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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will cut off similar triangles.

310 GEOMETRTCAL EXERCISES

65. If from any point, in the circumference of a circle perpen-

diculars be drawn to the sides, or sides produced, of an inscribed tri-

angle; shew that the three points of intersection will be in the same

straight line.

66. If through the middle point of any chord of a circle, two chords

be drawn, the lines joining their extremities shall intersect the first

chord at equal distances from its extremities.

67. If a straight line be divided into any two parts, to find the

locus of the point in which these parts subtend equal angles,

68. If the line bisecting the vertical angle of a triangle be divided

into parts which are to one another as the base to the sum of the sides,

the point of division is the center of the inscribed circle.

69. The rectangle contained by the sides of any triangle is to the

rectangle by the radii of the inscribed and circumscribed circles, as

twice the perimeter is to the base.

70. Shew that the locus of the vertices of all the triangles construct-

ed upon a given base, and having their sides in a given ratio, is a circle.

71. If from the extremities of the base of a triangle, perpen-

diculars be let fall on the opposite sides, and likewise straight lines

drawn to bisect the same, the intersection of the perpendiculars, that

of the bisecting lines, and the center of the circumscribing circle, will

be in the same straight line.

IX.

72. If a tangent to two circles be drawn cutting the straight line

which joins their centers, the chords are parallel which join the points

of contact, and the points where the line through the centers cuts the

circumferences.

73. If through the vertex, and the extremities of the base of a

triangle, two circles be described, intersecting one another in the base

or its continuation, their diameters are proportional to the sides of the

triangle.

74. If two circles touch each other externally and also touch a

straight line, the part of the line between the points of contact is a

mean proportional between the diameters of the circles.

75. If from the centers of each of two circles exterior to one

another, tangents be drawn to the other circles, so as to cut one another,

the rectangles of the segments are equal.

76. If a circle be inscribed in a right-angled triangle and another

be described touching the side opposite to the right angle and the

produced parts of the other sides, shew that the rectangle under the

radii is equal to the triangle, and the sum of the radii equal to the sum

of the sides which contain the right angle.

77. If a perpendicular be drawn from the right angle to the hy-

potenuse of a right-angled triangle, and circles be inscribed within the

two smaller triangles into which the given triangle is divided, their

diameters will be to each other as the sides containing the right angle.

X.

78. Describe a circle passing through two given points and touch-

ing a given circle.

ON BOOK VI. 311

79. Describe a circle which shall pass through a given point and

touch a given straight line and a given circle.

80. Through a given point di'aw a circle touching two given

circles.

81. Describe a circle to touch two given right lines and such that

a tangent drawn to it from a given point, may be equal to a given line.

82. Describe a circle which shall have its center in a given line,

and shall touch a cii'cle and a straight line given in position.

XI.

83. Given the perimeter of a right-angled triangle, it is required

to construct it, (1) If the sides are in arithmetical progression. (2) If

the sides are in geometrical progression.

84. Given the vertical angle, the perpendicular drawn from it to

tlie base, and the ratio of the segments of the base made by it, to

construct the triangle.

85. Apply (vi. c.) to construct a triangle; having given the

vertical angle, the radius of the inscribed circle, and the rectangle

contained by the straight lines drawn from the center of the circle to

the angles at the base.

86. Describe a triangle with a given vertical angle, so that the

line which bisects the base shall be equal to a given line, and the

angle which the bisecting line makes with the base shall be equal to

a given angle.

87. Given the base, the ratio of the sides containing the vertical

angle, and the distance of the vertex from a given point in the base ;

to construct the triangle.

88. Given the vertical angle and the base of a triangle, and also

a line drawn from either of the angles, cutting the opposite side in a

given ratio, to construct the triangle.

89.^ Upon the given base AÂ£ construct a triangle having its sides

in a given ratio and its vertex situated in the given indefinite line CD,

90. Describe an equilateral triangle equal to a given triangle.

91. Given the hypotenuse of a right-angled triangle, and the side

of an inscribed square. Required the two sides of the triangle.

92. To make a triangle, which shall be equal to a given triangle,

and have two of its sides equal to two given straight lines ; and shew

that if the rectangle contained by the two straight lines be less than

twice the given triangle, the problem is impossible.

XII.

93. Given the sides of a quadrilateral figure inscribed in a circle,

to find the ratio of its diagonals.

94. The diagonals AC, BD, of a trapezium inscribed in a circle,

cut each other at right angles in the point JE ;

the rectangle AB.BCi the rectangle AD. DC :: BE : ED.

XIII.

95. In any triangle, inscribe a triangle similar to a given triangle.

^ 96. Of the two squares which can be inscribed in a right-angled

triangle, which is the greater ?

97. From the vertex of an isosceles triangle two straight lines

SIZ GEOMETRICAL EXERCISES ON BOOK VI.

drawn to the opposite angles of the square described on the base, cut

the diagonals of the square in JE and F: prove that the line JEF is

parallel to the base.

98. Inscribe a square in a segment of a circle.

99. Inscribe a square in a sector of a circle, so that the angular

points shall be one on each radius, and the other two in the circum-

ference.

100. Inscribe a square in a given equilateral and equiangular

pentagon.

101. Inscribe a parallelogram in a given triangle similar to a

given parallelogram.

102. If any rectangle be inscribed in a given triangle, required the

locus of the point of intersection of its diagonals.

103. Inscribe the greatest parallelogram in a given semicircle,

104. In a given rectangle inscribe another, w^hose sides shall bear

to each other a given ratio.

105. In a given segment of a circle to inscribe a similar segment.

106. The square inscribed in a circle is to the square inscribed in

the semicircle : : 5 : 2.

107. If a square be inscribed in a right-angled triangle of which

one side coincides with the hypotenuse of the triangle, the extremities

of that side divide the base into three segments that are continued

proportionals.

108. The square inscribed in a semicircle is to the square inscribed

in a quadrant of the same circle : : 8 : 5.

109. Shew that if a triangle inscribed in a circle be isosceles,

having each of its sides double the base, the squares described upon the

radius of the circle and one of the sides of the triangle, shall be to each

other in the ratio of 4 : 15.

110. APB is a quadrant, SPT a straight line touching it at

P, PM perpendicular to CA ; prove that triangle SCT : triangle

ACB:: triangle A CB : triangle C3IP.

111. If through any point in the arc of a quadrant whose radius

is jK, two circles be drawn touching the bounding radii of the quadrant,

and r, r' be the radii of these circles : shew that rr'= H*.

112. If ^ be the radius of the circle inscribed in aright-angled

triangle ABC, right-angled at A ; and a perpendicular be let fall from

A on the hypotenuse BC, and if r, r' be the radii of the circles in-

scribed in the triangles ADB, ACD : prove that r*^ -i- r'* = -R\

XIV.

113. If in a given equilateral and equiangular hexagon another

be inscribed, to determine its ratio to the given one.

114. A regular hexagon inscribed in a circle is a mean propor-

tional betw^een an inscribed and circumscribed equilateral triangle.

115. The area of the inscribed pentagon, is to the area of the

circumscribing pentagon, as the square of the radius of the circle

inscribed within the greater pentagon, is to the square of the radius

of the circle circumscribing it.

116. The diameter of a circle is a mean proportional between the

sides of an equilateral triangle and hexagon which are described about

that circle.

GEOMETRICAL EXERCISES ON BOOK I,

HINTS, &c.

8. This is a particular case of Euc. i. 22. The triangle however ma>

be described by means of Euc. i. 1. Let AB be the given base, produce

AB both ways to meet the circles in D, E (fig. Euc. i. I.) ; with center A,

and radius AE, describe a circle, and with center B and radius BD, de-

scribe another circle cutting the former in G. Join GA, GB.

9. Apply Euc. I. 6, 8.

10. This is proved by Euc. i. 32, 13, 5.

1 1 . Let fall also a perpendicular from the vertex on the base.

12. Apply Euc. I. 4.

13. Let CAB be the triangle (fig. Euc. i. 10.) CD the line bisecting

tAe angle ACD and the base AB. Produce CD, and make DE equal to

CD, and join AE. Then CB may be proved equal to AE, also AE to AC

1 4. Let AB be the given line, and C, D the given points. Erom C

draw CE perpendicular to AB, and produce it making EF equal to CE,

join FD, and produce it to meet the given line in G, which will be the

point required.

15. Make the construction as the enunciation directs, then by Euc.

I. 4, BH is proved equal to CK : and by Euc. i. 13, 6, OB is shewn

to be equal to OC.

16. Let C, D be the two given points one on each side of the given line

AB, such that lines CE, DE drawn to any point E in tlie line are always

equal. Join CD cutting AB in F, then EC is also eqmil to FD. Then bv

Euc. I. 8.

17. Tlie angle BCD may be shewn to be equal to the sum of the

angles ABC, ADC.

18. The angles ADE, AED may be each proved to be equal to the

complements of the angles at tl.e base of the triangle.

19. The angles CAB, CBA, being equal, the angles CAD, CBE are

equal, Euc. i. 13. Then, by Euc. i. 4, CD is proved to be equal to CE.

And by Euc. i. 5, 32, the angle at the vertex is shewn to be four times

either of the angles at the base.

20. Let AB, CD be two straight lines intersecting each other in

E, and let P be the given point, within the angle AED. Draw EF

bisecting the angle AED, and through P draw PGH parallel to EF,

md cutting ED, EB in G, H. Then EG is equal to EH. And by

bisecting the angle DEB and drawing through P a line parallel to this

line, another solution is obtained. It will be found that the two lines

are at right angles to each other.

21. Let the two given straight lines meet in A, and let P be the

given point. Let PQR be the line required, meeting the lines AQ, All

in Q and R, so that PQ is equal to QB,. Through P draw PS parallel

to AR and join RS. Then APSR is a parallelogram and AS, PH the

diagonals. Hence the construction.

22. Let the two straight lines AB, AC meet in A. In AB take

any point D, and from AC cut off AE equal to AD, and join DE. On

DE, or DE produced, take DF equal to the given line, and through

F draw EG parallel to AB meeting AC in G, and through G draw Gil

parallel to DE meeting AB in H. Then GH is the line required.

314 GEOMETRICAL EXERCISES, &C.

23. The two given points may be both on the same side, or one point

may be on each side of the line. If thepoint required in the line be supposed

to be found, and lines be drawn joining this point and the given points,

an isosceles triangle is formed, and if a perpendicular be drawn on the

base Irom the point in the line : the construction is obvious.

24. The problem is simply this â€” to find a point in one side of a

triangle from which the perpendiculars drawn to the other two sides

shall be equal. If all the positions of these lines be considered, it will

readily be seen in what case the problem is impossible.

25. If the isosceles triangle be obtuse-angled, by Euc. i. 5, 32, the

truth will be made evident. If the triangle be acute- angled, the enun-

ciation of the proposition requires some modification.

26. Construct the figure and apply Euc. i. 5, 32, 15.

If the isosceles triangle have its vertical angle less than two-thirds of

a right- angle, the line ED produced, meets AB produced towards the

base, and then 3 . AEF = 4 right angles + AFE. If the vertical angle be

greater than two-thirds of a right angle, ED produced meets AB produced

towards the vertex, then 3 . AEF = 2 right angles + AFE.

27. Let ABC be an isosceles triangle, and from any point D in the

base BC, and the extremity B, let three lines DE, DF, JBG be drawn to

the sides and making equal angles with the base. Produce ED and make

DH equal to DF and join BH.

28. In the isosceles triangle ABC, let the line DFE which meets

the side AC in D and AB produced in E, be bisected by the bsise

in the point E. Then DC may be shewn to be equal to BE.

29. If two equal straight lines be drawn terminated by two lines

which meet in a point, they will cut off triangles of equal area. Hence

the two triangles have a common vertical angle and their areas and bases

equal. By Euc. i. 32 it is shewn that the angle contained by the bisecting

lines is equal to the exterior angle at the base.

30. (1) When the two lines are drawn perpendicular to the sides j

apply Euc. i. 26, 4. (2) The equal lines which bisect the sides of the

triangle may be shewn to make equal angles with the sides. (3) When the

two lines make equal angles with the sides ; apply Euc. i. 26, 4.

31. At C make the angle BCD equal to the angle ACB, and produce

AB to meet CD in D.

32. By bisecting the hypotenuse, and drawing a line from the vertex

to the point of bisection, it may be shewn that this line forms with the

shorter side and half the hypotenuse an isosceles triangle.

33. Let ABC be a triangle, having the right angle at A, and the angle

at C greater than the angle atB, also let AD be perpendicular to the base,

and AE be the line drawn to E the bisection of the base. Then AE may

be proved equal to BE or EC independently of Euc. iii. 31.

34. Produce EG, FG to meet the perpendiculars CE, BF, produced

if necessary. The demonstration is obvious.

35. If the given triangle have both of the angles at the base, acute

angles ; the difference of the angles at the base is at once obvious from

Euc. I. 32. If one of the angles at the base be obtuse, does the property

hold good r

36. Let ABC be a triangle having the angle ACB double of the angle

ABC, and let the perpendicular AD be drawn to the base BC. Take DB

equal to DC and join AE. Then AE may be proved to be equal to EB.

If ACB be an obtuse angle, then AC is equal to the sum of the seg-

ments of the base, made by the perpendicular from the vertex A.

37. Let the sides AB, AC of any triangle ABC be produced, the ex-

ON BOOK I. 315

tcrior angles bisected by two lines which meet in D, and let AD be joined,

then AD bisects the angle BAG. For draw DE perpendicular on BC,

also DF, DG perpendiculars on AB, AC produced, if necessary. TheuDF

may be proved equal to DG, and the squares on DF, DA are equal to the

squares on DG, GA, of which thesquareonFD is equal to the square onDG;

hence AF is equal to AG, and Euc. i. 8, the angle BAG is bisected by AD.

38. The line required will be found to be equal to half the sum

of the two sides of the triangle.

39. Apply Euc. i. 1, 9.

40. The angle to be trisected is one-fourth of a right angle. If an

equilateral triangle be described on one of the sides of a triangle which

contains the given angle, and a line be drawn to bisect that angle of the

equilateral triangle Avhich is at the given angle, the angle contained

between this line and the other side of the triangle will be one-twelfth

of a right angle, or equal to one-third of the given angle.

It may be remarked, generally, that any angle which is the half, fourth,

eighth, &c. part of a right angle, may be trisected by Plane Geometry.

41. Apply Euc. I. 20,

42. Let ABC, DBG be two equal triangles on the same base, of which

ABC is isosceles, fig. Euc. i. 37. By producing AB and making AG equal

to AB or AC, and joining GD, the perimeter of the triangle ABC may be

shewn to be less than the perimeter of the triangle DBG.

43. Apply Euc. i. 20.

44. For the first case, see Theo. 32, p. 76 : for the other two cases,

apply Euc. i. 19.

45. This is obvious from Euc. i. 26.

46. By Euc. i. 29, 6, FC may be shewn equal to each of the lines

EF, EG.

47. Join GA and AF, and prove GA and AF to be in the same

straight line,

48. Let the straight line drawn through D parallel to BC meet

the side AB in E, and AG in F. Then in the triangle EBD, EB is

equal to ED, by Euc. i. 29, 6. Also, in the triangle EAD, the angle

EAD may be shewn equal to the angle EDA, whence EA is equal

to ED, and therefore AlB is bisected in E. In a similar way it may

be shewn, by bisecting the angle G, that AC is bisected in F. Or

the bisection of AC in F may be proved when AB is shewn to be

bisected in E. ^

49. The triangle formed will be found to have its sides respectively

parallel to the sides of the original triangle.

50. If a line equal to the given line be drawn from the point where

the two lines meet, and parallel to the other given line ; a parallelogram

may be formed, and the construction effected.

51. Let ABC be the triangle; AD pei-pendicular to BC, AE drawn

to the bisection of BC, and AF bisecting the angle BAG. Produce AD

and make DA' equal to AD : join FA', EA'.

52. If the point in the base be supposed to be determined, and lines

Irawn from it parallel to the sides, it will be found to be in the line which

lisects the vertical angle of the triangle.

53. Let ABC be the triangle, at C draw CD perpendicular to CB and

jqual to the sum of the required lines, through D draw DE parallel to CB

neeting AC in E, and draw EF parallel to DC, meeting BC in F. Then

~F is equal to DC. Next produce CB, making GG equal to GE, and join

G cutting AB in H. From H draw HK perpendicular to EAC, and

p2

S\6 GEOMETRICAL EXERCISES, &C.

HL perpendicular to BC. Then HK and HL together are equal to DC.

The proof depends on Theorem 27, p. 75.

54. Let C' be the intersection of the circles on the other side of the

base, and ioin AC, BC. Then the angles CBA, CBA being equal, the

angles CBP, CBP are also equal, Euc.^i. 13 : next by Euc. i. 4, CP, PC

are proved equal ; lastly prove CC to be equal to CP or PC.

55. In the fig. Euc. i. 1, produce AB both ways to meet the circles

in D and E, join CD, CE, then CDE is an isosceles triangle, having each

of the angles at the base one-fourth of the angle at the vertex. At E

draw EG perpendicular to DB and meeting DC produced in G. Then

CEG is an equilateral triangle.

5G. Join CC, and shew that the angles CCF, CCG are equal to two

right angles ; also that the line FCG is equal to the diameter.

57. Construct the figure and by Euc. i. 32. If the angle BAC be

a right angle, then the angle BDC is half a right angle.

58. Let the lines which bisect the three exterior angles of the tri- â€¢

angle ABC form a new triangle A'B'C. Then each of the angles att

A', B', C may be shewn to be equal to half of the angles at A and B,

B and C, C and A respectively. And it will be found that half the

sums of every two of three unequal numbers whose sum is constant,

have less differences than the three numbers themselves.

59. The first case may be shewn by Euc. i. 4 : and the second by

Euc. I. 32, 6, 15.

60. At D any point in a line EF, draw DC perpendicular to EF and

equal to the given perpendicular on the hypotenuse. AVith centre C and

radius equal to the given base describe a circle cutting EF in B. At C

draw CA perpendicular to CB and meeting EF in A. Then ABC is the

triangle required.

61. Let ABC be the required triangle having the angle ACB a right

angle. In BC produced, take CE equal to AC, and with center B and

radius BA describe a circular arc cutting CE in D, and join AD. Then

DE is the difi"erence between the sum of the two sides AC, CB and the

hypotenuse AB ; also one side AC the perpendicular is given. Hence

the construction. On any line EB take EC equal to the given side, ED

equal to the given difference. At C, draw CA perpendicular to CB, and

equal to EC, join AD, at A in AD make the angle DAB equal to ADB,

and let AB meet EB in B. Then ABC is the triangle required.

62. (1) Let ABC be the triangle required, having ACB the right

angle. Produce AB to D making AD equal to AC or CB : then BD is

the sum of the sides. Join DC : then the angle ADC is one-fourth of a

right angle, and DBC is one-half of a right angle. Hence to construct :

at B in i3D make the angle DBM equal to half a right angle, and at D

the angle BDC equal to one-fourth of a riq;ht angle, and let DC meet BM

in C. At C draw CA at right angles to BC meeting BD in A : and ABC

is the triangle required.

(2) Let ABC be the triangle, C the right angle : from AB cut off

AD equal to AC ; then BD is the difference of the hypotenuse and one

side. Join CD; then the angles ACD, ADC are equal, and each is half

the supplement of DAC, which is half a right angle. Hence the con-

struction.

63. Take any straight line terminated at A. Make AB equal to

the difference of the sides, and AC equal to the hypotenuse. At B

make the angle CBD equal to half a right angle, and with center A

and radius AC describe a circle cutting BD in D : join AD, and draw

DE perpendicular to AC. Then ADE is the requiredtriangle.

I

ON BOOK I. oil

64. Let BC the given base be bisected in D. At D draw DE at

right angles to BC and equal to the sum of one side of the triangle

and the perpendicular from the vertex on the base : join DB, and at B

in BE make the angle EBA equal to the angle BED, and let BA meet

DE in A : join AC, and ABC is the isosceles triangle.

65 This construction may be effected by means of Prob. 4, p. 71.

66. Tlie perpendicular from the vertex on the base of an equilateral

triangle bisects the angle at the vertex, which is two-thirds of one right

angle.

67. Let ABC be the equilateral triangle of which a side is required

to be found, having given BD, CD the lines bisecting the angles atB, C.

Since the angles DBC, DCB are equal, each being one-third of a right

angle, the sides BD, DC are equal, and BDC is an isosceles triangle

having the angle at the vertex the supplement of a third of two right

angles. Hence the side BC may be found.

68. Let the given angle be taken, (1) as the included angle between

the given sides; and (2) as the opposite angle to one of the given sides.

In the latter case, an ambiguity will arise if the angle be an acute angle,

and opposite to the less of the two given sides.

69. Let ABC be the required triangle, BC the given base, CD the

given difference of the sides AB, AC : join BD, then DBC by Euc. t. 18,

can be shewn to be half the difference of the angles at the base, and AB

is equal to AD. Hence at B in the given base BC, make the angle CBD

equal to half the difference of the angles at the base. On CB take CE

equal to the difference of the sides, and Avith center C and radius CE,

describe a circle cutting BD in D : join CD and produce it to A, making

DA equal to DB. Then ABC is the triangle required.

70. On the line which is equal to the perimeter of the required tri-

angle describe a triangle having its angles equal to the given angles.

Then bisect the angles at the base ; and from the point where these lines

meet, draw lines parallel to the sides and meeting the base.

71. Let ABC be the required triangle, BC the given base, and the

j side AB greater than AC. Make AD equal to AC, and draw CD.

Then the angle BCD may be shewn to be equal to half the difference,

j and the angle DC A equal to half the sum of the angles at the base.

Hence ABC, ACB the angles at the base of the triangle are known.

72. Let the two given lines meet in A, and let B be the given point.

If BC, BD be supposed to be drawn making equal angles with AC,

and if AD and DC be joined, BCD is the triangle required, and the figure

ACBD may be shewn to be a parallelogram. Whence the construction.

73. It can be shewn that lines drawn from the angles of a triangle to

bisect the opposite sides, intersect each other at a point which is two-

thirds of their lengths from the angular points from which they are drawn.

Let ABC be the triangle required, AD, BE, CF the given lines from the

angles drawn to the bisections of the opposite sides and intersecting in G.

310 GEOMETRTCAL EXERCISES

65. If from any point, in the circumference of a circle perpen-

diculars be drawn to the sides, or sides produced, of an inscribed tri-

angle; shew that the three points of intersection will be in the same

straight line.

66. If through the middle point of any chord of a circle, two chords

be drawn, the lines joining their extremities shall intersect the first

chord at equal distances from its extremities.

67. If a straight line be divided into any two parts, to find the

locus of the point in which these parts subtend equal angles,

68. If the line bisecting the vertical angle of a triangle be divided

into parts which are to one another as the base to the sum of the sides,

the point of division is the center of the inscribed circle.

69. The rectangle contained by the sides of any triangle is to the

rectangle by the radii of the inscribed and circumscribed circles, as

twice the perimeter is to the base.

70. Shew that the locus of the vertices of all the triangles construct-

ed upon a given base, and having their sides in a given ratio, is a circle.

71. If from the extremities of the base of a triangle, perpen-

diculars be let fall on the opposite sides, and likewise straight lines

drawn to bisect the same, the intersection of the perpendiculars, that

of the bisecting lines, and the center of the circumscribing circle, will

be in the same straight line.

IX.

72. If a tangent to two circles be drawn cutting the straight line

which joins their centers, the chords are parallel which join the points

of contact, and the points where the line through the centers cuts the

circumferences.

73. If through the vertex, and the extremities of the base of a

triangle, two circles be described, intersecting one another in the base

or its continuation, their diameters are proportional to the sides of the

triangle.

74. If two circles touch each other externally and also touch a

straight line, the part of the line between the points of contact is a

mean proportional between the diameters of the circles.

75. If from the centers of each of two circles exterior to one

another, tangents be drawn to the other circles, so as to cut one another,

the rectangles of the segments are equal.

76. If a circle be inscribed in a right-angled triangle and another

be described touching the side opposite to the right angle and the

produced parts of the other sides, shew that the rectangle under the

radii is equal to the triangle, and the sum of the radii equal to the sum

of the sides which contain the right angle.

77. If a perpendicular be drawn from the right angle to the hy-

potenuse of a right-angled triangle, and circles be inscribed within the

two smaller triangles into which the given triangle is divided, their

diameters will be to each other as the sides containing the right angle.

X.

78. Describe a circle passing through two given points and touch-

ing a given circle.

ON BOOK VI. 311

79. Describe a circle which shall pass through a given point and

touch a given straight line and a given circle.

80. Through a given point di'aw a circle touching two given

circles.

81. Describe a circle to touch two given right lines and such that

a tangent drawn to it from a given point, may be equal to a given line.

82. Describe a circle which shall have its center in a given line,

and shall touch a cii'cle and a straight line given in position.

XI.

83. Given the perimeter of a right-angled triangle, it is required

to construct it, (1) If the sides are in arithmetical progression. (2) If

the sides are in geometrical progression.

84. Given the vertical angle, the perpendicular drawn from it to

tlie base, and the ratio of the segments of the base made by it, to

construct the triangle.

85. Apply (vi. c.) to construct a triangle; having given the

vertical angle, the radius of the inscribed circle, and the rectangle

contained by the straight lines drawn from the center of the circle to

the angles at the base.

86. Describe a triangle with a given vertical angle, so that the

line which bisects the base shall be equal to a given line, and the

angle which the bisecting line makes with the base shall be equal to

a given angle.

87. Given the base, the ratio of the sides containing the vertical

angle, and the distance of the vertex from a given point in the base ;

to construct the triangle.

88. Given the vertical angle and the base of a triangle, and also

a line drawn from either of the angles, cutting the opposite side in a

given ratio, to construct the triangle.

89.^ Upon the given base AÂ£ construct a triangle having its sides

in a given ratio and its vertex situated in the given indefinite line CD,

90. Describe an equilateral triangle equal to a given triangle.

91. Given the hypotenuse of a right-angled triangle, and the side

of an inscribed square. Required the two sides of the triangle.

92. To make a triangle, which shall be equal to a given triangle,

and have two of its sides equal to two given straight lines ; and shew

that if the rectangle contained by the two straight lines be less than

twice the given triangle, the problem is impossible.

XII.

93. Given the sides of a quadrilateral figure inscribed in a circle,

to find the ratio of its diagonals.

94. The diagonals AC, BD, of a trapezium inscribed in a circle,

cut each other at right angles in the point JE ;

the rectangle AB.BCi the rectangle AD. DC :: BE : ED.

XIII.

95. In any triangle, inscribe a triangle similar to a given triangle.

^ 96. Of the two squares which can be inscribed in a right-angled

triangle, which is the greater ?

97. From the vertex of an isosceles triangle two straight lines

SIZ GEOMETRICAL EXERCISES ON BOOK VI.

drawn to the opposite angles of the square described on the base, cut

the diagonals of the square in JE and F: prove that the line JEF is

parallel to the base.

98. Inscribe a square in a segment of a circle.

99. Inscribe a square in a sector of a circle, so that the angular

points shall be one on each radius, and the other two in the circum-

ference.

100. Inscribe a square in a given equilateral and equiangular

pentagon.

101. Inscribe a parallelogram in a given triangle similar to a

given parallelogram.

102. If any rectangle be inscribed in a given triangle, required the

locus of the point of intersection of its diagonals.

103. Inscribe the greatest parallelogram in a given semicircle,

104. In a given rectangle inscribe another, w^hose sides shall bear

to each other a given ratio.

105. In a given segment of a circle to inscribe a similar segment.

106. The square inscribed in a circle is to the square inscribed in

the semicircle : : 5 : 2.

107. If a square be inscribed in a right-angled triangle of which

one side coincides with the hypotenuse of the triangle, the extremities

of that side divide the base into three segments that are continued

proportionals.

108. The square inscribed in a semicircle is to the square inscribed

in a quadrant of the same circle : : 8 : 5.

109. Shew that if a triangle inscribed in a circle be isosceles,

having each of its sides double the base, the squares described upon the

radius of the circle and one of the sides of the triangle, shall be to each

other in the ratio of 4 : 15.

110. APB is a quadrant, SPT a straight line touching it at

P, PM perpendicular to CA ; prove that triangle SCT : triangle

ACB:: triangle A CB : triangle C3IP.

111. If through any point in the arc of a quadrant whose radius

is jK, two circles be drawn touching the bounding radii of the quadrant,

and r, r' be the radii of these circles : shew that rr'= H*.

112. If ^ be the radius of the circle inscribed in aright-angled

triangle ABC, right-angled at A ; and a perpendicular be let fall from

A on the hypotenuse BC, and if r, r' be the radii of the circles in-

scribed in the triangles ADB, ACD : prove that r*^ -i- r'* = -R\

XIV.

113. If in a given equilateral and equiangular hexagon another

be inscribed, to determine its ratio to the given one.

114. A regular hexagon inscribed in a circle is a mean propor-

tional betw^een an inscribed and circumscribed equilateral triangle.

115. The area of the inscribed pentagon, is to the area of the

circumscribing pentagon, as the square of the radius of the circle

inscribed within the greater pentagon, is to the square of the radius

of the circle circumscribing it.

116. The diameter of a circle is a mean proportional between the

sides of an equilateral triangle and hexagon which are described about

that circle.

GEOMETRICAL EXERCISES ON BOOK I,

HINTS, &c.

8. This is a particular case of Euc. i. 22. The triangle however ma>

be described by means of Euc. i. 1. Let AB be the given base, produce

AB both ways to meet the circles in D, E (fig. Euc. i. I.) ; with center A,

and radius AE, describe a circle, and with center B and radius BD, de-

scribe another circle cutting the former in G. Join GA, GB.

9. Apply Euc. I. 6, 8.

10. This is proved by Euc. i. 32, 13, 5.

1 1 . Let fall also a perpendicular from the vertex on the base.

12. Apply Euc. I. 4.

13. Let CAB be the triangle (fig. Euc. i. 10.) CD the line bisecting

tAe angle ACD and the base AB. Produce CD, and make DE equal to

CD, and join AE. Then CB may be proved equal to AE, also AE to AC

1 4. Let AB be the given line, and C, D the given points. Erom C

draw CE perpendicular to AB, and produce it making EF equal to CE,

join FD, and produce it to meet the given line in G, which will be the

point required.

15. Make the construction as the enunciation directs, then by Euc.

I. 4, BH is proved equal to CK : and by Euc. i. 13, 6, OB is shewn

to be equal to OC.

16. Let C, D be the two given points one on each side of the given line

AB, such that lines CE, DE drawn to any point E in tlie line are always

equal. Join CD cutting AB in F, then EC is also eqmil to FD. Then bv

Euc. I. 8.

17. Tlie angle BCD may be shewn to be equal to the sum of the

angles ABC, ADC.

18. The angles ADE, AED may be each proved to be equal to the

complements of the angles at tl.e base of the triangle.

19. The angles CAB, CBA, being equal, the angles CAD, CBE are

equal, Euc. i. 13. Then, by Euc. i. 4, CD is proved to be equal to CE.

And by Euc. i. 5, 32, the angle at the vertex is shewn to be four times

either of the angles at the base.

20. Let AB, CD be two straight lines intersecting each other in

E, and let P be the given point, within the angle AED. Draw EF

bisecting the angle AED, and through P draw PGH parallel to EF,

md cutting ED, EB in G, H. Then EG is equal to EH. And by

bisecting the angle DEB and drawing through P a line parallel to this

line, another solution is obtained. It will be found that the two lines

are at right angles to each other.

21. Let the two given straight lines meet in A, and let P be the

given point. Let PQR be the line required, meeting the lines AQ, All

in Q and R, so that PQ is equal to QB,. Through P draw PS parallel

to AR and join RS. Then APSR is a parallelogram and AS, PH the

diagonals. Hence the construction.

22. Let the two straight lines AB, AC meet in A. In AB take

any point D, and from AC cut off AE equal to AD, and join DE. On

DE, or DE produced, take DF equal to the given line, and through

F draw EG parallel to AB meeting AC in G, and through G draw Gil

parallel to DE meeting AB in H. Then GH is the line required.

314 GEOMETRICAL EXERCISES, &C.

23. The two given points may be both on the same side, or one point

may be on each side of the line. If thepoint required in the line be supposed

to be found, and lines be drawn joining this point and the given points,

an isosceles triangle is formed, and if a perpendicular be drawn on the

base Irom the point in the line : the construction is obvious.

24. The problem is simply this â€” to find a point in one side of a

triangle from which the perpendiculars drawn to the other two sides

shall be equal. If all the positions of these lines be considered, it will

readily be seen in what case the problem is impossible.

25. If the isosceles triangle be obtuse-angled, by Euc. i. 5, 32, the

truth will be made evident. If the triangle be acute- angled, the enun-

ciation of the proposition requires some modification.

26. Construct the figure and apply Euc. i. 5, 32, 15.

If the isosceles triangle have its vertical angle less than two-thirds of

a right- angle, the line ED produced, meets AB produced towards the

base, and then 3 . AEF = 4 right angles + AFE. If the vertical angle be

greater than two-thirds of a right angle, ED produced meets AB produced

towards the vertex, then 3 . AEF = 2 right angles + AFE.

27. Let ABC be an isosceles triangle, and from any point D in the

base BC, and the extremity B, let three lines DE, DF, JBG be drawn to

the sides and making equal angles with the base. Produce ED and make

DH equal to DF and join BH.

28. In the isosceles triangle ABC, let the line DFE which meets

the side AC in D and AB produced in E, be bisected by the bsise

in the point E. Then DC may be shewn to be equal to BE.

29. If two equal straight lines be drawn terminated by two lines

which meet in a point, they will cut off triangles of equal area. Hence

the two triangles have a common vertical angle and their areas and bases

equal. By Euc. i. 32 it is shewn that the angle contained by the bisecting

lines is equal to the exterior angle at the base.

30. (1) When the two lines are drawn perpendicular to the sides j

apply Euc. i. 26, 4. (2) The equal lines which bisect the sides of the

triangle may be shewn to make equal angles with the sides. (3) When the

two lines make equal angles with the sides ; apply Euc. i. 26, 4.

31. At C make the angle BCD equal to the angle ACB, and produce

AB to meet CD in D.

32. By bisecting the hypotenuse, and drawing a line from the vertex

to the point of bisection, it may be shewn that this line forms with the

shorter side and half the hypotenuse an isosceles triangle.

33. Let ABC be a triangle, having the right angle at A, and the angle

at C greater than the angle atB, also let AD be perpendicular to the base,

and AE be the line drawn to E the bisection of the base. Then AE may

be proved equal to BE or EC independently of Euc. iii. 31.

34. Produce EG, FG to meet the perpendiculars CE, BF, produced

if necessary. The demonstration is obvious.

35. If the given triangle have both of the angles at the base, acute

angles ; the difference of the angles at the base is at once obvious from

Euc. I. 32. If one of the angles at the base be obtuse, does the property

hold good r

36. Let ABC be a triangle having the angle ACB double of the angle

ABC, and let the perpendicular AD be drawn to the base BC. Take DB

equal to DC and join AE. Then AE may be proved to be equal to EB.

If ACB be an obtuse angle, then AC is equal to the sum of the seg-

ments of the base, made by the perpendicular from the vertex A.

37. Let the sides AB, AC of any triangle ABC be produced, the ex-

ON BOOK I. 315

tcrior angles bisected by two lines which meet in D, and let AD be joined,

then AD bisects the angle BAG. For draw DE perpendicular on BC,

also DF, DG perpendiculars on AB, AC produced, if necessary. TheuDF

may be proved equal to DG, and the squares on DF, DA are equal to the

squares on DG, GA, of which thesquareonFD is equal to the square onDG;

hence AF is equal to AG, and Euc. i. 8, the angle BAG is bisected by AD.

38. The line required will be found to be equal to half the sum

of the two sides of the triangle.

39. Apply Euc. i. 1, 9.

40. The angle to be trisected is one-fourth of a right angle. If an

equilateral triangle be described on one of the sides of a triangle which

contains the given angle, and a line be drawn to bisect that angle of the

equilateral triangle Avhich is at the given angle, the angle contained

between this line and the other side of the triangle will be one-twelfth

of a right angle, or equal to one-third of the given angle.

It may be remarked, generally, that any angle which is the half, fourth,

eighth, &c. part of a right angle, may be trisected by Plane Geometry.

41. Apply Euc. I. 20,

42. Let ABC, DBG be two equal triangles on the same base, of which

ABC is isosceles, fig. Euc. i. 37. By producing AB and making AG equal

to AB or AC, and joining GD, the perimeter of the triangle ABC may be

shewn to be less than the perimeter of the triangle DBG.

43. Apply Euc. i. 20.

44. For the first case, see Theo. 32, p. 76 : for the other two cases,

apply Euc. i. 19.

45. This is obvious from Euc. i. 26.

46. By Euc. i. 29, 6, FC may be shewn equal to each of the lines

EF, EG.

47. Join GA and AF, and prove GA and AF to be in the same

straight line,

48. Let the straight line drawn through D parallel to BC meet

the side AB in E, and AG in F. Then in the triangle EBD, EB is

equal to ED, by Euc. i. 29, 6. Also, in the triangle EAD, the angle

EAD may be shewn equal to the angle EDA, whence EA is equal

to ED, and therefore AlB is bisected in E. In a similar way it may

be shewn, by bisecting the angle G, that AC is bisected in F. Or

the bisection of AC in F may be proved when AB is shewn to be

bisected in E. ^

49. The triangle formed will be found to have its sides respectively

parallel to the sides of the original triangle.

50. If a line equal to the given line be drawn from the point where

the two lines meet, and parallel to the other given line ; a parallelogram

may be formed, and the construction effected.

51. Let ABC be the triangle; AD pei-pendicular to BC, AE drawn

to the bisection of BC, and AF bisecting the angle BAG. Produce AD

and make DA' equal to AD : join FA', EA'.

52. If the point in the base be supposed to be determined, and lines

Irawn from it parallel to the sides, it will be found to be in the line which

lisects the vertical angle of the triangle.

53. Let ABC be the triangle, at C draw CD perpendicular to CB and

jqual to the sum of the required lines, through D draw DE parallel to CB

neeting AC in E, and draw EF parallel to DC, meeting BC in F. Then

~F is equal to DC. Next produce CB, making GG equal to GE, and join

G cutting AB in H. From H draw HK perpendicular to EAC, and

p2

S\6 GEOMETRICAL EXERCISES, &C.

HL perpendicular to BC. Then HK and HL together are equal to DC.

The proof depends on Theorem 27, p. 75.

54. Let C' be the intersection of the circles on the other side of the

base, and ioin AC, BC. Then the angles CBA, CBA being equal, the

angles CBP, CBP are also equal, Euc.^i. 13 : next by Euc. i. 4, CP, PC

are proved equal ; lastly prove CC to be equal to CP or PC.

55. In the fig. Euc. i. 1, produce AB both ways to meet the circles

in D and E, join CD, CE, then CDE is an isosceles triangle, having each

of the angles at the base one-fourth of the angle at the vertex. At E

draw EG perpendicular to DB and meeting DC produced in G. Then

CEG is an equilateral triangle.

5G. Join CC, and shew that the angles CCF, CCG are equal to two

right angles ; also that the line FCG is equal to the diameter.

57. Construct the figure and by Euc. i. 32. If the angle BAC be

a right angle, then the angle BDC is half a right angle.

58. Let the lines which bisect the three exterior angles of the tri- â€¢

angle ABC form a new triangle A'B'C. Then each of the angles att

A', B', C may be shewn to be equal to half of the angles at A and B,

B and C, C and A respectively. And it will be found that half the

sums of every two of three unequal numbers whose sum is constant,

have less differences than the three numbers themselves.

59. The first case may be shewn by Euc. i. 4 : and the second by

Euc. I. 32, 6, 15.

60. At D any point in a line EF, draw DC perpendicular to EF and

equal to the given perpendicular on the hypotenuse. AVith centre C and

radius equal to the given base describe a circle cutting EF in B. At C

draw CA perpendicular to CB and meeting EF in A. Then ABC is the

triangle required.

61. Let ABC be the required triangle having the angle ACB a right

angle. In BC produced, take CE equal to AC, and with center B and

radius BA describe a circular arc cutting CE in D, and join AD. Then

DE is the difi"erence between the sum of the two sides AC, CB and the

hypotenuse AB ; also one side AC the perpendicular is given. Hence

the construction. On any line EB take EC equal to the given side, ED

equal to the given difference. At C, draw CA perpendicular to CB, and

equal to EC, join AD, at A in AD make the angle DAB equal to ADB,

and let AB meet EB in B. Then ABC is the triangle required.

62. (1) Let ABC be the triangle required, having ACB the right

angle. Produce AB to D making AD equal to AC or CB : then BD is

the sum of the sides. Join DC : then the angle ADC is one-fourth of a

right angle, and DBC is one-half of a right angle. Hence to construct :

at B in i3D make the angle DBM equal to half a right angle, and at D

the angle BDC equal to one-fourth of a riq;ht angle, and let DC meet BM

in C. At C draw CA at right angles to BC meeting BD in A : and ABC

is the triangle required.

(2) Let ABC be the triangle, C the right angle : from AB cut off

AD equal to AC ; then BD is the difference of the hypotenuse and one

side. Join CD; then the angles ACD, ADC are equal, and each is half

the supplement of DAC, which is half a right angle. Hence the con-

struction.

63. Take any straight line terminated at A. Make AB equal to

the difference of the sides, and AC equal to the hypotenuse. At B

make the angle CBD equal to half a right angle, and with center A

and radius AC describe a circle cutting BD in D : join AD, and draw

DE perpendicular to AC. Then ADE is the requiredtriangle.

I

ON BOOK I. oil

64. Let BC the given base be bisected in D. At D draw DE at

right angles to BC and equal to the sum of one side of the triangle

and the perpendicular from the vertex on the base : join DB, and at B

in BE make the angle EBA equal to the angle BED, and let BA meet

DE in A : join AC, and ABC is the isosceles triangle.

65 This construction may be effected by means of Prob. 4, p. 71.

66. Tlie perpendicular from the vertex on the base of an equilateral

triangle bisects the angle at the vertex, which is two-thirds of one right

angle.

67. Let ABC be the equilateral triangle of which a side is required

to be found, having given BD, CD the lines bisecting the angles atB, C.

Since the angles DBC, DCB are equal, each being one-third of a right

angle, the sides BD, DC are equal, and BDC is an isosceles triangle

having the angle at the vertex the supplement of a third of two right

angles. Hence the side BC may be found.

68. Let the given angle be taken, (1) as the included angle between

the given sides; and (2) as the opposite angle to one of the given sides.

In the latter case, an ambiguity will arise if the angle be an acute angle,

and opposite to the less of the two given sides.

69. Let ABC be the required triangle, BC the given base, CD the

given difference of the sides AB, AC : join BD, then DBC by Euc. t. 18,

can be shewn to be half the difference of the angles at the base, and AB

is equal to AD. Hence at B in the given base BC, make the angle CBD

equal to half the difference of the angles at the base. On CB take CE

equal to the difference of the sides, and Avith center C and radius CE,

describe a circle cutting BD in D : join CD and produce it to A, making

DA equal to DB. Then ABC is the triangle required.

70. On the line which is equal to the perimeter of the required tri-

angle describe a triangle having its angles equal to the given angles.

Then bisect the angles at the base ; and from the point where these lines

meet, draw lines parallel to the sides and meeting the base.

71. Let ABC be the required triangle, BC the given base, and the

j side AB greater than AC. Make AD equal to AC, and draw CD.

Then the angle BCD may be shewn to be equal to half the difference,

j and the angle DC A equal to half the sum of the angles at the base.

Hence ABC, ACB the angles at the base of the triangle are known.

72. Let the two given lines meet in A, and let B be the given point.

If BC, BD be supposed to be drawn making equal angles with AC,

and if AD and DC be joined, BCD is the triangle required, and the figure

ACBD may be shewn to be a parallelogram. Whence the construction.

73. It can be shewn that lines drawn from the angles of a triangle to

bisect the opposite sides, intersect each other at a point which is two-

thirds of their lengths from the angular points from which they are drawn.

Let ABC be the triangle required, AD, BE, CF the given lines from the

angles drawn to the bisections of the opposite sides and intersecting in G.

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