Euclid. # Euclid's Elements of geometry, the first six books, chiefly from the text of Dr. Simson, with explanatory notes; a series of questions on each book ... Designed for the use of the junior classes in public and private schools online

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equal to the other given line. On AD describe a semicircle, at B draw

BE perpendicular to AD. Bisect BD in O, and with center O and

radius OB describe a semicircle, join OE cutting the semicircle on BD

in F, at P draw FC perpendicular to OE and meeting AB in C. C is

the point of division, such that BC is a mean proportional between

AC and BD.

29. Find two squares in the given ratio, and if BF be the given line

(figure, Euc. vi. 4), draw BE at right angles to BF, and take BC, CE

respectively equal to the sides of the squares which are in the given ratio.

Join EF, and draw CA parallel to EF : then BF is divided in A as

required.

30. Produce one side of the triangle through the vertex and make

the part produced equal to the other side. Bisect this line, and with

the vertex of the triangle as center and radius equal to half the sum of

the sides, describe a circle cutting the base of the triangle.

31. If a circle be described about the given triangle, and another

circle upon the radius drawn from the vertex of the triangle to the center

of the circle, as a diameter, this circle will cut the base in two points, and

give two solutions of the problem. Give the Analysis.

32. This Problem is analogous to the preceding.

33. Apply Euc. vi. 8, Cor. ; 17. ^

34. Describe a circle about the triangle, and draw the diameter

through the vertex A, draw a line touching the circle at A, and meeting

the base BC produced in D. Then AD shall be a mean proportional

between DC and DB. Euc. in. 36.

35. In BC produced take CE a third proportional to BC and AC ;

on CE describe a circle, the center being O ; draw the tangent EF at

E equal to AC ; draw FO cutting the circle in T and T' ; and lastly

draw tangents at T, T meeting BC in P and P'. These points fulful the

conditions of the problem.

By combining the proportion in the construction with that from the

similar triangles ABC, DBP, and Euc. iii. 36, 37 : it may be proved

that CA.PD = CP^ The demonstration is similar for P'D'.

36. This property may be immediately deduced from Euc. vt. 8, Cor.

37. Let ABC be the triangle, right-angled at C, and let AE on AB

be equal to AC, also let the line bisecting the angle A, meet BC in D.

Join DE. Then the triangles ACD, AED are equal, and the triangles

ACB, DEB equiangular.

38. The segments cut off from the sides are to be measured from the

right angle, and by similar triangles are proved to be equal ; also by

similar triangles, either of them is proved to be a mean proportional

between the remaining segments of the two sides.

39. First prove AC- : AD^ :: BC : 2. BD : then 2. ACÂ«: AD^ : : BC : BD,

whence 2 . AC* - ADÂ« : AD^ : : BC - BD : BD,

and since 2. AC* - AD^ = 2. AC* - (AC* + DC*) = AC - CDÂ«,

the property is immediately deduced.

40. The construction is suggested by Euc. i. 47, and Euc. vi. 31.

41. See Note Euc. vi. A. p. 295. The bases of the triangles CBD,

ACD, ABC, CDE maybe shewn to be respectively equal to DB, 2.BD,

3.BD, 4.BD.

42. (1) Let ABC be the triangle which is to be bisected by a line

drawn parallel to the base BC. Describe a semicircle on AB, from the

center D draw DE perpendicular to AB meeting the circumference in

E, join EA, and with center A and radius AE describe a circle cutting

AB in F, the line drawn fron F parallel to BC, bisects the triangle. The

348 GEOMETRICAL EXERCISES, &C.

proof depends on Euc vi. 19 ; 20, Cor. 2. (2) Let ABC be the triangle,

13C being the base. Draw AD at right angles to BA meeting the base

produced in D. Bisect BC in E, and on ED describe a semicircle, from

B draw BP to touch the semicircle in P. From BA cut off BF equal

to BP, and from F draw FGr perpendicular to BC. The line FG bisects

the triangle. Then it may be proved that BFG : BAD :: BE : BD,

and that BAD : BAC :: BD : BC ; whence it follows that BFG : BAC

: : BE : BC or as 1 : 2.

43. Let ABC be the given triangle which is to be divided into two

parts havi'.;g a given ratio, by a Ime parallel to BC. Describe a semi-

circle on AB and divide AB in D in the given ratio ; at D draw DE

perpendicular to AB and meeting the circumference in E ; with center

A and radius AE describe a circle cutting AB in F : the line drawn

through F parallel to BC is the line required. In the same manner

a triangle may be divided into three or more parts having any given ratio

to one another by lines drawn parallel to one of the sides of the triangle.

44. Let these points be taken, one on each side, and straight lines be

drawn to them ; it may then be proved that these points severally bisect

the sides of the triangle.

4o. Let ABC be any triangle and D be the given point in BC, from

which lines are to be drawn which shall divide the triangle into any

number (suppose five) equal parts. Divide BC into five equal parts in

E, F, G, H, and draw AE, AF, AG, AH, AD, and through E, F, G, H

draw EL, FM, GN, HO parallel to AD, and join DL, DM, DN, DO;

these lines divide the triangle into five equal parts.

By a similar process, a triangle may be divided into any number of

parts which have a given ratio to one another.

46. Let ABC be the larger, abc the smaller triangle, it is required to

draw a line DE parallel to AC cutting off the triangle DBE equal to the

triangle abc. On BC take BG equal to be, and on BG describe the

triangle BGH equal to the triangle abc. Draw HK parallel to BC, join

KG ; then the triangle BGK is equal to the triangle abc. On BA, BC

take BD to BE in the ratio of BA to BC, and such that the rectangle

contained by BD, BE shall be equal to the rectangle contained by BK,

BG. Join DE, then DE is parallel to AC, and the triangle BDE is

equal to abc.

47. Let ABCD be any rectangle, contained by AB, BC,

Then AB* : AB . BC : : AB : BC,

andAB.BC:BC^:: AB:BC,

whence AB"^ : AB . BC : : AB . BC : BC^

or the rectangle contained by two adjacent sides of a rectangle, is a mean

proportional between their squares.

48. In a straight line at any point A, make Ac equal to Kd in the

given ratio. At A draw AB perpendicular to cKd, and equal to a side

of the given square. On cd describe a semicircle cutting AB in b ; and

join be, bd ; from B draw BC parallel to be, and BD parallel to bd\ then

AC, AD are the adjacent sides of the rectangle. For, CA is to AD

as cA to A.d, Euc. vi. 2 ; and CA. AD = AB', CBD being a right-angled

triangle.

49. From one of the given points two straight lines are to be drawn

perpendicular, one to each of any two adjacent sides of the parallelogram ;

and from the other point, two lines perpendicular in the same manner to

each of the two remaining sides. When these four lines are drawn to

intersect one another, the figure so formed may be shewn to be equi-

angular to the given parallelogram.

I

ON BOOK VI. 349

50. It is manifest that this is the general case of Prop. 4, p. 197.

If the rectangle to be cut off be two-thirds of the given rectangle ABCD.

Produce BC to E so that BE may be equal to a side of that square

which is equal to the rectangle required to be cut oif ; in this case, equal

to two-thirds of the rectangle ABCD. On AB take AF equal to AD or

BC ; bisect EB in G, and with center G and radius GE, describe a

semicircle meeting AB, and AB produced, in H and K. On CB take

CL equal to AH and draw HM, LM parallel to the sides, and HBLM

is two-thirds of the rectangle ABCD.

51. Let ABCD be the parallelogram, and CD be cut in P and BC

produced in Q. By means of the similar triangles formed, the property

may be proved.

52. The intersection of the diagonals is the common vertex of two

triangles which have the parallel sides of the trapezium for their bases.

53. Let AB be the given straight line, and C the center of the given

circle ; through C draw the diameter DCE perpendicular to AB. Place

in the circle a line EG which has to AB the given ratio ; bisect EG in

H, join CH, and on the diameter DCE, take CK, CL each equal to

CH ; either of the lines drawn through K, L, and parallel to AB is

the line required.

54. Let C be the center of the circle, CA, CB two radii at right angles

to each other ; and let DEEG be the line required which is trisected in

the points E, F. Draw CG perpendicular to DH and produce it to meet

the circumference in K ; draw a tangent to the circle at K : draw CG,

and produce CB, CG to meet the tangent in L, M, then MK may be

shewn to be treble of LK.

55. The triangles ACD, BCE are similar, and CE is a mean propor-

tional between AC and CB.

56. Let any tangent to the circle at E be terminated by AD, BC

tangents at the extremity of the diameter AB. Take O the center of the

circle and join OC, OD, OE ; then ODC is a right-angled triangle and

Gi^ is the perpendicular from the right angle upon the hypotenuse.

57. This problem only differs from problem 59, infra, in having the

given point without the given circle.

58. Let A be the given point in the circumference of the circle, C its

center. Draw the diameter ACB, and produce AB to D, taking AB to

BD in the given ratio : from D draw a line to touch the circle in E,

which is the point required. From A draw AF perpendicular to DE,

and cutting the circle in G.

59- Let A be the given point within the circle whose center is C, and

let BAD be the line required, so that BA is to AD in the given ratio.

Join AC and produce it to meet the circumference in E, F. Then EF

is a diameter. Draw BG, DH perpendicular on EF : then the triangles

BGA, DHA are equiangular. Hence the construction.

60. Through E one extremity of the chord EF, let a line be drawn

parallel to one diameter, and intersecting the other. Then the three

angles of the two triangles may be shewn to be respectively equal to one

another.

6 1 . Let AB be that diameter of the given circle which when produced

is perpendicular to the given line CD, and let it meet that line in C ; and

let P be the given point : it is required to find D in CD, so that DB

may be equal to the tangent DF. Make BC : CQ :: CQ : CA, and join

PQ ; bisect PQ in E, and draw ED perpendicular to PQ meeting CD in

D ; then D is the point required. Let O be the center of the circle, draw

the tangent DF ; and join OF, OD, QD, PD. Then QD may be shewn

350 GEOMETRICAL EXERCISES, &C.

to be equal to DF and to DP. When P coincides with Q, any point D

in CD fulfils the conditions of the problem ; that is, there are innume-

rable solutions.

62. It may be proved that the vertices of the two triangles which are

similar in the same segment of a circle, are in the extremities of a chord

parallel to the chord of the given segment.

63. For let the circle be described about the triangle EAC, then by

the converse to Euc, iii. 32 ; the truth of the proposition is manifest.

64. Let the figure be constructed, and the similarity of the two tri-

angles will be at once obvious from Euc iii. 32. ; Euc. i. 29.

65. In the arc AB (fig. Euc. iv. 2) let any point K be taken, and

from K let KL, KM, KN be drawn perpendicular to AB, AC, BC respec-

tively, produced if necessary, also let LM, LN be joined, then MLN may

be shewn to be a straight Ime. Draw AK, BK, CK, and by Euc. iii. 31,

22, 21 ; Euc. i. 14.

66. Let AB a chord in a circle be bisected in C, and DE, FG two

chords drawn through C; also let their extremities DG, FE be joined

intersecting CB in H, and AC in K ; then AK is equal to HB. Through

H draw MHL parallel to EF meeting FG in M, and DE produced in L.

Then by means of the equiangular triangles, HC may be proved to be

equal to CK, and hence AK is equal to HB.

67. Let A, B be the two given points, and let P be a point in the

locus so that PA, PB being joined, PA is to PB in the given ratio. Join

AB and divide it in C in the given ratio, and join PC. Then PC bisects

the angle APB. Euc. vi. 3. Again, in AB produced, take AD to AB

in the given ratio, join PD and produce AP to E, then PD bisects the

angle BPE. Euc. vi. A. Whence CPD is a right angle, and the point P

lies in the circumference of a circle whose diameter is CD.

68. Let ABC be a triangle, and let the line AD bisecting the vertical

angle A be divided in E, so that BC : BA+ AC :: AE : ED. By Euc.

VI. 3, may be deduced BC : BA + AC :: AC : AD. Whence may be

proved that CE bisects the angle ACD, and by Euc. iv. 4, that E is the

center of the inscribed circle.

69. By means of Euc. iv. 4, and Euc. vi. C. this theorem may be

shewn to be true.

70. Divide the given base BC in D, so that BD may be to DC in the

ratio of the sides. At B, D draw BB', DD' perpendicular to BC and

equal to BD, DC respectively. Join B'D' and produce it to meet BC

produced in O. With center O and radius OD, describe a circle. From

A any point in the circumference join AB, AC, AO. Prove that AB is

to AC as BD to DC. Or thus. If ABC be one of the triangles. Divide

the base BC in D so that BA is to AC as BD to DC. Produce BC and

take DO to OC as BA to AC : then O is the center of the circle.

71. Let ABC be any triangle, and from A, B let the perpendiculars

AD, BE on the opposite sides intersect in P : and let AF, BG drawn to

F, G the bisections of the opposite sides, intersect in Q. Also let FR,

GR be drawn perpendicular toBC, AC, and meet in K : then R is the

center of the circumscribed circle. Join PQ, QR ; these are in the

same line.

Join FG, and by the equiangular triangles, GRF, APB, AP is

proved double of FR. And AQ is double of QF, and the alternate

angles PAQ, QFR are equal. Hence the triangles APQ, RFQ are

equiangular.

72. Let C, C be the centers of the two circles, and let CC the line

joining the centers intersect the common tangent PP' in T. Let the

â€¢

ON BOOK VI. 351

line joining the centers cut the circles in Q, Q', and let PQ, P'Q' be

joined ; then PQ is parallel to P'Q'. Join CP, C'P', and then the angle

QPT may be proved to be equal to the alternate angle Q'P'T.

73. Let ABC be the triangle, and BC its base ; let the circles AFB,

!aPC be described intersecting the base in the point F, and their

diameters AD, AE, be drawn ; then DA : AE : : BA : AC. For join

rDB, DF, EF, EC, the triangles DAB, EAC may be proved to be similar.

; 74. If the extremities of the diameters of the two circles be joined

hy two straight lines, these lines may be proved to intersect at the

point of contact of the two circles; and the two right-angled triangles

thus formed may be shewn to be similar by Euc. iii. 34.

' 75. This follows directly from the similar triangles.

j 76. Let the figure be constructed as in Theorem 4, p. 162, the tri-

iangle EAD being right- angled at A, and let the circle inscribed in the

triangle ADE touch AD, AE, DE in the points K, L, M respectively.

Then AK is equal to AL, each being equal to the radius of the inscribed

icircle. Also AB is equal to GC, and AB is half the petimeter of the tri-

iangle AED.

Also if GA be joined, the triangle ADE is obviously equal to the

idifference of AGDE and the triangle GDE, and this diff'erence may be

proved equal to the rectangle contained by the radii of the other two

circles.

77. From the centers of the two circles let straight lines be drawn

to the extremities of the sides which are opposite to the right angles

in each triangle, and to the points where the circles touch these sides.

Euc. VI. 4.

78. Let A, B be the two given points, and C a point in the circum-

ference of the given circle. Let a circle be described through the points

A, B, C and cutting the circle in another point D. Join CD, AB, and

produce them to meet in E. Let EF be drawn touching the given

circle in F ; the circle described through the points A, B, F, will be

the circle required. Joining AD and CB, by Euc. in. 21, the tri-

angles CEB, AED are equiangular, and by Euc. vi. 4, 16, iii. 36, 37,

the given circle and the required circle each touch the line EF in the

same point, and therefore touch one another. When does this solution

fail ?

Various cases will arise according to the relative position of the two

points and the circle.

79. Let A be the given point, BC the given straight line, and D the

center of the given circle. Through D draw CD perpendicular to BC,

meeting the circumference in E, F. Join AF, and take FG to the

diameter FE, as FC is to FA. The circle described passing through the

two points A, G and touching the line BC in B is the circle required.

Let H be the center of this circle ; join HB, and BF cutting the

circumference of the given circle in K, and join EK. Then the tri-

angles FBC, FKE being equiangular, by Euc. vi. 4, 16, and the con-

struction, K is proved to be a point in the circumference of the circle

passing through the points A, G, B. And if DK, KH be joined, DKH

may be proved to be a straight line :â€” the straight line which joins the

centers of the two circles, and passes through a common point in their

circumferences.

80. Let A be the given point, B, C the centers of the two given

circles. Let a line drawn through B, C meet the circumferences of

the circles in G, F ; E, D, respectively. In GD produced, take the

point H, 80 that BH is to CH as the radius of the circle whose center

S52 GEOMETRICAL EXERCISES, &C.

is B to the radius of the circle whose center is C. Join AH, and take

KH to DH as GH to AH. Through A, K describe a circle ALK touch-

ing the circle whose center is B, in L. Then M may be proved to be a

point in the circumference of the circle whose center is C. For by join-

ing HL and producing it to meet the circumference of the circle whose

center is B in N ; and joining BN, BL, and drawing CO parallel to BL,

and CM parallel to BN, the line HN is proved to cut the circumference

of the circle whose center is B in M, O ; and CO, CM are radii. By

joining GL, DM, M may be proved to be a point in the circumference

of the circle ALK. And by producing BL, CM to meet in P, P is

proved to be the center of ALK, and BP joining the centers of the two

circles passes through L the point of contact. Hence also is shewn that

PMC passes through M, the point where the circles whose centers are P

and C touch each other.

Note. If the given point be in the circumference of one of the circles,

the construction may be more simply effected thus :

Let A be in the circumference of the circle whose center is B. Join

BA, and in AB produced, if necessary, take AD equal to the radius of the

circle whose center is C ; join DC, and at C make the angle DCE equal to

the angle CDE, the point E determined by the intersection of D A pro-

duced and CE, is the center of the circle.

81. Let AB, AC be the given lines and P the given point. Then if

O be the center of the required circle touching AB, AC, in R, S, the line

AO will bisect the given angle BAC. Let the tangent from P meet the

circle in Q, and draw OQ, OS, OP, AP. Then there are given AP and

the angle OAP. Also since OQP is a right angle, we have OP'â€” QO^

= 0P^ â€” OS* = PQ- a given magnitude. Moreover the right-angled tri-

angle AOS is given in species, or OS to OA is a given ratio. Whence

in the triangle AOP there is given, the angle AOP, the side AP,

and the excess of OP^ above the square of a line having a given

ratio to OA, to determine OA. Whence the construction is obvious.

82. Let the two given lines AB, BD meet in B, and let C be the cen-

ter of the given circle, and let the required circle touch the line AB, and

have its center in BD. Draw CFE perpendicular to HB intersecting the

circumference of the given circle in F, and produce CE, making EF

equal to the radius CF. Through G draw GK parallel to AB, and

meeting DB in K. Join CK, and through B, draw BL parallel to KC,

meeting the circumference of the circle whose center is C in L ; join

CL and produce CL to meet BD in O. Then O is the center of the

circle required. Draw OM perj^endicular to AB, and produce EC tc

meet BD in N. Then by the similar triangles, OL may be proved

equal to OM.

83. (1) In every right-angled triangle when its three sides are in

Arithmetical progression, they may be shewn to be as the numbers 5, 4,

3. On the given line AC describe a triangle having its sides AC, AD,

DC in this proportion, bisect the angles at A, C by AE, CE meeting in E,

and through E draw EF, EG parallel to AD, DC meeting in F and G.

(2) Let AC be the sum of the sides of the triangle, tig. Euc. vi. 13.

Upon AC describe a triangle ADC whose sides shall be in continued

proportion. Bisect the angles at A and C by two lines meeting in

E. From E draw EF, EG parallel to DA, DC respectively.

84. Describe a circle with any radius, and draw within it the straight

line MN cutting off a segment containing an angle equal to the given

angle, Euc. iii. 34, Divide MN in the given ratio in P, and at P draw

PA perpendicular to MN and meeting the circumference in A. Join

il

ON BOOK VI. 353

AM, AX, and on AP or AP produced, take AD equal to the given per-

pendicular, and through D draw BC parallel to MN meeting AM, AN,

or these lines produced. Then ABC shall be the triangle required.

85. Let PAQ be the given angle, bisect the angle A by AB, in

AB find D the center of the inscribed circle, and draw DC perpen-

dicular to AP. In DB take DE such that the rectangle DE, DC is

equal to the given rectangle. Describe a circle on DE as diameter

meeting AP in F, G ; and AQ in F', G'. Join FG', and AFG' will

be the triangle. Draw DH perpendicular to FG' and join G'D.

Bv Euc. VI. C, the rectangle FD, DG' is equal to the rectangle ED,

DK or CD, DE.

86. On any base BC describe a segment of a circle BAC containing

an angle equal to the given angle. From D the middle point of BC draw

DA to make the given angle ADC with the base. Produce AD to E so

that AE is equal to the given bisecting line, and through E draw FG

parallel to BC. Join AB, AC and produce them to meet FG in F and G.

87. Employ Theorem 70, p. 310, and the construction becomes

obvious.

88. Let AB be the given base, ACB the segment containing the

vertical angle ; draw the diameter AB of the circle, and divide it in E,

in the given ratio ; on AE as a diameter, describe a circle AFE ; and with

center B and a radius equal to the given line, describe a circle cutting

AFE in F. Then AF being drawn and produced to meet the circum-

scribing circle in C, and CB being joined, ABC is the triangle required.

For AF is to FC in the given ratio.

89. The line CD is not necessarily parallel to AB. Divide the base

AB in C, so that AC is to CB in the ratio of the sides of the triangle.

Then if a point E in CD can be determined such that when AE, CE,

EB, are joined, the angle AEB is bisected by CE, the problem is solved.

90. Let ABC be any triangle having the base BC. On the same

base describe an isosceles triangle DBC equal to the given triangle.

Bisect BC in E, and join DE, also upon BC describe an equilateral

triangle. On FD, FB, take EG to EH as EF to FB : also take EK

equal to EH and join GH, GK ; then GHK is an equilateral triangle

equal to the triangle ABC.

91. Let ABC be the required triangle, BC the hypotenuse, and

FHKG the inscribed square : the side HK being on BC. Then BC may

be proved to be divided in H and K, so that HK is a mean proportional

between BH and KC.

92. Let ABC be the given triangle. On BC take BD equal to one

of the given lines, through A draw AE parallel to BC. From B draw

BE to meet AE in E, and such that BE is a fourth proportional to BC,

BD, and the other given line. Join EC, produce BE to F, making BF

equal to the other given line, and join FD ; then FBD is the triangle

required.

93. By means of Euc. vi. C, the ratio of the diagonals AC to BD

may be found to be as AB . AD + BC. CD to AB.BE + AD.DC,

figure, Euc. vi. D.

94. This property follows directly from Euc. vr. C.

95. Let ABC be any triangle, and DEF the given triangle to which

the inscribed triangle is required to be similar. Draw any line de

terminated by AB, AC, and on de towards AC describe the triangle def

similar to DEF, join B/, and produce it to meet AC in F\ Through F'

draw FD' parallel to fd, F'E' parallel to /e, and join D'E', then the

triangle DEF' i.^ similar to DEF.

354 GEO lETRICAL EXERCISES, &C.

96. The square inscribed in a right-angled triangle which has one

of its sides coinciding with the hypotenuse, may be shewn to be less than

that which has two of its sides coinciding with the base and perpendicular.

97. Let BCDE be the square on the side EC of the isosceles triangle

ABC. Then by Euc. vi. 2, FG is proved parallel to ED or BC.

98. Let AB be the base of the segment ABD, fig. Euc. iii. 30.,

Bisect AB in C, take any point E in AC and make CF equal to CE:

upon EF describe a square EFGH : from C draw CG and produce it to

meet the arc of the segment in K.

99. Take two points on the radii equidistant from the center, and

BE perpendicular to AD. Bisect BD in O, and with center O and

radius OB describe a semicircle, join OE cutting the semicircle on BD

in F, at P draw FC perpendicular to OE and meeting AB in C. C is

the point of division, such that BC is a mean proportional between

AC and BD.

29. Find two squares in the given ratio, and if BF be the given line

(figure, Euc. vi. 4), draw BE at right angles to BF, and take BC, CE

respectively equal to the sides of the squares which are in the given ratio.

Join EF, and draw CA parallel to EF : then BF is divided in A as

required.

30. Produce one side of the triangle through the vertex and make

the part produced equal to the other side. Bisect this line, and with

the vertex of the triangle as center and radius equal to half the sum of

the sides, describe a circle cutting the base of the triangle.

31. If a circle be described about the given triangle, and another

circle upon the radius drawn from the vertex of the triangle to the center

of the circle, as a diameter, this circle will cut the base in two points, and

give two solutions of the problem. Give the Analysis.

32. This Problem is analogous to the preceding.

33. Apply Euc. vi. 8, Cor. ; 17. ^

34. Describe a circle about the triangle, and draw the diameter

through the vertex A, draw a line touching the circle at A, and meeting

the base BC produced in D. Then AD shall be a mean proportional

between DC and DB. Euc. in. 36.

35. In BC produced take CE a third proportional to BC and AC ;

on CE describe a circle, the center being O ; draw the tangent EF at

E equal to AC ; draw FO cutting the circle in T and T' ; and lastly

draw tangents at T, T meeting BC in P and P'. These points fulful the

conditions of the problem.

By combining the proportion in the construction with that from the

similar triangles ABC, DBP, and Euc. iii. 36, 37 : it may be proved

that CA.PD = CP^ The demonstration is similar for P'D'.

36. This property may be immediately deduced from Euc. vt. 8, Cor.

37. Let ABC be the triangle, right-angled at C, and let AE on AB

be equal to AC, also let the line bisecting the angle A, meet BC in D.

Join DE. Then the triangles ACD, AED are equal, and the triangles

ACB, DEB equiangular.

38. The segments cut off from the sides are to be measured from the

right angle, and by similar triangles are proved to be equal ; also by

similar triangles, either of them is proved to be a mean proportional

between the remaining segments of the two sides.

39. First prove AC- : AD^ :: BC : 2. BD : then 2. ACÂ«: AD^ : : BC : BD,

whence 2 . AC* - ADÂ« : AD^ : : BC - BD : BD,

and since 2. AC* - AD^ = 2. AC* - (AC* + DC*) = AC - CDÂ«,

the property is immediately deduced.

40. The construction is suggested by Euc. i. 47, and Euc. vi. 31.

41. See Note Euc. vi. A. p. 295. The bases of the triangles CBD,

ACD, ABC, CDE maybe shewn to be respectively equal to DB, 2.BD,

3.BD, 4.BD.

42. (1) Let ABC be the triangle which is to be bisected by a line

drawn parallel to the base BC. Describe a semicircle on AB, from the

center D draw DE perpendicular to AB meeting the circumference in

E, join EA, and with center A and radius AE describe a circle cutting

AB in F, the line drawn fron F parallel to BC, bisects the triangle. The

348 GEOMETRICAL EXERCISES, &C.

proof depends on Euc vi. 19 ; 20, Cor. 2. (2) Let ABC be the triangle,

13C being the base. Draw AD at right angles to BA meeting the base

produced in D. Bisect BC in E, and on ED describe a semicircle, from

B draw BP to touch the semicircle in P. From BA cut off BF equal

to BP, and from F draw FGr perpendicular to BC. The line FG bisects

the triangle. Then it may be proved that BFG : BAD :: BE : BD,

and that BAD : BAC :: BD : BC ; whence it follows that BFG : BAC

: : BE : BC or as 1 : 2.

43. Let ABC be the given triangle which is to be divided into two

parts havi'.;g a given ratio, by a Ime parallel to BC. Describe a semi-

circle on AB and divide AB in D in the given ratio ; at D draw DE

perpendicular to AB and meeting the circumference in E ; with center

A and radius AE describe a circle cutting AB in F : the line drawn

through F parallel to BC is the line required. In the same manner

a triangle may be divided into three or more parts having any given ratio

to one another by lines drawn parallel to one of the sides of the triangle.

44. Let these points be taken, one on each side, and straight lines be

drawn to them ; it may then be proved that these points severally bisect

the sides of the triangle.

4o. Let ABC be any triangle and D be the given point in BC, from

which lines are to be drawn which shall divide the triangle into any

number (suppose five) equal parts. Divide BC into five equal parts in

E, F, G, H, and draw AE, AF, AG, AH, AD, and through E, F, G, H

draw EL, FM, GN, HO parallel to AD, and join DL, DM, DN, DO;

these lines divide the triangle into five equal parts.

By a similar process, a triangle may be divided into any number of

parts which have a given ratio to one another.

46. Let ABC be the larger, abc the smaller triangle, it is required to

draw a line DE parallel to AC cutting off the triangle DBE equal to the

triangle abc. On BC take BG equal to be, and on BG describe the

triangle BGH equal to the triangle abc. Draw HK parallel to BC, join

KG ; then the triangle BGK is equal to the triangle abc. On BA, BC

take BD to BE in the ratio of BA to BC, and such that the rectangle

contained by BD, BE shall be equal to the rectangle contained by BK,

BG. Join DE, then DE is parallel to AC, and the triangle BDE is

equal to abc.

47. Let ABCD be any rectangle, contained by AB, BC,

Then AB* : AB . BC : : AB : BC,

andAB.BC:BC^:: AB:BC,

whence AB"^ : AB . BC : : AB . BC : BC^

or the rectangle contained by two adjacent sides of a rectangle, is a mean

proportional between their squares.

48. In a straight line at any point A, make Ac equal to Kd in the

given ratio. At A draw AB perpendicular to cKd, and equal to a side

of the given square. On cd describe a semicircle cutting AB in b ; and

join be, bd ; from B draw BC parallel to be, and BD parallel to bd\ then

AC, AD are the adjacent sides of the rectangle. For, CA is to AD

as cA to A.d, Euc. vi. 2 ; and CA. AD = AB', CBD being a right-angled

triangle.

49. From one of the given points two straight lines are to be drawn

perpendicular, one to each of any two adjacent sides of the parallelogram ;

and from the other point, two lines perpendicular in the same manner to

each of the two remaining sides. When these four lines are drawn to

intersect one another, the figure so formed may be shewn to be equi-

angular to the given parallelogram.

I

ON BOOK VI. 349

50. It is manifest that this is the general case of Prop. 4, p. 197.

If the rectangle to be cut off be two-thirds of the given rectangle ABCD.

Produce BC to E so that BE may be equal to a side of that square

which is equal to the rectangle required to be cut oif ; in this case, equal

to two-thirds of the rectangle ABCD. On AB take AF equal to AD or

BC ; bisect EB in G, and with center G and radius GE, describe a

semicircle meeting AB, and AB produced, in H and K. On CB take

CL equal to AH and draw HM, LM parallel to the sides, and HBLM

is two-thirds of the rectangle ABCD.

51. Let ABCD be the parallelogram, and CD be cut in P and BC

produced in Q. By means of the similar triangles formed, the property

may be proved.

52. The intersection of the diagonals is the common vertex of two

triangles which have the parallel sides of the trapezium for their bases.

53. Let AB be the given straight line, and C the center of the given

circle ; through C draw the diameter DCE perpendicular to AB. Place

in the circle a line EG which has to AB the given ratio ; bisect EG in

H, join CH, and on the diameter DCE, take CK, CL each equal to

CH ; either of the lines drawn through K, L, and parallel to AB is

the line required.

54. Let C be the center of the circle, CA, CB two radii at right angles

to each other ; and let DEEG be the line required which is trisected in

the points E, F. Draw CG perpendicular to DH and produce it to meet

the circumference in K ; draw a tangent to the circle at K : draw CG,

and produce CB, CG to meet the tangent in L, M, then MK may be

shewn to be treble of LK.

55. The triangles ACD, BCE are similar, and CE is a mean propor-

tional between AC and CB.

56. Let any tangent to the circle at E be terminated by AD, BC

tangents at the extremity of the diameter AB. Take O the center of the

circle and join OC, OD, OE ; then ODC is a right-angled triangle and

Gi^ is the perpendicular from the right angle upon the hypotenuse.

57. This problem only differs from problem 59, infra, in having the

given point without the given circle.

58. Let A be the given point in the circumference of the circle, C its

center. Draw the diameter ACB, and produce AB to D, taking AB to

BD in the given ratio : from D draw a line to touch the circle in E,

which is the point required. From A draw AF perpendicular to DE,

and cutting the circle in G.

59- Let A be the given point within the circle whose center is C, and

let BAD be the line required, so that BA is to AD in the given ratio.

Join AC and produce it to meet the circumference in E, F. Then EF

is a diameter. Draw BG, DH perpendicular on EF : then the triangles

BGA, DHA are equiangular. Hence the construction.

60. Through E one extremity of the chord EF, let a line be drawn

parallel to one diameter, and intersecting the other. Then the three

angles of the two triangles may be shewn to be respectively equal to one

another.

6 1 . Let AB be that diameter of the given circle which when produced

is perpendicular to the given line CD, and let it meet that line in C ; and

let P be the given point : it is required to find D in CD, so that DB

may be equal to the tangent DF. Make BC : CQ :: CQ : CA, and join

PQ ; bisect PQ in E, and draw ED perpendicular to PQ meeting CD in

D ; then D is the point required. Let O be the center of the circle, draw

the tangent DF ; and join OF, OD, QD, PD. Then QD may be shewn

350 GEOMETRICAL EXERCISES, &C.

to be equal to DF and to DP. When P coincides with Q, any point D

in CD fulfils the conditions of the problem ; that is, there are innume-

rable solutions.

62. It may be proved that the vertices of the two triangles which are

similar in the same segment of a circle, are in the extremities of a chord

parallel to the chord of the given segment.

63. For let the circle be described about the triangle EAC, then by

the converse to Euc, iii. 32 ; the truth of the proposition is manifest.

64. Let the figure be constructed, and the similarity of the two tri-

angles will be at once obvious from Euc iii. 32. ; Euc. i. 29.

65. In the arc AB (fig. Euc. iv. 2) let any point K be taken, and

from K let KL, KM, KN be drawn perpendicular to AB, AC, BC respec-

tively, produced if necessary, also let LM, LN be joined, then MLN may

be shewn to be a straight Ime. Draw AK, BK, CK, and by Euc. iii. 31,

22, 21 ; Euc. i. 14.

66. Let AB a chord in a circle be bisected in C, and DE, FG two

chords drawn through C; also let their extremities DG, FE be joined

intersecting CB in H, and AC in K ; then AK is equal to HB. Through

H draw MHL parallel to EF meeting FG in M, and DE produced in L.

Then by means of the equiangular triangles, HC may be proved to be

equal to CK, and hence AK is equal to HB.

67. Let A, B be the two given points, and let P be a point in the

locus so that PA, PB being joined, PA is to PB in the given ratio. Join

AB and divide it in C in the given ratio, and join PC. Then PC bisects

the angle APB. Euc. vi. 3. Again, in AB produced, take AD to AB

in the given ratio, join PD and produce AP to E, then PD bisects the

angle BPE. Euc. vi. A. Whence CPD is a right angle, and the point P

lies in the circumference of a circle whose diameter is CD.

68. Let ABC be a triangle, and let the line AD bisecting the vertical

angle A be divided in E, so that BC : BA+ AC :: AE : ED. By Euc.

VI. 3, may be deduced BC : BA + AC :: AC : AD. Whence may be

proved that CE bisects the angle ACD, and by Euc. iv. 4, that E is the

center of the inscribed circle.

69. By means of Euc. iv. 4, and Euc. vi. C. this theorem may be

shewn to be true.

70. Divide the given base BC in D, so that BD may be to DC in the

ratio of the sides. At B, D draw BB', DD' perpendicular to BC and

equal to BD, DC respectively. Join B'D' and produce it to meet BC

produced in O. With center O and radius OD, describe a circle. From

A any point in the circumference join AB, AC, AO. Prove that AB is

to AC as BD to DC. Or thus. If ABC be one of the triangles. Divide

the base BC in D so that BA is to AC as BD to DC. Produce BC and

take DO to OC as BA to AC : then O is the center of the circle.

71. Let ABC be any triangle, and from A, B let the perpendiculars

AD, BE on the opposite sides intersect in P : and let AF, BG drawn to

F, G the bisections of the opposite sides, intersect in Q. Also let FR,

GR be drawn perpendicular toBC, AC, and meet in K : then R is the

center of the circumscribed circle. Join PQ, QR ; these are in the

same line.

Join FG, and by the equiangular triangles, GRF, APB, AP is

proved double of FR. And AQ is double of QF, and the alternate

angles PAQ, QFR are equal. Hence the triangles APQ, RFQ are

equiangular.

72. Let C, C be the centers of the two circles, and let CC the line

joining the centers intersect the common tangent PP' in T. Let the

â€¢

ON BOOK VI. 351

line joining the centers cut the circles in Q, Q', and let PQ, P'Q' be

joined ; then PQ is parallel to P'Q'. Join CP, C'P', and then the angle

QPT may be proved to be equal to the alternate angle Q'P'T.

73. Let ABC be the triangle, and BC its base ; let the circles AFB,

!aPC be described intersecting the base in the point F, and their

diameters AD, AE, be drawn ; then DA : AE : : BA : AC. For join

rDB, DF, EF, EC, the triangles DAB, EAC may be proved to be similar.

; 74. If the extremities of the diameters of the two circles be joined

hy two straight lines, these lines may be proved to intersect at the

point of contact of the two circles; and the two right-angled triangles

thus formed may be shewn to be similar by Euc. iii. 34.

' 75. This follows directly from the similar triangles.

j 76. Let the figure be constructed as in Theorem 4, p. 162, the tri-

iangle EAD being right- angled at A, and let the circle inscribed in the

triangle ADE touch AD, AE, DE in the points K, L, M respectively.

Then AK is equal to AL, each being equal to the radius of the inscribed

icircle. Also AB is equal to GC, and AB is half the petimeter of the tri-

iangle AED.

Also if GA be joined, the triangle ADE is obviously equal to the

idifference of AGDE and the triangle GDE, and this diff'erence may be

proved equal to the rectangle contained by the radii of the other two

circles.

77. From the centers of the two circles let straight lines be drawn

to the extremities of the sides which are opposite to the right angles

in each triangle, and to the points where the circles touch these sides.

Euc. VI. 4.

78. Let A, B be the two given points, and C a point in the circum-

ference of the given circle. Let a circle be described through the points

A, B, C and cutting the circle in another point D. Join CD, AB, and

produce them to meet in E. Let EF be drawn touching the given

circle in F ; the circle described through the points A, B, F, will be

the circle required. Joining AD and CB, by Euc. in. 21, the tri-

angles CEB, AED are equiangular, and by Euc. vi. 4, 16, iii. 36, 37,

the given circle and the required circle each touch the line EF in the

same point, and therefore touch one another. When does this solution

fail ?

Various cases will arise according to the relative position of the two

points and the circle.

79. Let A be the given point, BC the given straight line, and D the

center of the given circle. Through D draw CD perpendicular to BC,

meeting the circumference in E, F. Join AF, and take FG to the

diameter FE, as FC is to FA. The circle described passing through the

two points A, G and touching the line BC in B is the circle required.

Let H be the center of this circle ; join HB, and BF cutting the

circumference of the given circle in K, and join EK. Then the tri-

angles FBC, FKE being equiangular, by Euc. vi. 4, 16, and the con-

struction, K is proved to be a point in the circumference of the circle

passing through the points A, G, B. And if DK, KH be joined, DKH

may be proved to be a straight line :â€” the straight line which joins the

centers of the two circles, and passes through a common point in their

circumferences.

80. Let A be the given point, B, C the centers of the two given

circles. Let a line drawn through B, C meet the circumferences of

the circles in G, F ; E, D, respectively. In GD produced, take the

point H, 80 that BH is to CH as the radius of the circle whose center

S52 GEOMETRICAL EXERCISES, &C.

is B to the radius of the circle whose center is C. Join AH, and take

KH to DH as GH to AH. Through A, K describe a circle ALK touch-

ing the circle whose center is B, in L. Then M may be proved to be a

point in the circumference of the circle whose center is C. For by join-

ing HL and producing it to meet the circumference of the circle whose

center is B in N ; and joining BN, BL, and drawing CO parallel to BL,

and CM parallel to BN, the line HN is proved to cut the circumference

of the circle whose center is B in M, O ; and CO, CM are radii. By

joining GL, DM, M may be proved to be a point in the circumference

of the circle ALK. And by producing BL, CM to meet in P, P is

proved to be the center of ALK, and BP joining the centers of the two

circles passes through L the point of contact. Hence also is shewn that

PMC passes through M, the point where the circles whose centers are P

and C touch each other.

Note. If the given point be in the circumference of one of the circles,

the construction may be more simply effected thus :

Let A be in the circumference of the circle whose center is B. Join

BA, and in AB produced, if necessary, take AD equal to the radius of the

circle whose center is C ; join DC, and at C make the angle DCE equal to

the angle CDE, the point E determined by the intersection of D A pro-

duced and CE, is the center of the circle.

81. Let AB, AC be the given lines and P the given point. Then if

O be the center of the required circle touching AB, AC, in R, S, the line

AO will bisect the given angle BAC. Let the tangent from P meet the

circle in Q, and draw OQ, OS, OP, AP. Then there are given AP and

the angle OAP. Also since OQP is a right angle, we have OP'â€” QO^

= 0P^ â€” OS* = PQ- a given magnitude. Moreover the right-angled tri-

angle AOS is given in species, or OS to OA is a given ratio. Whence

in the triangle AOP there is given, the angle AOP, the side AP,

and the excess of OP^ above the square of a line having a given

ratio to OA, to determine OA. Whence the construction is obvious.

82. Let the two given lines AB, BD meet in B, and let C be the cen-

ter of the given circle, and let the required circle touch the line AB, and

have its center in BD. Draw CFE perpendicular to HB intersecting the

circumference of the given circle in F, and produce CE, making EF

equal to the radius CF. Through G draw GK parallel to AB, and

meeting DB in K. Join CK, and through B, draw BL parallel to KC,

meeting the circumference of the circle whose center is C in L ; join

CL and produce CL to meet BD in O. Then O is the center of the

circle required. Draw OM perj^endicular to AB, and produce EC tc

meet BD in N. Then by the similar triangles, OL may be proved

equal to OM.

83. (1) In every right-angled triangle when its three sides are in

Arithmetical progression, they may be shewn to be as the numbers 5, 4,

3. On the given line AC describe a triangle having its sides AC, AD,

DC in this proportion, bisect the angles at A, C by AE, CE meeting in E,

and through E draw EF, EG parallel to AD, DC meeting in F and G.

(2) Let AC be the sum of the sides of the triangle, tig. Euc. vi. 13.

Upon AC describe a triangle ADC whose sides shall be in continued

proportion. Bisect the angles at A and C by two lines meeting in

E. From E draw EF, EG parallel to DA, DC respectively.

84. Describe a circle with any radius, and draw within it the straight

line MN cutting off a segment containing an angle equal to the given

angle, Euc. iii. 34, Divide MN in the given ratio in P, and at P draw

PA perpendicular to MN and meeting the circumference in A. Join

il

ON BOOK VI. 353

AM, AX, and on AP or AP produced, take AD equal to the given per-

pendicular, and through D draw BC parallel to MN meeting AM, AN,

or these lines produced. Then ABC shall be the triangle required.

85. Let PAQ be the given angle, bisect the angle A by AB, in

AB find D the center of the inscribed circle, and draw DC perpen-

dicular to AP. In DB take DE such that the rectangle DE, DC is

equal to the given rectangle. Describe a circle on DE as diameter

meeting AP in F, G ; and AQ in F', G'. Join FG', and AFG' will

be the triangle. Draw DH perpendicular to FG' and join G'D.

Bv Euc. VI. C, the rectangle FD, DG' is equal to the rectangle ED,

DK or CD, DE.

86. On any base BC describe a segment of a circle BAC containing

an angle equal to the given angle. From D the middle point of BC draw

DA to make the given angle ADC with the base. Produce AD to E so

that AE is equal to the given bisecting line, and through E draw FG

parallel to BC. Join AB, AC and produce them to meet FG in F and G.

87. Employ Theorem 70, p. 310, and the construction becomes

obvious.

88. Let AB be the given base, ACB the segment containing the

vertical angle ; draw the diameter AB of the circle, and divide it in E,

in the given ratio ; on AE as a diameter, describe a circle AFE ; and with

center B and a radius equal to the given line, describe a circle cutting

AFE in F. Then AF being drawn and produced to meet the circum-

scribing circle in C, and CB being joined, ABC is the triangle required.

For AF is to FC in the given ratio.

89. The line CD is not necessarily parallel to AB. Divide the base

AB in C, so that AC is to CB in the ratio of the sides of the triangle.

Then if a point E in CD can be determined such that when AE, CE,

EB, are joined, the angle AEB is bisected by CE, the problem is solved.

90. Let ABC be any triangle having the base BC. On the same

base describe an isosceles triangle DBC equal to the given triangle.

Bisect BC in E, and join DE, also upon BC describe an equilateral

triangle. On FD, FB, take EG to EH as EF to FB : also take EK

equal to EH and join GH, GK ; then GHK is an equilateral triangle

equal to the triangle ABC.

91. Let ABC be the required triangle, BC the hypotenuse, and

FHKG the inscribed square : the side HK being on BC. Then BC may

be proved to be divided in H and K, so that HK is a mean proportional

between BH and KC.

92. Let ABC be the given triangle. On BC take BD equal to one

of the given lines, through A draw AE parallel to BC. From B draw

BE to meet AE in E, and such that BE is a fourth proportional to BC,

BD, and the other given line. Join EC, produce BE to F, making BF

equal to the other given line, and join FD ; then FBD is the triangle

required.

93. By means of Euc. vi. C, the ratio of the diagonals AC to BD

may be found to be as AB . AD + BC. CD to AB.BE + AD.DC,

figure, Euc. vi. D.

94. This property follows directly from Euc. vr. C.

95. Let ABC be any triangle, and DEF the given triangle to which

the inscribed triangle is required to be similar. Draw any line de

terminated by AB, AC, and on de towards AC describe the triangle def

similar to DEF, join B/, and produce it to meet AC in F\ Through F'

draw FD' parallel to fd, F'E' parallel to /e, and join D'E', then the

triangle DEF' i.^ similar to DEF.

354 GEO lETRICAL EXERCISES, &C.

96. The square inscribed in a right-angled triangle which has one

of its sides coinciding with the hypotenuse, may be shewn to be less than

that which has two of its sides coinciding with the base and perpendicular.

97. Let BCDE be the square on the side EC of the isosceles triangle

ABC. Then by Euc. vi. 2, FG is proved parallel to ED or BC.

98. Let AB be the base of the segment ABD, fig. Euc. iii. 30.,

Bisect AB in C, take any point E in AC and make CF equal to CE:

upon EF describe a square EFGH : from C draw CG and produce it to

meet the arc of the segment in K.

99. Take two points on the radii equidistant from the center, and

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