangle BFC i s half the triangle BA C .

A gain the triangle DAC is equal to the triangle EAB, each being onethird of the triangle ABC . Take away from each the figure A EFD ; thusthe triangle FEC is equal to the triangle FDB. But the triangle FAC wasshewn to be equal to the triangle FAB , therefore the triangle AFF is equalto the triangle A DE. Therefor e the fi gure AD EE IS twice the triangle AD E.

But the triang le BDFis twice the triangle A DE. Therefore the figure ADEFis equal to the triangle BDE.

484 . On AB take AF equal to AE ,and on BF take BG equal to BD .

Then the triangle A GE i s equal to the triangle A GE , and the triangle BOGis equa l to the triangle BOD , by I. 4 .

We shall now show that the triangle FOG is equal in area to the triangle D OE .

The angle E0A is equal to the angles OAB and OBA,by I . 32 ; that is to

half the sum of the angles BAC and ABC,that is to half a right angle.

Therefore the angle A CE i s half a right angle . Similarly the angle BODis half a r ight angle , and also the angle BOG . Hence the angle GOF is halfa right angle

,and the angle D OE i s three halves of a right angle . On 0A

take OH equal to OD , and j oin EH . Then in the triangles EOH and FOGthe side E0 i s equal to the side F0 ; the side H0 is equal to the side G0 ;and the angle H OE i s equal to the angle GOF, each being half a right angle .Therefore the triangles EOH and FOG are equal in all respects. But thetriangle EOH i s equal in area to the triangle E CB, by I. 38 therefore thetriangle FOG is equal in area to the triangle EOD .

Thus the triangle A OF is equal to the triangle A OE ,the triangle BOG

i s equal to the triangle BOD , and the triangle GOF is equal to the triangleD OE ; therefore the triangle A OB is half the quadr ilateral ABD E.

485 . Let ABC be the scalene triangle . If possible let BD be thedividing straight line . Th e angle BCD is not equal to the angle BAD ,

since by supposition the tri angle is scalene ; the angle BCD cannot be


equal to the angle BD A , by I. 32 ; and thus the only possible case is tha tthe angle BCD should be equal to the angle ABD ,

and the angle BADequal to the angle OED . This requires AB to be equal to BC

,and is con

trary to the supposition that the triangle is scalene.

486 . Let AE and CD intersect at G . Then CG is equal to GD ,by

Exercise 78 . Therefore the triangle GGFi s equal to the triangle D GF,and

the triangle CGE is equal to the triangle D GE , by I . 38 . Therefore thetriangle FCE is equal to the triangle DFE . But BC is equal to CE ,

bysupposition ; therefore the triangle BCF is equal to the triangle ECF,

byI . 38 . Therefore the triangle BFE is double the triangle DEE . Ther eforeBF is twice D E.

487 . BE and AD are each equal and parallel to CZ ; therefore BEand AD are equal and parallel ; therefore AB and ED are equal and parallel,by I . 33, and AD EB is a parallelogram .

Produce ZG to meet AB at L . Then the parallelogram D L is equal tothe parallelogram AZ by I . 35 , which is equal to the parallelogram FC .

Six

éi ila rly EL i s equal to KC ; thus ADEB is equal to the sum of FC

an OK.

488 . In the quadrilateral ABCD suppose that AB is parallel to D C .

Let AC and BD intersect at E . Through E draw a straight line parallelto AB meeting AD at M

,and BC at N. Then MN will be bisected at E .

For if ME be not equal to NE one of them must be the greater ; suppose MEgreater than NE . Then the triangle MAE will be greater than the triangleNBE , and the triangle MDE will be greater than the triangle NCE see

I . 38 . Therefore the triangle ABD i s greater than the triangle BEC.

A gain, the triangle ABC is equal to the triangle ABD , by I. 37 ; thereforethe triangle AED i s equal to the triangle BE0 . But the triangle AED wasshewn to be greater than the triangle BEC . Therefore ME and NE cannotbe unequal ; that is , they are equal .

489 . Let ABC and D EFbe two triangles ; let the bases AB and DE beequal, and in the same straight line , and let CE be parallel to thi s straightline. Let a straight line be drawn parallel to CE , meeting CA at G , CBat H,

ED at K, and EFat L . Then GH will be equal to KL .

For if GH be not equal to KL one of them must be the greater ; supposeGH the greater

,and from it cut off GM equal to KL . Join AM, BM, CM.

Then the triangle GOM is equal to the triangle KEL , the triangle GMAis equal to the triangle KL D , and the triangle AMB is equal to the triangleD LF, by I. Therefore the triangle D EF is equal to the sum of thetriangles GCM , GMA , and AMB ; but the triangle A GB is equal to thetriangle D EF; therefore the triangle A CB is equal to the sum of the trianglesG CM , GMA ,

and AMB ; that i s the whole is equal to a part , which is absurd.

Therefore GE i s not unequal to KL , that is GH is equal to KL .

490. Let 0 be the middle point of BC ; then 0A , OB, and 00 are allequal

,by Exercise 59. Now AB is equal to half A C, and is therefore less

than OB,which is half BC ; therefore the angle OBA is greater than the

angle BOA,by I . 18. But the angle BOA is equal to the angles OCA and

OA C ,by I . 32 ; therefore the angle BOA is equal to twice the angle OCA ,

by I. 5 . Therefore the angle OBA is greater than twice the angle OCA .


491 . L et ABCD be the parallelogram. Trisect DC at E and F ; andtrisect BC at H and G : see Exercise 70. Then the triangles ADC and ABCare equal

,by I. 34. Thus each of the - triangles D AE

,EAF, FA C , CA G ,

GAH,HAB is one - sixth of the parallelogram ABCD . Thus the triangle

D AF,the quadrilateral FA G C , and the triangle GAB are all equal , each

being one- thi rd of the parallelogram ABCD .

492 . The angles DAB and ABH are together equal to two right angles,by I . 29 ; therefore the angles DAB and ABB are together equal to tworight angles , by I. 5 . But the angles AHC and AHB are together equal totwo right angles

,by I . 13 ; therefore the angle DAB is equal to the angle

AH C . Similarly the angle DAB is equal to th e angle AKO. A lso theangle DAB is equal to the angle HCK

,by I . 34 . Therefore three times the

angle BAD is equal to the sum of the angles AH C,AKC

,H CK ; that is to

the sum of the angles of the triangle HKO together with the angles AHK'

and AKH ; that is to two right angles together with two angles of an equilateral triangle ; that is to ten - thirds of a right angle . Therefore the angleDAB is ten - ninths of a right angle.

493 . Let ABC be the given triangle ; D a given point in the side A C.

D raw a straight line D E m eeting AC at D and AB at E , so that the triangleD A E may be one - third of the given triangle : this can be done by aid ofI. 44, DA being the given straight line , A the given angle, and D E thediagonal of the parallelogram so described, which will be two - thirds of thetriangle ABC . Then bisect the quadrilateral D ED G by a straight line drawnfrom D

,by Exercise 123. Thus the triangle ABC is divided into three

equal parts by straight lines drawn from D .

If the triangle DAB be less than a third of the triangle ABC , then DCand the angle C must be used in the first part of the construction instead ofDA and the angle A .

494. Let A C and BF intersect at H,let BE and CD intersect at G.

Let GH,produced if necessary

,meet BC atK. Then will BK be equal to CK.

Thr ough H draw a straight line parallel to BC, meeting BE at M andCD at N . The triangles A CD and EBF are equal , by I . 38 . ThereforeHN i s equal to HM, by Exercise 489 . Therefore the triangle CHN i sequal to the triangle C HM

,by I . 38 . Therefore BK is equal to CK

,by

Exercise 489 .

II. 1 to 14.

495 . Let ABC be the given triangle , BC the side to be produced.

Suppose that CA is not les s thanBA . From A draw AD perpendicular toBC or BC produced . Produce CD to E so that D E may be equal to DC.

Then the difference of the squares on CA and BA is equal to the differenceof the squares on CD and BD , by I. 47 ; that is to the rectangle of thesum and difference of CD and BD , by pa ge 269 of the Euc lid. N ow if theperpendicular fall s within the tri angle ABC the sum of CD and BD is BC

,

and the difference is BE and if the perpendicul ar falls wi thout the triangleABC the sum of CD and BD is BE

,and the difference is BC . Thus in

each case the difi erenc e of the squares on CD and BD i s equal to therectangle BC , BE ; so that BE is the produced part required.


496 . Let AB be the g iven straight line ; let BC represent the producedpart. Then we require that the squares on AB and BC shall be equal totwice the rectangle A C

,BC ; so that the square on BC together with twice

the square on AB must be!equal to the square on AB together with twice

the rectangle AC, BC ; that is the square on BC together with twice thesquare on AB must be equal to the squares on AC and BC , by II . 7 ;therefore the square on A C must be equal to twice the square on AB . Thisdetermines A C

,and shews that A C must be equal to the diagonal of a

square described on the side AB .

497. Let AB be the given straight line ; let BC represent the producedpart . Then we require that the squares on AB and AC shall be equal totwi ce the rectangle A C

,BC ; so that the square on AC together with twice

the square on AB must be equal to the square on AB together with twicethe rectangle A C

,BC ; that is the square on AC together with twice the

square on AB m ust be equal to the squares on A C and BC,by II. 7 ; there

fore the square on BC must be equal to twice the square on AB . Thisdetermines BC

,and shews that BC must be equal to the di agonal of a

square described on the side AB .

498 . In the diagram and proof of I I . 11 it is shewn that the rectangleCF, FA is equal to the square on CA . If then CA be the given straightline descri be a square on CA , and p roceed as in II. 11 ; then F will be thepoint required.

499. Take any straight line AD ; produce AD to B so that DBmay b eequal

‘

to twice AD . From A draw a straight line AE at right angles toAB. From the centre D with radius equal to DB describe a circle cuttingAE at C . Then DBC will b e

'

such a triangle as is required.

For the square on BC is equal to the squares on CD , DB and twice therectangle AD , DB . But twice the rectangle AD , DB i s equal to the square onDB , because DB is equal to twice AD . A lso CD is equal to DB . Thereforethe square on BC is equal to three times the square on DB .

500. Let ABC be a triangle , and AD a perpendi cular to the base BCproduced ; and suppose that the square on AB exceeds the squares on BC,CA by the rectangle BC

,CA .

The square on AB exceeds the squares on BC ,CA by twice the rectangle

BC , CD ; therefore th e rectangle BC, CA is equal to twice the rectangleBC, CD ; therefore CA is equal to twice CD . Produce CD to E

,so that

D E i s equal to CD ; and join AE . Then AC is equal to CE . Al so A E isequal to A C, by I. 4. Therefore A CE i s an equi lateral triangle . Thereforethe angle A CE is one - third of two right angles

,by I . 32 . Therefore the

angle A OB is two - thirds of two right angles,by I . 13.

501 . Let the straight line AB be the sum of two adjacent sides of therectangle ; let CD represent the d if erenc e of the two sides. Then AB isknown, and we proceed to find CD.

The difference of the squares on AB and CD is equal to the rectanglecontained by the sum and the difference of AB and CD ; that is to fourtimes the rectangle whi ch i s required ; that is to four times the given square ;that is to a known quantity. But the square on AB is known ; thus thesquare on CD is known ; and therefore CD i s known . Hence the sides

90 EXERC ISES IN EUCLID ;

of the required rectangle are found ; for one side is half the sum of AB'and

CD, and the oth er is half the difference of AB and CD .

502 . U se the same notation as. in Exercise 501 . Then the differenceof the squares on AB and CD is equal to four times the given square . ButCD is known ; hence the square on AB is known, and therefore AB isknown . Hence the sides of the required rectangle are found.

503. Let ABCD be the given square ; let H, E ,F, G be the middle

points of the successive sides , so that HEEG is a square . Let PQRS be anyother inscribed square . In each square the diagonals intersect at the samepoint , say T : see Exercise 36 .

Then the square on P Q is equal to the squares on TP and TO, and thesquare on HE is equal to the squares on TH and TE , by I . 47 . But TP isgreater than TH , and TQ i s greater than TE , by I. 19 therefore the squareon PQ is greater than the square on HE .

504. Let AB' be the given straight line ; it is required to divide it at C,so that the squares on AB and BC may be equal to twice the square on AC.

We require then that three times the square on AB,together with the

square on BC , may be equal to twice the square on AB together with twicethe square on AC. Produce BA to D

,so that AD may be equal to A C.

Then DB is the sum of AB and A C,and GB i s the difference ; so that the

square on D B together with the square on CB is equal to twice the squareon AB together with twice the square on AC

,by page 269 of the Euc lid.

Hence we must have the square on DB equal to three times the square onAB . Thus DB is known ; then we take AC equal to AD ; and the straightline AB is divided at C in the mann er required.

505 . Let ABD O and A GFE .b e rectangles of equal areas and per im a ers.

Place them so a s to have a common angle at A , and let AE fall on AC, andAB on AG. Let BD and EFintersect at H.

Then since the perimeters are equal we have twice EC equal to twiceHF,

and therefore EC is equal to HF. Al so since the areas are equal th erectangle BGFH i s equal to the rectangle E CD H . But E C is equal to HF.

Therefore CD is equal to FG . Therefore by using either the condition thatthe areas are equal or that the perimeters are equal we find that AC is equalto A G. Thus the two rectangles are equal in all respects .

506. Let 0 be the centre of the rectangle. Since the sum of PA and

PC is equal to the sum of PB and PD , the squares on PA and PC togetherwith twice the rectangle PA

,PC are equal to the squares on PB and PI

together with twice the rectangle PB,PD

,by II. 4 . But the squares or

PA and PC are equal to twice the square on PO and twice the square orAO ; and the squares on PB and PD are equal to twice the square on PCand twice the square on B0, see the Euc lid , p . 293 : so that the squares orPA and PC are equal to the squares on PB and PD . Hence the rec tangLPA

,PC is equal to the rectangle PB , PD .

Thus these two rectangles have equal areas and equal perimeters ; amtherefore

,by Exercise 505 , must be equal in all respects . Hence PA mus

be equal either to PB or PD . If we take PA equal to PB the point P fallon th e straight line which i s parallel to BC and passes through the centrof the rectangle. If we take PA equal to PD the point P falls on th


Therefore a circle will go round FDCE , by page 276 of the Euc lid ; therefore the angle ED E is equal to the angle FOE , by III. 21 .

The proof is very similar for other forms of the diagram.

5 13. Let ABCD be a quadrilateral . Bisect the angles A , B , C , .D .by

straight lines which meet and form a quadrilateral : then a circle can bedescribed round this quadrilateral.Let AP and BP be two of the straight lines, and C Q and D Q the other

two straight lines . Then the angle APB , half the angle A , and half theangle B are together equal to two right angles

,by I . 32 . So also the angle

C QD , half the angle C, and half the angle D are together equal to two rightangles. Therefore the angles APB and D QC together with half the sum

of the angles A,B

,C,D are together equal to four right angles . But half

the sum of the angles A , B,C, D is equal to two right angles . Hence

the angles APB and D QC are together equal to two right angles . Hence acircle can be described round the quadrilateral of which P and Q are opposite corners , by page 276 of the Euc lid .

5 14. Let A be the centre of one circle,B the centre of the other ; join

AB meeting the first circle at G and the second at H. Then GH is theshortest distance between the circles .For draw the straight line CD from any point C on the first circle to

any point D on the second . Join AC,AD

,and DB . Then AC and CD are

together greater than AD , by I . 20 ; therefore AC , CD,and DB are together

greater than AD and DB together . But AD and DB together are greaterthan AB ; therefore AC , CD and DB together are greater than AB. ButAC is equal to AG , and BD is equal to BH ; therefore CD is greater thanGH .

5 15 . Suppose the straight line drawn ; let BA C denote it . Join B and Cwith D the other extremity of the common chord . Then the angles ABDand A CD , being the angles in known segments of circles are known ; also theside BC is known . Hence the triangle BCD can be constructed . Then theangle BAD becomes known , being the angle subtended by a known chord BDin a known circle. This determines the situation of the straight line BAC.

There will be in general two solutions,in one of which the segment BAD

will be greater than a semicircle , and in the other, less .

5 16 . Suppose the polygon to be a quadr ilateral . Then th e sum of thealternate angles is equal to two right angles

,by I I I . 22 ; and the sum of the

alternate angles,together with two right angles , is equal to four right angles.

Suppose the polygon to be a hexagon , as ABCD EF. D raw AD. Thesum of the angles ABC and ADC is two right angles , by III . 22 ; so also isthe sum of the angles AD E and AFE . Therefore the sum of the anglesABC , CDE ,

EFA is equal to four right angles ; and the sum of these angles ,together with two right angles , is equal to six right angles .

A gain , suppose the polyg on to be an octagon , as ABCDEFGH . D rawthe straight lines AD and AF. Then the angles ABC and ADC are equa lto two right angles , so are the angles ADE and AFE , so are the angles AFGand AHG. Then as before we obtain the required result.In this way the proposition may be established for a polygon of any even

number of sides.


517 . Let C be the centre of the g iven circle , A the given point on thec ircumference, HK the given chord . On AC as diameter describe a semicircle , and let it cut HK at B. Join AB and produce it to meet the givencircle at D .

The angle ABC is a right angle , by III . 31 ; therefore AD is bisectedat B , by 111 . 3.

There will be two solutions in general , as the semicircle may cut HK attwo points ; but no solution if the semicircle does not meet HK.

5 18 . Let A,B

,C, D , be the successive angular points of the

polygon ; O the c entre of the circle .The angles at A

,B

,C are bisected respectively by the straight

lines OA , OB , C C , OD,see the note on III . 17 on page 275 of the

Euc lid . Then in the triangles OBA,OBC the side OB is common , the angle

OBA equal to the angle GEG, and the side BA equal to the side BC, bysupposition : therefore the angle BAO is equal to the angle BC O by I . 4 .

Therefore the doubles of these are equal ; so that the whole angle at A isequal to the whole angle at C.

In this way we shew that the alternate angles are equal ; and so if thenumber of them is odd they are all equal . If the number of the angles iseven they are not necessarily equal ; for instance, a circle m i ght be inscribedin any rhombus .

5 19 . If AE and BD intersect within the circle the angle APB is measured by half the sum of the arc D E and a semicircumference, by page 294of the Euc lid, and is therefore constant . Sim ilarly if AE and BD intersectwi thout the circle the angle APB is measured by half the diflerence of thearc D E and a semicircumference and is therefore constant.

520. Let ABC be one of the triangles . The four angles of the quadrilateral AD are together equal to four right angles ; hence the angle BBC isequal to the excess of two right angles above the angle BAC and is therefore a constant angle. Hence the locus of D is a segment of a circle bypage 276 of the Euc lid . The straight lines whi ch bisect the angle BBC allmeet at a point, by Exercise 230.

521 . The angle OBA is equal to the angle OAB , by I. 5 ; that is theangle OBC is a constant angle, by III . 21 . Hence the locus of B is a circle,by page 276 of theEuc lid . This circle is made up of two segments each equalto the corresponding segment of the given circle cut offby 00 : see III. 24 .

522 . Suppose PE drawn perpendi cular to AB , and produce EP to meetDC then these straight lines will meet at right angles by I. 29 . Thus PEand PG form one straight line.

Since the angles PFB and PEB are r ight angles a circle will go roundPFBE ,

by page 276 of the Euc lid ; therefore the angle PEE is equal to theangle PBF. Similarly the angleFGH is equal to the angle PD H . But BC

is parallel to AD ; and therefore the angle PBFis equal to the angle PDH ,

b I . 29 .yTherefore the angle PEFis equal to the angle PGH . Therefore EFis

parallel to GH , by I. 27

523.Let AB be a fixed chord of a circle, C a fixed point in it . Let

PC Q be any other chord of the circle . Let D be the middle point of AB and.R the middle point of PQ . Let 0 be the centre of the circle.

94 EXERCISES I N EUCLID .

The angles OD‘

C and ORC are right angles, by I II . 3 ; therefore a circlewill go round ORCD , by page 276 of the Euc lid ; therefore the angle GRDis equal to the angle COD , by III. 21. Thus the angle GRD is a constantangle.

524. The angle BFA is equal to the angle BEC , because the segmentsare simila r. For the same reason the angle ADB is equal to the angle CDB;therefore the angle BDE is equal to the angle BDF. by I. 13 . Therefore theangle BCE is equal to the angle BAF, by III . 22 . Thus the angle BEAi s equal to the angle BEC

,and the angle BAF is equal to the angle BCE ;

therefore the angle ABF i s equal to the angle CBE ,by I. 32 . Thus the .

triangle BEC is equiangular to the triangle BEA .

A l so the angle BEA is equal to the angle BD A , by III. 21 ; th at is to theangle BCE

,by I . 1 3 and III . 22 ; that i s to the angle BAF, as shewn above .

Thus the triangle BAF is isosceles by I. 6 . Therefore also the triangle BCEis isosceles.

525 . Let A and B be the centres of the circles, A C , BD perpendiculars

on the common tangent . From 0 the middle point of AB draw OQ perpendi cular to CD. Then CO i s equal to half the sum of CA and BD

,by

Exercise 89 ; that is OQ is equal to half AB , and is therefore constant.Hence CD touches at Q the circle described from 0 as centre with radiusequal to half AB.

526 . Suppose A the given point,and BC the given straight line . Sup

pose that P and Q are two points in BC, such that PQ is of the given lengthand the angle PAQ equal to the given angle . Suppose that a circle goesround PAQ , and that O is the centre of the circle. The radius of this circleis known , for it i s the circle in which a chord of given length subtends agiven angle at the circumference . Since the radius and the length of PQare known the distance of 0 from the fixed straight line becomes known.

D raw a straight line parallel to BC, and at a distance from it equal to thatjust determ i ned . With A as a centre , and radius equal to that already

'

determined describe a circle. The intersection of the circle and this straight“

line will determine the position of O ; and then the positions of AP and A Obecome known .

527 . We shall first shew that a certain straight line can be found thetangents drawn from any point of which to the two circles are equal : thenthe intersection of this straight line with the given straight line determinesthe required point .Let A and B be the centres of the given circles . D raw any circle inter

sec ting the two given circles ; let EFbe the common chord of the first circleand the third circle ; let CD be the common chord of the second circle andthe third circle. LetFE and D C produced meet at G ; from G draw GKperpendicular to AB . Then the tangents drawn to the two given circlesfrom any point in GK will be equal . For take any point P in that straightline . A lso from G draw GS to touch the first given circle , and GT to touchthe second. The rectangle GE

,GF is equal to the rectangle GC , GD,

byI II . 36 C orollary. Therefore the square on GS i s equal to the square on GT

,

by III . 36 ; that is the excess of the square on GA over the square on SAis equal to the excess of the square on GB over the square on TB

,by I . 47.

Therefore the squares on GK and KA dimin ished by the square on SA are


from A as centre with a known radius . When D is known, by joining AD

and producing it to meet the given circle C can be found.

533. Let AB denote the base of the triangle , and CD the perpendicularfrom the vertex on the base . Suppose DA greater than DB , and from DAcut off D E equal to DB

,and j oin EC . Then we have given AE which is

the difference of DA and DB ; also the sum of CA and CE , for this isequal to the sum of CA and CB ; and the angle A CE ,

for this is the difference of the angles CED and CAB,

by I . 32,that is the difference of the

angles CBD and CAB. Hence the triangle A CE can be constructed byExercise 268 ; and then the triangle A CB.

534 . Suppose the segment APB to fall within the segment AOB. LetA T with in

'

the segment AOB be a tangent at A to APB,and produce TA

to R . Let AS be a tangent at A to A QB.

Then the angle PA T is equal to the angle PBA , and the angle BAS isequal to the angle BQA by 111 . 32 . The angle PAR will be equal to the sumof the angles PAB and APB , by I. 1 3, III. 32 and I. 32 . That is the anglesPAB , BAS, SAR are together equal to the angles PAB , and APB . Therefore-the angles BA S and SA R together are equal to the angle APB , that is toPAQ and AOB . But the angle BAS was shewn to be equal to the angleA OB therefore the angle SAR is equal to the angle PAQ . That is theangle PA Q is equal to. the angle between the tangents.

535 . Let C be the centre of the given circle. Let T be th e middle pointof KL , M the middle point of PQ , and N the middle point of BS. Theangles CN A

, GTA , OMA are all right angles by III. 3 ; therefore N, T ,and

M are on a circle described on AC as diameter,by page 276 of the Euc lid .

The angle between MN and AL is measured by half the sum of the arcs ANand MT

,by page 294 of the Euc li d ; that is by half the sum of the arcs

AN and N T , by III. 26 ; that is by half the arc A T , which i s a fixed arc .

Thus the angle between MN and AL is constant, so that MN always remainsparallel to itself.

536. Let EFGH be a quadrilateral , such that round it the quadrilateralABCD can be described , so that the angle BEFi s equal to the angle BEE,

the angle CFG to the angle GGF,the angle D GH to the angle DHG, and the

angle AHE to the angle AEH . Then a cir cle may be described about thequadrilateral EFGH .

The angles HEF,AEH , BEF are together equal to two right angles ,

by I . 13 ; and so are the angles HGF, D GB, GGF. Therefore the anglesHEF, AEH , BEF,

HGF,BGH GGFare together equal to four right angles.N ow in th e four triangles AEH , BFE,

FCG , and GDH the sum of all theangles is eight right angles also the sum of the four angles at A , B , C, Di s four right angles ; therefore the sum of the remaining angles is four rightangles . Hence we find that the sum of the four angles AEH ,

BEF,D GH ,

C GFi s two right angles. Therefore the angles HEE and HGFare togetherequal to two right angles ; and therefore a circle would go round HEFG, bypage 276 of the Euc lid .

537. From A draw the straight line ABC passing through the centres ofthe two circles , meeting the inner circle at B , and the outer c ir c le at C .

Suppose that the straight line AFD is such as is required , meeting the inner


circle at E,and the outer circle at D , and making ED of the given length.

Join CD , EB,and from B draw BF perpendicular to CD .

The angles AEB and ADC are right angles , by II I . 31 ; and the angleBFD is a right angle , by construction . Therefore EBED is a rectangle, andBF is equal to ED .

Thus to solve the problem we describe a semicircle on BC and in it placethe straight line BF equal to the given straight line , and then draw ABDparallel to BF.

538 . CD is parallel to AB,and therefore CD produced will cut AE at

right angles , by I . 29 : similarly AD produced will cut CE at right angles .Hence the perpendicular from E on AC will pass through D

,by page 313

of the Euc lid ; that is if ED be joined and produced it will cut A C at rightangles.

539 . Let ABC be a triangle ; let A D , BE , CF be the perpendiculars fromthe angular: points on the Opposite sides. By the Euc lid, page 313, theseperpendiculars meet at a point ; denote the point by 0 .

Since the angles AFC and ADC are right angles a circle wi ll go roundAFD C , by the Euc lid , page 276 . Then the rectangle A O

, OD is equal to therectangle CO , OF, by III. 35 . Similarly the rectangle A O

,OD is equal to

the rectangle BO , OE .

540. Let ABC be a triangle ; let BF bisect the angle ABC , and meetA C at F ; let CG bisect the angle A CB,

and meet AB at G. Let BF and CGintersect at 0 . From A draw AD perpendicular to BF

,and AE perpendicular

to CG. Then ED will be parallel to BC , and ED , produced if necessary ,will bisect A C and AB .

The straight line A O will bisect the angle BA C , by page 312 of theEuc lid . The angle OAF i s half the angle BA C ; the angle AFB is equalto the anglesFCB, and EBO,

by I . 32 ; that is to the angle FCB and halfthe angle ABC therefore the angle A CE is equal to the difference of a rightangle and half the angle A OB. But a cir cle would go round AEOD therefore the angle AED i s equal to the angle A OF,

that is to the difference of aright angle and half the angle A CB. But the angle AEC is a right angle ;therefore the angle - D EC is equal to half the angle A OB ; that is to theangle E CB. Therefore ED is parallel to BC , by I. 27.

Produce ED to meet A C at K. The angle KEC has been shewn equalto the angle KCE ; and AEC is a right angle ; therefore the angle KEA isequal to the angle KAE

,by I . 32 ; therefore KA and K0 are each equal

to EX , by I . 6 therefore KA is equal to KC.

Similarly D E , produced if necessary, will bisect AB .

541 . Let AB be the diameter of the g iven circle. On AB describe atriangle ABD equal to half the given rectilineal figur e see I. 45 , C orollary.

Through D draw a straight line parallel to AB,meeting the circle at C .

Then the triangle A GB is equal to the triangle ADB,by I . 37 and is there

fore equal to half the given rectilineal figure. Al so the angle A CB is a rightangle by I I I . 31 .

From A dr aw the chord AE parallel to CB ; and join BE . Then theangle BAE is equal to the angle ABC , by I. 29 ; the right angle AEB isequal to the right angleBCA therefore the angle ABE is equal to the angleBA C . Thus the triangle ABE is equal to the triangle BAC in all respects ;and therefore the figure A CBE is equa l to the g iven rectilinea l figure.

T . EX. EUC. 7


A lso the angle CAE is a right angle ; for it is equal to the two anglesCAB and BAE that is to the two angles CAB and CBA , by I . 29 that isto a right angle . Similarly the angle CBE is a right angle . Therefore thefigure A CBE is a rectangle .

542 . Let 0 be th e point of intersection of AD and BE ; join C O andproduce it to meet AB at L . Then CL is perpendicular to AB , by page 313of the Euc lid . The two circles both pass through L

,by III. 31 ; thus the

rectangles H0 , 0K, and CO , 0L , and F0 , 0G are all equal. Therefore acircle will go through F, G, H,

K by page 277 of the Euc lid .

543. Let AB be one diameter ; and CD another diameter, at right anglesto the former . L et AE , CG, BF, DH be four parallel chords. Then thearcs EA G , GDF, FEH , HCE wi ll all be equal.Join AC, the angle BA C i s equal to the angle A C G , by I. 29 ; therefore

the arc EC is equal to the arc A G , by III. 26 . Therefore the arc EA G i sequal to the arc AEC, that i s to a quarter of the cir cumference. Similarlythe arc HBF is a quarter of the circumference.In like manner the arcs H CE and FD G are equal ; and as they are

together equal to half the circumference each of them is a quarter of thecir cumference.

544. From A draw AM perpendicular to EC , from B draw BN perpendicular to EC . Then we must shew that AM and BN are together equalto EC . For then the square desc ribed on EC will be equal to the rectang leE C ,

AM together with the rectangle EC , BN that is to twice the triangleAEC together with twice the triangle EBO that i s to twice the figureAEBC .

Through 0 the centre of the circle draw the diameter POOparallel to EC ;letBN intersect PQ at R ; and from O draw OS perpendicular to E C . ThenOS i s equal to RN, by I . 34 ; therefore the sum of BN and AM i s twiceBR ; and EC is twice ES. Therefore we have to shew that ES is equalto BR .

Now the angle ED O is equal to the angle ROB, by I . 29 the right angleEGD is equal to the right angle BRO therefore the angle D EO is equal tothe angle OBR .

In th e two triangles EOS and BOR the angle OES is equal to the angleOBR ,

as just shewn ; the right angles OSE and ORB are equal ; and theside OE is equal to the side 0B therefore ES is equal to BR , by I. 26.

545 . Let 0 be the centre of the circle . Bisect BC at D and draw DEparallel to OC , meeting 0B at E . Then E is the middle point of 0B, andED is half of 00 : see Exercises 106 and 109 . And D is the middle pointof the diagonals of the parallelogram , by Exercise 78 .

Thus the requi red locus is the circle having its centre aequal to half OC , that is to half the radius of the given circle .

546 . D escribe the circle which is obfor the required locus. Join AE and produceof this circle at

”

H Then AH is the requireddi agonal may have its greatest possible length

547. Let A and B be the centres of thetouches the circle having its centre at B , andh aving its centre at A , where C and D are on

100 EXERCISES IN EUCLID :

the former join BC,and on It describe a segment containing an angle equalto the la tter. Then the intersection of the two segments will determine thepoint

‘

P .

552. Let A be any point on the circumference of the circle. From A drawAB perpendicular to 0B one of the given straight lines , and AC perpendicular to 00 the other given straight line. A lso through A draw a straight linemeeting OB at E, and 00 at D ; and equally inclined to the two givenstraight lines .Then the angle OED i s equal to the angle ODE by construction

,and

therefore the angle CAD is equal to the angle CD A , by I . 29 ; therefore A Cis equal to DC. Hence the sum of AB and AC is equal to the sum of OCand CD , that is equal to OD .

Thus if the straight line DA cuts the c ircle we can obtain a less value ofthe sum of AB and AC by taking instead of A some point on the are betweenD E and 0 . In th is way we see that when the sum of AB and AC is least thepoint A must be such that the tang ent to the circle there is equally inclinedto the fixed straight lines , and is between 0 and the circle .Similarly in order that the sum of AB and A C may be greatest the

tangent at A must be equally inclined to the fixed straight lines , and thec ircle and the point 0 be on the same side of the tangent .

5 53. Let the segments described on A C and AB intersect at D. Thenthe angles ADB and A CB are together equal to two right angles

,so are the

angles ADC and ABC ; therefore the angles ED O and BAC are togetherequal to two right angles by I . 15 , C orollary 2 . Therefore the segmentdescribed on B0 will pass through D : thus the segments all pass thr oughone point.The angles ADB and A CB are together equal to two right angles therefore

the angle A CB is equal to the angle contained by the remaining part of thecircle of which ADB is a segment

,by III . 22. Hence it follows that the

circle of which ADB is a segment is equal to the circle which could be described round the triangle ABC . Similarly this holds for the circles of whichAD C and BD C are segments. Hence the three circles are equal . ProduceBD to meet AC at M

,and produce CD to meet AB at N. Since the circles

ADB“

and ADC are equal the angles NBD and MCD are equal , by III. 28and I II . 27 ; the angles N DB and MDC are equal, by I . 1 5 ; therefore theangle D NB i s equal to the angle DMC , by I . 32 therefore the angle AND isequal to the angle AMD . The angles NDM and NAM are together equal tothe angles BD C and NAM, that is to two right angles. Hence the anglesAND and AMD are together equal to two right angles ; and as they are equaleach of them is a right angle. Similarly AD produced meets BC at rightangles.

IV. 1 to 16.

554. The three perpendiculars meet at a point , by page 313 of theEuc lid denote this poin t by 0 . In the triangles AFC and OEC the anglea t C is common ; the right angle AFC is equal to the right angle OEC thereq

fore the angle FAC is equal to the angle EOC .

Since the angles BEC and ADC are right angles a circle will go roundOECD , by the Euc lid , page 276 ; therefore the angle ED C is equal to theangle EOC therefore the angle ED O is equal to the angle BA C .

EXERC ISES I N EUCLID . 101

Sim ilarly the angle FDB is equal to the angle BAC . Therefore D E and

DFare equally inclined to BC , and therefore to AD.

5 55 . Let ABC be “a triangle ; suppose that the inscribed circle touchesBC at D , CA at E , and AB at F. C onstruct the triangle FD E ; from Ddraw DP perpendicular to EF

,from E draw EQ perpendicular to FD,

and

from F draw FR perpendicular to ED . Then will PQ , QR , RP be parallelto AB , BC, CA respectively.

AB touches the inscribed circle D EF; therefore the angle AFP is byIII . 32 equal to the angle FD E . But in the solution of Exercise 5 54 it isshewn that the angleFPQ i s equal to the angle FDE . Therefore the angleEPO is equal to the angle AFP ; therefore P0 i s parallel to ,

AB , by I. 27.Similarly QR is parallel to BC, and PR is parallel to CA .

556. D raw the circumscribing circle ; then a s one angle is given theside opposite this angle is g iven in magnitude . For if from any point on thecircumference of the circumscribing circle we draw two straight lines containing an angle equal to the given angle , the chord which they interceptwill be equal in magnitude to the side. Thus the problem is reduced to thatof Exercise 293.

557. Let AB be the base ; then the vertices of all the triangles lie on asegment of a circle described on AB ; see the Euc lid , p . 276.

Let A CB be one of th e triangles. Produce AB , AC to H and K, andbisect the angles KCB

,CBH , by the straight lines CE,

BE meeting at E .

Then as in Exercise 282,E i s the centre of the circle touching BC and

AB , AC produced ; and the straight line AE bisects the angle CAB as inExercise 280.

Since the angle AEB is the difference of the angles EBH and EAB byI . 32 ; and CBH is double of EBH , and CAB double of EAB, therefore theangle A CB is double of the angle AFB , and the point E lies on the segmentof a circle described on AB and containing an angle AEB equal to half thegiven angle A CB. Let this segment AEB be described.

Bisect AB at G , and draw GDFat right angles to AB meeting the segment A CB at D,

and the segment AEB at F, and join AD, DB,FA

,FB.

’

Then the angle ADB is double of the angle AFB by construction ; also byI . 4 these angles are bisected by GF, therefore the angle A D G is double ofAFD and AD is equal to DE. Similarly DB is equal to DE,

and thereforeD is the centre of the segment AEB, by III. 9.

558 . Let AD meet BC at M. The angles A OM and CAM are togetherequal to a right angle ; the angles A CB and CBE are together equal to a.right angle ; therefore the angle CAD is equal to the angle CBE . But theangle CAD is equal to the angle CBD , by III. 21 . Therefore the angleMBD i s equal to the angle MBL . A lso the right angle DME is equal to therigh t angle LMB ; and BM is common to the two triangles BMD and BMLtherefore MD is equal to ML, by I . 2 6.

559 . A circle may be described round the regular pentagon ; then theangle D A C standing on DC, i s equal to the angle ADB standing on AB

,

which is equal to DC. Therefore AO is equal to OD, by I. 6 .

A gain,the angle C OD is equal to the sum of the angles CBO and B00 ,

that is to twi ce the angle CBO ; and the angle A CD is equal to the sum ofthe angles A CE and E CD , that is to twice the angle A CE : therefore the

102 EXERC ISES I N EUCLID .

angle COD is equal to the angle OCD ; therefore 0D is equal to DC ; andtherefore 0A i s equal to CD .

Thus the triangle A C D is exactly like the triangle ABD of IV. 10 ; thesides AC and AD are equal ; the angles ADC and A CD are each double ofthe angle D A C ; and AO is equal to DC. Therefore as in IV. 10 we havethe rectangle AB , BC equal to the square on BD , so here we have therectangle A C

,CO equal to the square on D C .

560. The angles C QR and CPR are right angles ; therefore a circledescribed on CR as diameter will go round C ORP. This circle will be ofconstant magnitude ; for PQ i s of constant length

,and the angle PCQ is

constant . Thus the di stance of R from C is constant .A gain , let PM be perpendicular to C Q ,

and ON perpendicular to CP .

Then a circle will go round QMNP ; and this circle will be of constantmagnitude , for its diameter P0 is of constant magnitude. A l so MN Wl ll beof constant magnitude , for the angle MQN is the difference between a rightangle and the angle C, and 1s therefore constant .Finally a cir cle described on CS as diameter will go round CMSN ; and

it will be of constant magnitude ; for MN is of constant length , and theangle MCN is constant. Thus the di stance of S from C is constant.

561 . Let AB be the hypotenuse , A CB one of the right - angled triangles.It follows from the solution of Exerc ise 293 that the required locus is asegment of a circle ADB containing an angle equal to a right angle anda h alf. C omplete the circle of which thi s segment is part ; let 0 be thecentre . Then the angle contained in the remaining part of the circle is halfa right angle

,by III . 22 ; therefore the angle A OB is a right angle, by

II I . 20. Therefore the arc ADB is a quarter of the circumference.

5 62 . D is the centre of the circle described round ABC,by IV. 5 ; there

fore the locus of D is the straight line which bisects AB at right angles :see III. 1 .

563. Let AB be the given base ; let A CE represent the triangle, 0 thecentre of the inscribed circle , P th e centre of the circle which touch es AB ,

and touches CA and CB produced . Then OP is a kn own length . Supposethe angle CAB given .

0A bisects the angle CAB, and PA bisects the angle between AB and CAproduced ; therefore OAP IS a right angle. Similarly OBB is a right angle .A circle may be described on the known length OP as a diameter . Thenthe angle A OB subtended in such a circle by a known chord AB becomesknown ; also the angle OAB is known , for it is half the angle CAB. Thusall the angles of the triangle OAB are known

,and the triangle can be

constructed . Then make the angle CAB equal to twice the angle OAB, andthe angle CBA equal to twice the angle OBA : thus we obtain the requiredtr iangle A CE .

5 64. L et A be the given point in the given straight line AE then thecentre of the required circle must be on the straight lin e drawn through A atright angles to AE . Let C be the centre of the given circle

,and suppose the

required circle to cut it at the points L and M then by supposition L,C,

and M are in one straight line ; suppose AC produced to meet the requiredcircle ag ain at D : then the rectangle A C

,CD is equal to the rectangle

LC, CM,by III. 35 ; that is to the square on LC . N ow as LC and AC are

104 EXERC ISES I N EUC LID .

569 . L et ABC be the given triangle,B the vertex. D escribe a circle

about it . Let F be the centre of this circle. Join FB. On FB as diameterdescribe a circle cutting AC at D . Join BD . This shall be the requiredstraight line. Produce BD to meet the circumference of the circum scribedc ircle at E . Join FD . ThenFD B is a right angle by III . 31 , therefore EDi s equal to DB by III . 3. Th e rectangle AD

, D C is equal to the rectangleED

,DB by III. 35

,that is to the square on DB .

5 70. D raw a quadrilateral DBEFand let BD , EFproduced intersect atA

,and let BE , D E produced intersect at C. D escribe a circle about the

triangle AEB and another about the triangle BD C . Let B0 be the commonchord of the circles. J oin AO and DO . Then the angles A OB and AEBare equal by III. 21 ; and the angle A OB is equal to the sum of the anglesA OD and D OB ; also the angle AEB is equal to the sum of the angles BCD

‘

and D EA by I . 32 and I . 1 5 . Therefore the angle AEB is equal to the sumof the angles DOB and DFA by III. 21 .

Therefore the angle A OD is equal to the angle DFA . Therefore thecircle descri bed about the triangle AFD passes through the point 0 by page276 of the Euc lid .

Sim ilarly it may be shewn that the circle which circumscribes EFC alsopasses thr ough 0 .

5 71 . Let ABC be a triangle. From A draw a perpendicular on BC .

From B draw a perpendicular on AC ; let these perpendiculars cut at D .

From C draw CG perpendicular to AB. This wi ll pass through D by p . 313

of the Euc lid.

The rectangle GA , GE is equal to the rectangle C C , GD and thereforeequal to the rectangle GB, GFby III . 36 C orollary.

The angle DAG is equal to the angle EC G by III . 21 . Th e angle DAGis also equal to the angle BCG by I . 32. Therefore the angle EC G is equalto the angle BCG . Therefore the triangles EC G and BCG are similar

,and

CG is common,therefore EG is equal to BG by I . 26 . Now the rectangle

GA, GE is equal to the rectangle GB,

GF, therefore GA is equal to GF,

therefore AE is equal toFB.

572 . Let 0 be the centre of the circle inscribed in the triangle ABCand let D , E , F be the centres of the escribed circles . Then AO is perpendicular to FE see Exercise 282 . The result required follows as in Exercise5 53.

573. Let ABD O be the quadrilateral . Let EA bisect the angle BACand let EB bisect the angle ABD . Then E i s the centre of the circle whichtouches internally the Straight lines DB , BA , AC. Similarly let F be thecentre of the circle which touches internally A C ,

CD , DB. In like mannerlet G

,H be the centres of the other two circles. Then it can be easily shewn

that the angles at E and F are together equal to two right angles. Thereforea circle can be described round the quadr ilateral EFGH by page 276 of the

Euc lid . See Exercise 513.

5 74. Take the point P in the arc AB . Because DP,PFare respectively

perpendicular to BC, BA , therefore th e angle ABC is equal to the angleD PF,

and the angle D PF i s equal to the angle DEF by III. 21 , thereforethe angle ABC is equal to the angle D EF. Similarly the angle A CE isequal to the angle DEE and the angle BAC is equal to the angle EDF.

EXERC ISES IN EUCLID . 105 .

Therefore the sides of the triangle ABC are equal to the sides of the triangleD EF respectively by III. 26 and III. 29. A gain

,the arc BC is equal to the

arc EF; take away if necessary the common are FC,then BE is parallel to

22byExercise 219 . Similarly it may be shewn that AD is parallel to either

or F.

5 75 . We will take the case in which the point D falls within the givencircle ; the case in which D falls without the given circle can be treated insubstantially the same manner . Let C be the centre of the described cir cle .Join QB , BC, CA, QC, CD . The two angles QBC , QA C are together equalto two right angles by III. 22 . Now since CD is equal to CA ,

the angle CDAis equal to the angle CAD : therefore the two angles QD C , QA C are togetherequal to two right angles ; therefore the angle QD C i s equal to the angleQBC and the triangle DCB is equilateral

,therefore the angle QED is equal

to the angle QDB, therefore the side QD i s equal to the side QB by I. 6 ;therefore Q C bisects the angle DCB by I. 8 , therefore QB subtends at thecentre of the given circle an angle equal to two - thirds of a right angle byI II . 20, therefore QB is equal to the radius of this circle by IV . 15 C orollary,but QD has been shewn to be equal to QB . Therefore QD is equal to theradius of the circle .

5 76 . Let ABCD be the given square, and P a point without it such thatthe angles APB , BPC , CPD are a ll equal . Then PAB wil l be greater thana right angle and PCB less than a right angle .D raw BM,

BN perpendiculars on PA , PC produced if necessary ; thensince BP bisects the angle A PC ,

BM will be equal to BN . Hence as BA isequal to BC the triangles BAM, BCN will be equal in every respect . Thusthe angles BAP

,BCP are together equal to two right angles

,and therefore

P lies on the cir cle described so as to pass through the points A,B

,C ; that

is the circle described about the square ABCD.

577. Let ABC be a triangle . D raw AP perpendicular to BC . Let thecircle inscribed in the triangle APB touch AP

,BP , AB at M,

N, E respec

tively . Then the sum of the straight lines AP,BP is equal to the sum of

the straight lines AM, MP ,BN ,

NP, but AM is equal to A E , PM is equal toPN andBN is equal to BE byE xerc ise 176 , therefore the sum of the straightlines AP and BP is equal to the sum of the straight lines AE

,BE together

with twice the straight line MP , that is equal to the sum of AB and twicethe radius of the circle . Hence if X denote the diameter of the circle inscribedin ABP, the sum of X and AB is equal to the sum of AP andBP . Similarlyif Y denote the diameter of the circle inscribed in A CP ,

th e sum of Y andAC is equal to the sum of AP and CP . Therefore the sum of the straightlines X

,Y,AB , AC is equal to the straight line BC together with twice the

straight line AP. Two other results like thi s can be found ; and then byaddition the required result can be obtained .

578 . Let 0 be the common centre. Take any point P on the circumference of the middle circle ; j oin OP and produce it to Q making PQ equal toOP. With centre Q and radius equal to th at of the smallest cir cle , describea circle , and let one of the points at which it meets the outermost circle be S.

Join QS from O draw OR a radius of the inner circle parallel to QS. JoinSP PR .’

By construction SQ is equal to R0 and QP is equal to PO . Then inth e two triangles SPQ, RPO the two sides SQ , QP are equal to the two sides


R0 , OP each to each , and the included angle SQP is equal to the includedangle ROP by I. 29, therefore the base SP is equal to the

,

base PR and thetriangles are equal in all respects by I . 4 , therefore the angle SPQ i s equalto the angle OPE, therefore SP , PR are in the same straight line ; See I . 15 .

VI. 1 to D . .

579. D raw PM perpendicular to AB. Then CD is to PM as A C is toAM, and CE is to PM as BC i s to BM by VI. 4 . Therefore the rectangleCD,

CE i s to the square on PM as the rectangle AC,BC is to the rect

angle AM, MB. But the square on PM is equal to the rectangle AM,BM by

III. 35 , therefore the rectangle CD , CE is equal to the rectangle AC, BC,which i s equal to the square on CF by III. 35 . Therefore since the rectangleCD , CE is equal to the square on CF, CD is a .third proportional to CE andCF by VI . 17 .

580. From the middle point P of BD draw a straight line at right anglesto BD meeting AB produced through 0 at the point 0 then OB is equal toOD by I . 4 . Therefore the angle ODB i s equal to the angle OBD by I . 5 ,which is equal to the sum of the angles OAD and ADB vb y I . 32 . But theangle ADB is equal to th e angle OBB, therefore the angle ODB i s equal toth e sum of the angles OAD and CDB, therefore the angle OD O is equal tothe angle OAD . A lso the angle A OD is common to the two triangles OCD ,

ODA . Hence these triangles are similar by I . 32 ; therefore OD is to 00 a s

AD is to DC,therefore OD is to 00 as AB is to BC by VI. 3 . But OD is

equal to OB,therefore OB is to 00 as AB is to BC , which shews that O

i s a fixed point . “ Hence the locus of D is a circle whose centre is O andr adius OB .

581 . Let ABCD be a square . Take BE equal to a fourth of BD andtherefore equal to a third of D E . Join AE and produce it to meet BCat F. Then by the similar triangles A D E and BEE we have BF a third ofAD or BC. Let the straight lines Bbc

, C cd , D d d be similarly drawn fromB

,C,D . Then by the symmetry of the construction it is evident that the

figure a bcd is both equilateral and equiangular, that is it is a square . And

since the straight line BC is equal to three times the straight line BF th e

straight line Bc i s equal to three times the straight line Bb and therefore thestraight hne b c is equal to twice the straight line Bb . And the square onBC is equal to the sum of the squares on Bc

,C c by I . 47 . But the square

on C c is equal to the square on Bb , therefore the square on BC is equal tothe sum of the squares onBc

,Eh, which is equal to ten times the square on

Bb . Therefore the square a bc d i s equal to four times the square on Bb orfour- tenths of the square on BC which is equal to two - fifth s of ABCD.

582 . AF bisects the angle A of the triangle by VI. 2 , 3.

583. On the straight line AC describe a semicircle ; from B draw BQmeeting the semicircle at Q so that the angle ABQ may be equal to half ari ght angle. From Q dr aw QP perpendicular to A C .

Since the angle PBQ is half a right angle, and the angle QPB is a rightangle

,therefore the remaining angle PQB i s half a right angle, and is there

fore equal to the angle AEQ ; therefore the side PB is equal to the side PQ ,

by I. 6. Therefore the square on PB is equal to the square on PQ , that is


Hence if th e points where they meet the given straight lines he joined weh ave a triangle with known angles .Make such a triangle pqr : on qr describe a segment of a c ircle having an

angle equal to the angle 0 , and on p g a segment having an angle equal tothe angle POB . Let 0 be the point where these segments meet . Join op .

Then draw PQ making the angle OPQ equal to the ang le op q and PR makingthe angle OPR equal to the angle 01 m

590. Produce PE to meet the arc BO at G. Join AG .

Then the rectangle D E , EA is equal to the rectangle BE , EC ,that is to

the rectangle PE,EG by III. 35 , therefore DE is to EP a s GE is to EA by

VI. 16 ; therefore the angle GA G is equal to the angle D PE by VI. 6 . Butthe angle GA G is equal to twice the angle OPG by I I I . 20, therefore theangle D PE is bisected by the straight line OP, therefore the angles DPO , EPOa re equal to one another.

591 . Join CF and produce it to meet AB at H. Join HG.

N ow HF I s equal to FC ; see VI. 1 ; thus HG is equal and parallel to ECb y I. 4 , 27 . A gain the ratio ofB0 toBE 13 half that ofBG toBE ,

that Is halfthat of HG to AE by VI. 4, that i s half that of HB to AB by VI . 2 . D rawHK parallel to DC ; then DFis equal to FK by I . 26. Therefore the ratioof D E to DA is half that of DK to DA , that is half that of HB to AB byVI. 2. Therefore D E is to DA as B0 i s to BE .

592 . Join AB , AF. First we shall shew that the square on AB is equalto the rectangle AE

,AD . We have the square on AB equal to the square on

A E together with the rectangle BE , E C by the note on III . 35 and III. 36on page 277 of the Euc lid . A lso the rectangle BE

,EC is equal to the rect

angle AE,ED by III . 35 , therefore the square on AB is equal to the square

on AE together with the rectangle AE , ED , that i s it is equal to the rectangleAE , AD . But AB is equal to AF : therefore the square on AF is equal tothe rectangle AE

,AD, therefore by the converse of VI. 8 we have the angle

AEF a right angle. Similarly the angle AEG i s a right angle, thereforeGE,

EFare in one straight line by I. 14.

593. The triangle ADE i s equal to the triangle BFE by I . 38 ,the triangle ADE together with the triangle ABE is equal to theAFE ; therefore by I. 38 the triangle A DE together with the trianis equal to the triangle FOG . Take away the common part , theGHC ; therefore the triangle ADE together with the triangle BHCto the triangle EHG.

594. Let CPQ be the straight line drawn from the angle C m eeti

intersection of the two straight lines at P and the side AB at Q .

The triangle PEC i s equal to onethird of the triangle PAE byalso the triangle BEC is equal to one - third of the triangle BAEthe triangle BPA is equal to thr ee times the triangle BBC , that istimes the triangle BPD ,

therefore the straight line PA is equaltimes the straight line PD , therefore the triangletwelfth of the triangle PAC byVI . 1 , therefore the tone thirty- sixth of the triangle PA C : therefore the tone - ninth of the tri angle PA C . Therefore this is a l


595 .The tangents so drawn are parallel to the sides of the inscribed

figure,as may be shewn by drawing radi i to the points of contact. Then the

required result follows by Exercise 35 .

5 96. Let ABC , CBD be the two right - angled triangles , then the angleA CD i s a right angle. Make BE equal to BC. Join E C : then BEC is therequired triangle

,for the three triangles are as AB , BD ,

EB by VI. 1 , andEB i s a mean proportional between AB and BD by c onstruc tion

‘

and VI. 8C orollary.

597 .Join D E . Th en DE is parallel to AB by VI. 2 . From the similar

triangles D OE , CAB,CD is to D E a s CA is to AB , but CD is equal to one

third of CA,therefore ED is equal to onethird of AB. From the similar tri

angles OED ,BOA , OE is to ED as 0A i s to BA , but ED is equal to one

thir d of BA ,therefore OE i s equal to onethird of OA ,

that is to one- fourthof EA . Similarly 0D is equal to one - fourth of DB .

598 . Let 0 and P be the centres of the two circles. Join CD, PE .

Then the angle CDA is a right angle by III. 31 ; also the angle PED i s aright angle by I II . 18 , therefore PE i s parallel to CD by I. 28 , thereforeAD is to D E as AC is to CP by VI. 2 , therefore AD is to twice DE as ACis to twice CP by V. 4 . Similarly BF i s to twice FG as twice CP i s totwice CO. Hence AD is to twice D E , as twice GE is to FB, by V. 1 1

therefore the rectangle AD , FB is equal to four times the rectangle FG ,

DE by VI. 16.

599 . Wh en circles cut at right angles the tangents at a point of intersect ion are at right angles to each other ; and thus the radius of each circle is atang ent to the other circle at this point. Let D be the point where thecircles whi ch have B and C as centres meet ; we have only to shew that theangle BD C i s a right angle .Since BD is equal to the tangent from B to the circle A C , the square

on BD is equal to'

the rectangle BA , BC by III. 36 . Similarly the square onCD is equal to the rectangle CA , CB. Therefore by II . 2 the square on BCis equal to the sum of the squares on BD , D C , therefore the angle BD C isa right angle by I . 48.

600. Let BF cut AD_at P. Then it may be shewn that the angle APB

is a right angle and that the angle PAB is two - th irds of a right angle .Hence it follows that PA is one -half of AB . A lso AD is twice AB : See

IV. 15 C orollary. Hence AP is one- four th of AD and therefore AP is onethird of PD .

601 . Since the angle A is equal to the angle D , and AB is equal to DE,

therefore the perpendi cular from B on AC is equal to the perpendicula r fromF on D E . Th erefore by VI. 1 , the triangles are as A C to D E .

602 . Let E be the centre of the inscribed , D of the escribed circle , thenB

, E ,D , are in one straight line . D raw EF, D C perpendicular to BA .

Then BE i s to BD as FE is to D G , that is as EM i s to DP, therefore PDis parallel to EM by VI. 7 and I . 28 . And ND is parallel to EM. Therefore PD and D N must lie in a straight line .

603. The triangle DMA is similar to the triangle BA C ; therefore AM is

to MD as CA is to AB by VI. 4. Therefore by V. 16, AM is to CA as MD is

1 10 EXERCISES IN EUCLID .

to AB, that is as NA is to AB,therefore the triangle CAM i s similar to

the triangle BAN, by VI. 6, therefore the angle AMC is equal to the ang leANB , therefore the angle BMC is equal to the angle BN C by

604. Let B be the middle point of the arc ABC ‘ From B draw anytwo straight lines BF, BG , meeting the circumference at F and G ,

and thechord AC at D and E , respectively.

The angle BED is equal to the sum of the ang les BCE ,EBO, that is to

th e sum ofo

the angles BFC , CFG by III. 27 and III. 21 , that i s to the angleBFG . Now BED and D EG are equal to two right angles

,therefore BEG

and D EG are equal to two right angles,therefore the points F

,D

,E ,G are

on the circumference of a circle . See page 276 of the Euc li d .

605 . Let H be the centre of the inscribed K of the escribed circle . Theangle HAK is equal to the angle HBK, that i s to a right angle by IV. 4 ;therefore a circle would go round HAKB : see page 276 of the Euc lid .

Produce HD to meet the circumference of this circle again at L ; then theangle KLH is a right angle by III. 31 therefore D EKL i s a. rectangle andEX i s equal to DL . And the rectangle AD

,DB is equal to the rectangle

HD,DL , that is to HD , EX . Similarly the rectangle AE , EB is equal to

th e rectangle HD , EX .

606. L et ABC be any triangle. Let D E be drawn parallel to the baseBC . Let F be the middl e point of DE . Join AF and produce it to meetBC at G.

Then AB is to AD as CA i s to AE by VI. 2 ; but BA is to AD as BC isto DFand CA is to AE as C G is to FE by VI. 4 , therefore DF is to FE a s

BG is to C C , but DFis equal to FE , therefore BC i s equal to GO. Hencethe locus of F is the straight line drawn from A to the middle point of

607. Let ABC be a triangle , AC the base and D the m iddle point ofA C ; then BD bisects every straight line parallel to AC : see Exercise 606 .

Let EHL G be the parallelogram having the side HL in AC , and the sidesEH , GL parallel to the fixed direction . Then F, the middle point of EG , i son BD . D raw FK parallel to the fixed direction ; then P the middle pointof FK is the intersection of the di agonals of the parallelogram EHL G .

N ow the straight line EX moves always parallel to itself and so the locusof its middl e point is a fixed straight line passing through D .

608 . The three bisectors meet at a point , see page 311 of theE uc li d

the locus is a circle having its centre at the.m i ddle point of AB and its .

radius equal to one - sixth of AB . For let G be th e right angle . Let thebisectors be BD , AE ,

CF and let them intersect at G . Because AF is equalto FB, the triangle A GE is equal to the triangle GFB. A gain becauseAD i s equal to DC the triangle ADB is equal to the triangle DCB and the

“

triangle A D G is equal to the triangle D OG by I . 38 , therefore the triangleAGB is equal to the triangle C GB, therefore the triangle FGB is equa lto half the triangle C GB,

therefore GF13 equal to half C C , that is to onethird ofFC . See VI . 1 . But CFI S equal to FA by Exercise 59, thereforethe radius

‘FG is equal to one - sixth of AB.

609 . D raw a common tangent ; let this meet the straight line whichjoins the centres at A . Then a straight line drawn through A and the

1 12 EXERC ISES IN EUCLID .

on PQ . But the rectangle PA,PB is equal to the square on the tangent

from P , and the rectangle QC, QB is equal to the square on the tangent fromQ by III. 36 .

615 . Let H be the centre of the given circle , G the middle point of EF.

J oin HA , HE , HG , HF.

Then twice the square on EG together with twice the square on GHis equal to the sum of the squares on EH , EH by page 293 of the Euc lid ,that is to the rectangle EA , ED together with the rectangle FB, FA ,

together with twi ce the square on HA ; that is to the square on EF, togetherwith twice the square on HA by Exercise 614

,that is to four times the

square on EG together with twice the square on HA . Therefore the squareon GE is equal to the sum of the squares on HA , EG, and thus the radiidrawn from H and G to a point of intersection of the two circles are at righ tangles by I. 48 .

616 . Let ABC be a. right - angled triangle having the angle at B a rightangle . Let BD be drawn from the angle at B perpendicular to A C : from Dlet D E , D E be drawn perpendicular to CB , BA . Join FE . ThenFED is atriangle having the perpendiculars D E

,D E as two of its sides. A lso the

angle EDFis a right angle since EDEE is a right - angled parallelogram.

Since D is on a circle of which BC is a diameter by III. 31 , D E is notgreater than one -half of BC. Similarly D E i s not greater than one - halfof AB .

617 . Let PQ , RS be the two straight lines. Let them intersect at 0 .

Let A be the middle point of RP, B of RS,C of QS, D of PQ . Then ABCD

shall be a parallelogram . For AB and DC are both parallel to PS byExercise 106. Therefore they are parallel to one another. Similarly ADand BC are parallel .Now DC is half PS by Exercise 109 ; and by VI . 4 it may be shewn that

the perpendicular from D on PS i s hal f the perpendicular from Q on PS,

and that the perpendicular from B on PS is half the perpendicular from Ron PS. The parallelogram ABCD is equal to the difference of two parallelograms , each having

'

D'

C for base , one having for'

h eight the perpendicularfrom B on PS, and the other havi ng for height the perpendicular fromD on PS. Thus the parallelogram ABCD i s h alf the difference of twoparallelograms each having DC for base , one having for height the perpendicular from R on PS,

and the other having for h eight the perpendicularfrom Q on PS. Thus the parallelogram ABCD is half the differencetriangles RPS and QPS, that is half the difference of the trianglesand QOS.

618 . Let Q be the m iddle” point of AB , and S the middle point of

Let 0 be the centre of the circle . Join AO cutting QS at P.

The difference of the'

squares on Q0 and OB is equal to the squarQB by I . 47, that is to the square on QA ,

that is to the squares on AP,

therefore the difference of the squares on PO and Oon PA ; therefore the difference of the

'

squares on R0 , OB issquare on RA , that is the square on the tangent from R issquare on RA : see II I . 36 . Therefore "RA is equal to the tange

619. By the preceding Exercise the rectangle RQ ,RP is

square on. RA , therefore RQ is to RA as RA is to RP by VI

EXERCISES IN EUC LID . 1 13

the tri angle QRA is similar to the triangle ARP by VI . 6, therefore theangle A QR is equal to the angle RAP .

620. Let ABCD be the quadrilateral . L et a,b,c,d be the middle

points of AB,BC , CD , DA , respectively ; let E , F be the middle points

of AC,DB respectively. Let 0 be the intersection of the circles round the

triangles aFd , aEb ; we will prove that a circle will pass through (lcEO .

The angle FOa. is the difference of two right angles and the angle a c ;

that is the difference of two right angles a nd the angle ABD . The anglea OE is the difference of two right angles and the angle a bE , that is thedifference of two right angles and the angle BA C . We shall obtain theangle dOE by subtracting from four right angles , the excess of the anglesa OE , and a OF above the angle FOd : thus the angle dOE i s equal to thesum of the angles ABD . BAC, FOd that is to the sum of the angles ABD ,

BA C , Fa d ; that is to the sum of the angles ABD , BAC , ADB ; that is tothe excess of two right angles and the angle BAC over the angle BA D ;

that is to the difference of two right angles and the angle D A C ; that i s tothe difference of two right angles and th e angle c . Therefore a circlecan be described round c 0 , by page 276 of the Euc lid .

Similarly a circle can be describ ed round cFOb .

621 . Let OA , OB ,00 be the three straight lines which bisect the angles

of an equilateral triangle. They meet at one point 0,by page 312 of the

Euc lid . Let P be the given point from which PA ,PB , PC are drawn

perpendicular to OA , OB , 00 respectively . Then the straight line PA shallbe equal to the sum of the straight lines PB and PC, supposing PA to bethe longest perpendicular . D raw PD parallel to OB meeting 00 at D .

D raw DH from D perpendicular to OA , and DFperpendicular to PA . ThenDH is equal to BB because 00 bisects the angle BOH.

From the triangles POD,PFD , PC is equal to PF. A lso BP h a s been

shewn to be equal to DH, that is to AF. But AF,FP are together equal to

AP , therefore BP , PC are together equal to AP.

622 . D raw the straight line AF bisecting the exterior angle between ACand AD , and m eeting the circle ABC at E and draw the tangents TE

,TF

meeting at T. Then the angles EA 0 , FAD are equal ; therefore EBD ,EBC

are equal by III. 21 . Therefore the complements of these angles are equal,that is ABFi s equal to ABE , and thus by II I . 32 the angles TEFand TFEare equal . Therefore T is on the line AB produced ; see Exercise 527 .

If the interior angle between AC and AD be b isec ted'

the proof is substantially the same.

623 . Let ABC be the given triangle. Through E and D the middlepoints of A C and AB respectively draw parallel straight lines EF, D G

meeting BC at the given angle . Through A draw a straight line parallel toBC and let GD,

FE produced meet this straight line at K and H respec

tively . Then FGKH is the parallelogram required ; for it is obvious thatthe triangle AEH is equal to the triangle FEC, and that the triangle KADis equal to the triangle ~BD G.

624. D raw the di agonal s AC and BD intersecting at 0 . Suppose P tobe within the triangle D OC .

T . EX. EUC .

EXERCISES IN EUC LID .

We have to shew that the triangle PAC is equal to the difference of thetriangles PAB and PA D ; this we will do by shewing that the sum of thetriangles PAD and PAC is equal to the tri angle PAB .

The sum of the triangles PAD and PAC is equal to the sum of thetriangles POD

,D OA , POC . The triangle PAB is equal to the sum of the

triangles POB , BOA ,POA . N ow AC and BD bisect each other by Exercise

78 , therefore by I . 38 the triangle POD is equal to the triangle POB ; thetriangle DOA is equal to the triangle BOA and the triangle POC is equal tothe triangle POA . Thus the requir ed result is obtained .

In a sim ilar way any other case can be tr eated .

625 . Let R, Q be the two points where the circles cut one another .

The rectangle A C, CD is equal to the rectangle RC, C Q, that is to therectangle EC , CB by III . 35 ; therefore AC is to CE as CB is to CD by VI.16 therefore by V . 18 AE i s to CE a s BD is to CD therefore AE is to BDas CE is to CD by V. 16. Similar ly AE is to BD as CA is to CB ; therefore the square on AE is to the square on BD as the rectangle AC

, CE

is to the rectangle BC, CD , therefore the square on BD is to the square onAE as the rectangle BC, CD is to the rectangle AC , CE .

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