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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the perpendicular.

29. Since the base and area are given, the altitude of the triangle is

known. Hence the problem is reduced to ; â€” Given the base and altitude

of a triangle, and the line drawn from the vertex to the bisection of the

base, construct the triangle.

30. This follows immediately from Euc. i. 47.

31. Apply Euc. II. 13.

32. The truth of this property depends on the fact that the rectangle

contained by AC, CB is equal to that contained by AB, CD.

33. Let P the required point in the base AB be supposed to be known.

Join CP. It may then be shewn that the property stated in the Prob-

lem is contained in Theorem 3. p. 114.

34. This may be shewn from Euc. i. 47 ; ii- 5. Cor.

35. Erom C let fall CF pd-pendicular on AB. Then ACE is an ob-

tuse-angled, and BEC an acute-angled triangle. Apply Euc. ii. 12, 13 ,

and by Euc. i. 47, the squares on AC and CB are equal to the square

on AB.

36. Apply Euc. i. 47, ii. 4 ; and the note p. 102, on Euc. ii. 4.

37. Draw a perpendicular from the vertex to the base, and apply

Euc. I. 47 ; II. 5, Cor. Enunciate and prove the proposition when the

straight line drawn from the vertex meets the base produced.

38. This follows directly from Euc. ii. 13, Case 1.

39. The truth of this proposition may be shewn from Euc. i. 47 ; ii. 4.

40. Let the square on the base of the isosceles triangle be described.

Draw the diagonals of the square, and the proof is obvious.

41. Let ABC be the triangle required, such that the square on AB

is three times the square on AC or BC. Produce BC and draw AD per-

pendicular to BC. Then by Euc. ii. 12, CD may be shewn to be equal

to one half of BC. Hence the construction.

42. Apply Euc. ii. 12, and Theorem 38, p. 118.

43. Draw EF parallel to AB and meeting the base in E ; draw also

EG perpendicular to the base. Then by Euc. i. 47 ; n. 5, Cor.

44. Bisect the angle B by BD meeting the opposite side in D, and

draw BE perpendicular to AC. Then by Euc. i. 47 ; ii. 5, Cor.

45. This follows directly from Theorem 3, p. 114.

46. Draw the diagonals intersecting each other in P, and join OP.

By Theo. 3, p. 114.

47. Draw from any two opposite angles, straight lines to meet in the

bisection of the diagonal joining the other angles. Then by Euc. ii. 12, 13.

48. Draw two lines from the point of bisection of either of the bi-

sected sides to the extremities of the opposite side ; and three triangles

will be formed, two on one of the bisected sides and one on the other, in

S26 GEOMETRICAL EXERCISES, &C.

each of which is a line drawn from the vertex to the bisection of the base.

Then by Theo. 3, p, 114.

49. If the extremities of the two lines which bisect the opposite sides

of the trapezium be joined, the figure formed is a parallelogram which

has its sides respectively parallel to, and equal to, half the diagonals of

the trapezium. The sum of the squares on the two diagonals of the tra-

pezium may be easily shewn to be equal to the sum of the squares on

the four sides of the parallelogram.

50. Draw perpendiculars from the extremities of one of the parallel

sides, meeting the other side produced, if necessary. Then from the four

right-angled triangles thus formed, may be shewn the truth of the pro-

position.

51. Let AD be parallel to BC in the figure ABCD. Draw the diagonal

AC, then the sum of the triangles ABC, ADC may be shewn to be equal to

the rectangle contained by the altitude and half the sum of AD and BC.

52. Let ABCD be the trapezium, having the sides AB, CD, parallel,

and AD, BC equal. Join AC and draw AE perpendicular to DC. Then

by Euc. II. 13.

53. Let ABC be any triangle ; AHKB, AGFC, BDEC, the squares

upon their sides ; EF, GH, KL the lines joining the angles of the squares.

Produce GA, KB, EC, and draw HN, DQ, FR perpendiculars upon them

respectively: also draw AP, BM, CS perpendiculars on the sides of the

triangle. Then AN may be proved to be equal to AM ; CR to CP ; and

BQ to BS ; and by Euc. ii. 12, 13.

54. Convert the triangle into a rectangle, then Euc. ii. 14.

55. Find a rectangle equal to the two figures, and apply Euc. ii. 14.

56. Find the side of a square which shall be equal to the given

rectangle. See Prob. i. p. 113.

57. On any line PQ take AB equal to the given difference of the

sides of the rectangle, at A draw AC at right angles to AB, and equal to

the side of the given square ; bisect AB in O and join OC ; with center

O and radius OC describe a semicircle meeting PQ in D and E. Then

the lines AD, AE have AB for their difierence, and the rectangle con~

tained by them is equal to the square on AC,

58. Apply Euc. ii. 14.

GEOMETRICAL EXERCISES ON BOOK III.

HINTS, &c.

7. Euc. III. 3, suggests the construction.

8. The given point may be either within or without the circle. Find

the center of the circle, and join the given point and the center, and upon

this line describe a semicircle, a line equal to the given distance may be

drawn from the given point to meet the arc of the semicircle. When

the point is without the circle, the given distance may meet the diameter

produced.

9. This may be easily shewn to be a straight line passing through

the center of the circle.

10. The two chords form by their intersections the sides of two isos-

celes triangles, of which the parallel chords in the circle are the bases.

ON BOOK III. S21

n. The angles in equal segments are equal, and by Euc. i. 29. If

the chords are equally distant from the center, the lines intersect the

diameter in the center of the circle.

12. Construct the figure and the arc BC may be proved equal to the

arc B'C

13. The point determined by the lines drawn from the bisections of

the chords and at right angles to them respectively, will be the center of

the required circle.

14. Construct the figures : the proof offers no difficulty.

15. From the centre C ot the circle, draw CA, CB at right angles to

each other meeting the circumference ; join AB, and draw CD perpendicular

to AB.

16. Join the extremities of the chords, then Euc. i. 27 ; iii. 28.

17. Take the center O, and join AP, AO, &c. and apply Euc. i. 20.

18. Draw any straight line intersecting two parallel chords and meet-

ing the circumference.

19. Produce the radii to meet the circumference.

20. Join AD, and the first equality follows directly from Euc. in.

20, T. 32. Also by joining AC, the second equality may be proved in a

similar way. If however the line AD do not fall on the same side of the

center O as E, it will be found that the difference^ not the sum of the two

angles, is equal to 2 . AED. See note to Euc. in. 20, p. 155.

21. Let DKE, DBO (fig. Euc. in. 8) be two lines equally inclined

to DA, then KE may be proved to be equal to BO, and the segments cut

off" by equal straight lines in the same circle, as well as in equal circles,

are equal to one another.

22. Apply Euc. i. 15, and in, 21.

23. This is the same as Euc. in. 34, with the condition, that the line

must pass through a given point.

24. Let the segments AHB, AKC be externally described on the

given lines AB, AC, to contain angles equal to BAC. Then by the con-

verse to Euc. ITT. 32, AB touches the circle AKC, and AC the circle AHB.

25. Let ABC be a triangle of which the base or longest side is BC,

and let a segment of a circle be described on BC. Produce BA, CA to

meet the arc of the segment in D, E, and join BD, CE. If circles be de-

scribed about the triangles ABD, ACE, the sides AB, AC shall cut off

segments similar to the segment described upon the base BC.

26. This is obvious from the note to Euc. in. 26, p. 156.

27. The segment must be described on the opposite side of the pro-

duced chord. By converse of Euc. in. 32.

28. If a circle be described upon the side AC as a diameter, the cir-

cumference will pass through the points D, E. Then Euc. in. 21.

29. Let AB, AC be the bounding radii, and D any point in the arc

BC, and DE, DF, perpendiculars from D on AB, AC. The circle de-

scribed on AD will always be of the same magnitude, and the angle EAP

in it, is constant : â€” whence the arc EDF is constant, and therefore its

chord EF.

30. Construct the figure, and let the circle with center O, described

on AH as a diameter, intersect the given circle in P, Q, join OP, PE, and

prove EP at right angles to OP.

31. If the tangent be required to be perpendicular to a given line:

draw the diameter parallel to this line, and the tangent drawn at the ex-

tremity of this diameter will be perpendicular to the given line.

32. The straight line which joins the center and passes through the

intersection of two tangents to a circle, bisects the angle f^ntained by

the tangents.

328 GEOMETRICAL EXERCISES, &C.

33. Draw two radii containing an angle equal to the supplement of

the given angle ; the tangents drawn at the extremities of these radii will

contain the given angle,

34. Since the circle is to touch two parallel lines drawn from two

given points in a third line, the radius of the circle is determined by the

distance between the two given points.

35. It is sufficient to suggest that the angle between a chord and a

tangent is equal to the angle in the alternate segment of the circle. Euc.

m. 32.

36. Let AB be the given chord of the circle whose center is O. Draw

DE touching the circle at any point E and equal to the given line ; join

DO, and with center O and radius DO describe a circle: produce the

chord AB to meet the circumference of this circle in F : then F is the

point required.

37. Let D be the point required in the diameter BA produced, such

that the tangent DP is half of DB. Join CP, C being the ct-nter. Then

CPD is a right-angled triangle, having the sum of the base PC and hypo-

tenuse CD double of the perpendicular PD.

38. If BE intersect DF in K (fig. Euc. iii. 37). Join FB, FE, then

by means of the triangles, BE is shewn to be bisected in K at right angles.

39. Let AB, CD be any two diameters of a circle, O the center, and

let the tangents at their extremities form the quadrilateral figure EFGII.

Join EO, OF, then EO and OF may be proved to be in the same straight

line, and similarly HO, OK.

Note. â€” This Proposition is equally true if AB, CD be any two chords

v.'hatever. It then becomes equivalent to the following proposition: â€”

The diagonals of the circumscribed and inscribed quadrilaterals, intersect

in the same point, the points of contact of the former being the angles of

the latter figure.

40. Let C be the point without the circle from which the tangents

CA, CB are drawn, and let DE be any diameter, also let AE, BD be

joined, intersecting in P, then if CP be joined and produced to meet DE

in G : CG is perpendicular to DE. Join DA, EB, and produce them to

meet in F.

Then the angles DAE, EBD being angles in a semicircle, are right

ambles ; or DB, EA are drawn perpendicular to the sides of the triangle

DEF : whence the line drawn from F through P is perpendicular to the

third side DE.

41. Let the chord AB, of which P is its middle point, be produced

both ways to C, D, so that AC is equal to BD. From C, D, draw the

tangents to the circle forming the tangential quadrilateral CKDR, the

points of contact of the fides, being E, II, F, G. Let O be the center of

the circle. Join EH, GF, CO, GO, FO, DO. Then EH and GF may

be proved each parallel to CD, they are therefore parallel to one another.

"Whence is proved that both EF and DG bisect AB.

42. This is obvious from Euc. i. 29, and the note to iii. 22. p. 156.

43. From any point A in the circumference, let any chord AB and

tangent AC be drawn. Bisect the arc AB in D, and from D draw DE,

DC perpendiculars on the chord AB and tangent AC. Join AD, the

triangles ADE, ADC may be shewn to be equal.

44. Let A, B, be the given points. Join AB, and upon it describe a

segment of a circle which shall contain an angle equal to the given angle.

If the circle cut the given line, there will be two points; if it only touch

the line, there will be one ; and if it neither cut nor touch the line, the

problem is impossible.

I

JN BOOK III. 329

45. It may be shewn that the point required is determined by a per-

pendicular drawn from the center of the circle on the <;ivcn line.

46. Let two lines AP, BP be drawn from the given points A, B,

making equal angles with the tangent to the circle at the point of contact

P, take any other point Q in the convex circumference, and join QA,

QB : then by Prob. 4, p. 71, and Euc. i. 21.

47. Let C be the center of the circle, and E the point of contact of

DF with the circle. Join DC, CE, CF.

48. liCt the tangents at E, F meet in a point K. Produce HE, RF

to meet the diameter AB produced in S, T. Then RST is a triangle,

and the quadrilateral RFOE maybe circumscribed by a circle, and RPO

may be proved to be one of the diagonals,

49. Let C be the middle point of the chord of contact : produce AC,

BC to meet the circumference in B', A', and join AA', BB'.

50. Let A be the given point, and B the given point in the given line

CD. At B draw BE at right angles to CD, join AB and bisect it in F,

and from F draw FE perpendicular to AB and meeting BE in E. E is

the center of the required circle.

51. Let O be the center of the given circle. Draw OA perpendicular

to the given straight line ; at O in OA make the angle AOP equal to the

given angle, produce PO to meet the circumference again in Q. Then P,

Q are two points from which tangents may be drawn fulfilling the re-

quired condition.

52. Let C be the center of the given circle, B the given point in the

circumference, and A the other given point through which the required

circle is to be made to pass. Join CB, the center of the circle is a point

in CB produced. The center itself may be found in three ways.

53. Euc. III. 11 suggests the construction.

54. Let AB, AC be the two given lines which meet at A, and let D

be the given point. Bisect the angle BAC by AE, the center of the circle

is in AE. Through D draw DF perpendicular to AE, and produce DF

to G, making FG equal to FD. Then DG is a chord of the circle, and

the circle which passes through D and touches AB, will also pass through

G and touch AC. ^

55. As the center is given, the line joining this point and center of

the given circle, is perpendicular to that diameter, through the extremi-

ties of which the required circle is to pass.

56. Let AB be the given line and D the given point in it, through

which the circle is required to pass, and AC the line which the circle is

to touch. From D draw DE perpendicular to AB and meeting AC in C.

Suppose O a point in AD to be the centre of the required circle. Draw

OE perpendicular to AC, and join OC, then it may be shown that CO

bisects the angle ACD.

57. Let the given circle be described. Draw a line through the

center and intersection of the two lines. Next draw a chord pei-pendi-

cular to this line, cutting off a segment containing the given angle. The

circle described passing through one extremity of the chord and touch-

ing one of the straight lines, shall also pass through the other extremity

Of the chord and touch the other line.

58. The line drawn through the point of intersection of the two

circles parallel to the line which joins their centers, may be shewn to be

double of the line which joins their centers, and greater than any other

straight line drawn through tlie same point and terminated by the cir-

cumferences. The greatest line therefore depends on the distance be-

tween the centers ot the two circles.

330 GEOMETRICAL EXERCISES, &C.

59. Apply Euc. m. 27. i. 6.

60. Let two unequal circles cut one another, and let the line ABC

drawn throuj^h B, one of the points of intersection, be the line required,

such that AB is equal to BC. Join O, O' the centers of the circles, and

draw OP, O'P' perpendiculars on ABC, then PB is equal to BP' ; through

O' draw O'D parallel to PP' ; then ODO' is a right-angled triangle, and

a semicircle described on 00' as a diameter will pass through the point

D. Hence the synthesis. If the line ABC be supposed to move round

the point B and its extremities A, C to be in the extremities of the two

circles, it is manifest that ABC admits of a maximum.

61. Suppose the thing done, then it will appear that the line joining

the points of intersection of the two circles is bisected at right angles by

the line joining the centers of the circles. Since the radii are known,

the centers of the two circles may be determined,

62. Let the circles intersect in A, B ; and let CAD, EBF be any

parallels passing through A, B and intercepted by the circles. Join

CE, AB, DF. Then the figure CEFD may be proved to be a parallelo-

gram. Whence CAD is equal to EBF.

63. Complete the circle whose segment is ADB ; AHB being the

other part. Then since the angle ACB is constant, being in a given

segment, the sum of the arcs DE and AHB is constant. But AHB is

given, hence ED is also given and therefore constant.

64. From A suppose ACD drawn, so that when BD, BC are joined,

AD and DB shall together be double of AC and CB together. Then

the angles ACD, ADB are supplementary, and hence the angles BCD,

BDC are equal, and the triangle BCD is isosceles. Also the angles

BCD, BDC are given, hence the triangle BDC is given in species.

Again AD + DB = 2.AC + 2.BC, or CD = AC + BC.

Whence, make the triangle bdc having its angles at c?, c equal to that

in the segment BDA ; and make ca â€” cd â€” ch, and join ah. At A make

the angle BAD equal to bad^ and AD is the line required.

65. The line drawn from the point of intersection of the two lines

to the center of the given circle may be shewn to be constant, and the

center of the given circle is a fixed point.

66. This is at once obvious from Euc. iii. 36.

67. This follows directly from Euc. iii. 36.

68. Each of the lines CE, DF may be proved parallel to the common

chord AB.

69. By constructing the figure and joining AC and AD, by Euc.

III. 27, it may be proved that the line BC falls on BD.

70. By constructing the figure and applying Euc i. 8, 4, the truth

is manifest.

71. The bisecting line is a common chord to the two circles ; join the

other extremities of the chord and the diameter in each ch'cle, and the

angles in the two segments may be proved to be equal.

72. Apply Euc. m. 27 ; i. 32, 6.

73. Draw a common tangent at C the point of contact of the circles,

and prove AC and CB to be in the same straight line.

74. Let A, B, be the centers, and C the point of contact of the two

circles ; D, E the points of contact of the circles with the common tangent

DE, and CF a tangent common to the two circles at C, meeting DF in E.

Join DC, CE. Then DF, FC, FE may be shewn to be equal, and FC

to be at right angles to AB.

75. The line must be drawn to the extremities of the diameters which

are on opposite sides of the line joining the centers.

ON BOOK III. 331

76. The sum of the distances of the center of the third circle from

the centers of the two given circles, is equal to the sum of the radii of

the given circles, which is constant.

77. Let the circles touch at C either externally or internally, and

their diameters AC, BC through the point of contact will either coincide

or be in the same straight line. CDE any line through C will cut off

similar segments from the two circles. For joining AD, BE, the angles

in the segments DAC, EBC are proved to be equal.

The remaining segments are also similar, since they contain angles

which are supplementary to the angles DAC, EBC.

78. Let the line which joins the centers of the two circles be pro-

duced to meet the circumferences, and let the extremities of this line

and any other line from the point of contact be joined. From the center

of the larger circle draw perpendiculars on the sides of the right- angled

triangle inscribed within it.

79. Li general, the locus of a point in the circumference of a circle

which rolls within the circumference of another, is a curve called the

Hypocycloid ; but to this there is one exception, in which the radius of

one of the circles is double that of the other: in this case, the locus is

a straight line, as may be easily shewn from the figure.

80. Let A, B be the centers of the circles. Draw AB cutting the

circumferences in C, D. On AB take CE, DF each equal to the radius

of the required circle : the two circles described with centers A, B, and

radii AE, BF, respectively, will cut one another, and the point of inter-

section will be the center of the required circle.

81. Apply Euc. III. 31.

82. Apply Euc. iii. 21.

83. (1) When the tangent is on the same side of the two circles.

Join C, C their centers, and on CC describe a semicircle. With center

C and radius equal to the difference of the radii of the two circles, describe

another circle cutting the semicircle in D : join DC and produce it to

meet the circumference of the given circle in B. Through C draw CA

parallel to DB and join BA ; this line touches the two circles.

(2) When the tangent is on the alternate sides. Having joined C,

C ; on CC describe a semicircle ; with center C, and radius equal to the

sum of the radii of the two circles describe another circle cutting the

semicircle in D, join CD cutting the circumference in A, through C

draw CB parallel to CA and join AB.

84. The possibility is obvious. The point of bisection of the segment

intercepted between the convex circumferences will be the center of one

of the circles : and the center of a second circle will be found to be the

point of intersection of two circles described from the centers of the

given circles with their radii increased by the radius of the second circle.

The line passing through the centers of these two circles will be the locus

of the centers of all the circles which touch the two given circles.

85. At any points P, R in the circumferences of the circles, whose

centers are A, B, draw PQ, RS, tangents equal to the given lines, and

join AQ, BS. These being made the sides of a triangle of which AB

is the base, the vertex of the triangle is the point required.

86. In each circle draw a chord of the given length, describe circles

concentric with the given circles touching these chords, and then draw

a straight line touching these circles.

87. Within one of the circles draw a chord cutting off a segment

equal to the given segment, and describe a concentric circle touching

the chord : then draw a straight line touching this latter circle and the

other given circle.

8S2 GEOMETRICAL EXERCISES, &C.

88. The tangent may intersect the line joining the centers, or the line

produced. Prove that the angle in the segment of one circle is equal to

the angle in the corresponding segment of the other circle.

89. Join the centers A, B ; at C the point of contact draw a tangent,

and at A draw AF cutting the tangent in F, and making with OF an

angle equal to one-fourth of the given angle. From F draw tangents

to the circles.

90. Let C be the center of the given circle, and D the given point in

the given line AB. At D draw any line DE at right angles to AB, then

the center of the circle required is in the line AE. Through C draw a

diameter FG parallel to DE, the circle described passing through the

points E, F, G will be the circle required.

91. Apply Euc. III. 18.

92. Let A, B, be the two given points, and C the center of the given

circle. Join AC, and at C draw the diameter DCE perpendicular to AC,

and through the points A, D, E describe a circle, and produce AC to

meet the circumference in F. Bisect AF in G, and AB in H, and draw

GK, HK, perpendiculars to AF, AB respectively and intersecting in K.

Then K is the center of the circle which passes through the points A, B,

and bisects the circumference of the circle whose center is C.

93. Let D be the given point and EF the given straight line. (fig.

Euc. III. 32.) Draw DB to make the angle DBF equal to that contained

in the alternate segment. Draw BA at right angles to EF, and DA at

right angles to DB and meeting BA in A. Then AB is the diameter of

the circle.

94. Let A, B be the given points, and CD the given line. From E

the middle of the line AB, draw EM perpendicular to AB, meeting CD

in M, and draw MA. In EM take any point F ; draw FII to make the

given angle with CD ; and draw FG equal to FH, and meeting MA

produced in G. Through A draw AP parallel to FG, and CPK parallel

to FH. Then P is the center, and C the third defining point of the

circle required : and AP mav be proved equal to CP by means of the

triangles GMF, AMP; and HMF, CMP, Euc. vi. 2. Also CPK the

diameter makes with CD the angle KCD equal to FHD, that is, to the

29. Since the base and area are given, the altitude of the triangle is

known. Hence the problem is reduced to ; â€” Given the base and altitude

of a triangle, and the line drawn from the vertex to the bisection of the

base, construct the triangle.

30. This follows immediately from Euc. i. 47.

31. Apply Euc. II. 13.

32. The truth of this property depends on the fact that the rectangle

contained by AC, CB is equal to that contained by AB, CD.

33. Let P the required point in the base AB be supposed to be known.

Join CP. It may then be shewn that the property stated in the Prob-

lem is contained in Theorem 3. p. 114.

34. This may be shewn from Euc. i. 47 ; ii- 5. Cor.

35. Erom C let fall CF pd-pendicular on AB. Then ACE is an ob-

tuse-angled, and BEC an acute-angled triangle. Apply Euc. ii. 12, 13 ,

and by Euc. i. 47, the squares on AC and CB are equal to the square

on AB.

36. Apply Euc. i. 47, ii. 4 ; and the note p. 102, on Euc. ii. 4.

37. Draw a perpendicular from the vertex to the base, and apply

Euc. I. 47 ; II. 5, Cor. Enunciate and prove the proposition when the

straight line drawn from the vertex meets the base produced.

38. This follows directly from Euc. ii. 13, Case 1.

39. The truth of this proposition may be shewn from Euc. i. 47 ; ii. 4.

40. Let the square on the base of the isosceles triangle be described.

Draw the diagonals of the square, and the proof is obvious.

41. Let ABC be the triangle required, such that the square on AB

is three times the square on AC or BC. Produce BC and draw AD per-

pendicular to BC. Then by Euc. ii. 12, CD may be shewn to be equal

to one half of BC. Hence the construction.

42. Apply Euc. ii. 12, and Theorem 38, p. 118.

43. Draw EF parallel to AB and meeting the base in E ; draw also

EG perpendicular to the base. Then by Euc. i. 47 ; n. 5, Cor.

44. Bisect the angle B by BD meeting the opposite side in D, and

draw BE perpendicular to AC. Then by Euc. i. 47 ; ii. 5, Cor.

45. This follows directly from Theorem 3, p. 114.

46. Draw the diagonals intersecting each other in P, and join OP.

By Theo. 3, p. 114.

47. Draw from any two opposite angles, straight lines to meet in the

bisection of the diagonal joining the other angles. Then by Euc. ii. 12, 13.

48. Draw two lines from the point of bisection of either of the bi-

sected sides to the extremities of the opposite side ; and three triangles

will be formed, two on one of the bisected sides and one on the other, in

S26 GEOMETRICAL EXERCISES, &C.

each of which is a line drawn from the vertex to the bisection of the base.

Then by Theo. 3, p, 114.

49. If the extremities of the two lines which bisect the opposite sides

of the trapezium be joined, the figure formed is a parallelogram which

has its sides respectively parallel to, and equal to, half the diagonals of

the trapezium. The sum of the squares on the two diagonals of the tra-

pezium may be easily shewn to be equal to the sum of the squares on

the four sides of the parallelogram.

50. Draw perpendiculars from the extremities of one of the parallel

sides, meeting the other side produced, if necessary. Then from the four

right-angled triangles thus formed, may be shewn the truth of the pro-

position.

51. Let AD be parallel to BC in the figure ABCD. Draw the diagonal

AC, then the sum of the triangles ABC, ADC may be shewn to be equal to

the rectangle contained by the altitude and half the sum of AD and BC.

52. Let ABCD be the trapezium, having the sides AB, CD, parallel,

and AD, BC equal. Join AC and draw AE perpendicular to DC. Then

by Euc. II. 13.

53. Let ABC be any triangle ; AHKB, AGFC, BDEC, the squares

upon their sides ; EF, GH, KL the lines joining the angles of the squares.

Produce GA, KB, EC, and draw HN, DQ, FR perpendiculars upon them

respectively: also draw AP, BM, CS perpendiculars on the sides of the

triangle. Then AN may be proved to be equal to AM ; CR to CP ; and

BQ to BS ; and by Euc. ii. 12, 13.

54. Convert the triangle into a rectangle, then Euc. ii. 14.

55. Find a rectangle equal to the two figures, and apply Euc. ii. 14.

56. Find the side of a square which shall be equal to the given

rectangle. See Prob. i. p. 113.

57. On any line PQ take AB equal to the given difference of the

sides of the rectangle, at A draw AC at right angles to AB, and equal to

the side of the given square ; bisect AB in O and join OC ; with center

O and radius OC describe a semicircle meeting PQ in D and E. Then

the lines AD, AE have AB for their difierence, and the rectangle con~

tained by them is equal to the square on AC,

58. Apply Euc. ii. 14.

GEOMETRICAL EXERCISES ON BOOK III.

HINTS, &c.

7. Euc. III. 3, suggests the construction.

8. The given point may be either within or without the circle. Find

the center of the circle, and join the given point and the center, and upon

this line describe a semicircle, a line equal to the given distance may be

drawn from the given point to meet the arc of the semicircle. When

the point is without the circle, the given distance may meet the diameter

produced.

9. This may be easily shewn to be a straight line passing through

the center of the circle.

10. The two chords form by their intersections the sides of two isos-

celes triangles, of which the parallel chords in the circle are the bases.

ON BOOK III. S21

n. The angles in equal segments are equal, and by Euc. i. 29. If

the chords are equally distant from the center, the lines intersect the

diameter in the center of the circle.

12. Construct the figure and the arc BC may be proved equal to the

arc B'C

13. The point determined by the lines drawn from the bisections of

the chords and at right angles to them respectively, will be the center of

the required circle.

14. Construct the figures : the proof offers no difficulty.

15. From the centre C ot the circle, draw CA, CB at right angles to

each other meeting the circumference ; join AB, and draw CD perpendicular

to AB.

16. Join the extremities of the chords, then Euc. i. 27 ; iii. 28.

17. Take the center O, and join AP, AO, &c. and apply Euc. i. 20.

18. Draw any straight line intersecting two parallel chords and meet-

ing the circumference.

19. Produce the radii to meet the circumference.

20. Join AD, and the first equality follows directly from Euc. in.

20, T. 32. Also by joining AC, the second equality may be proved in a

similar way. If however the line AD do not fall on the same side of the

center O as E, it will be found that the difference^ not the sum of the two

angles, is equal to 2 . AED. See note to Euc. in. 20, p. 155.

21. Let DKE, DBO (fig. Euc. in. 8) be two lines equally inclined

to DA, then KE may be proved to be equal to BO, and the segments cut

off" by equal straight lines in the same circle, as well as in equal circles,

are equal to one another.

22. Apply Euc. i. 15, and in, 21.

23. This is the same as Euc. in. 34, with the condition, that the line

must pass through a given point.

24. Let the segments AHB, AKC be externally described on the

given lines AB, AC, to contain angles equal to BAC. Then by the con-

verse to Euc. ITT. 32, AB touches the circle AKC, and AC the circle AHB.

25. Let ABC be a triangle of which the base or longest side is BC,

and let a segment of a circle be described on BC. Produce BA, CA to

meet the arc of the segment in D, E, and join BD, CE. If circles be de-

scribed about the triangles ABD, ACE, the sides AB, AC shall cut off

segments similar to the segment described upon the base BC.

26. This is obvious from the note to Euc. in. 26, p. 156.

27. The segment must be described on the opposite side of the pro-

duced chord. By converse of Euc. in. 32.

28. If a circle be described upon the side AC as a diameter, the cir-

cumference will pass through the points D, E. Then Euc. in. 21.

29. Let AB, AC be the bounding radii, and D any point in the arc

BC, and DE, DF, perpendiculars from D on AB, AC. The circle de-

scribed on AD will always be of the same magnitude, and the angle EAP

in it, is constant : â€” whence the arc EDF is constant, and therefore its

chord EF.

30. Construct the figure, and let the circle with center O, described

on AH as a diameter, intersect the given circle in P, Q, join OP, PE, and

prove EP at right angles to OP.

31. If the tangent be required to be perpendicular to a given line:

draw the diameter parallel to this line, and the tangent drawn at the ex-

tremity of this diameter will be perpendicular to the given line.

32. The straight line which joins the center and passes through the

intersection of two tangents to a circle, bisects the angle f^ntained by

the tangents.

328 GEOMETRICAL EXERCISES, &C.

33. Draw two radii containing an angle equal to the supplement of

the given angle ; the tangents drawn at the extremities of these radii will

contain the given angle,

34. Since the circle is to touch two parallel lines drawn from two

given points in a third line, the radius of the circle is determined by the

distance between the two given points.

35. It is sufficient to suggest that the angle between a chord and a

tangent is equal to the angle in the alternate segment of the circle. Euc.

m. 32.

36. Let AB be the given chord of the circle whose center is O. Draw

DE touching the circle at any point E and equal to the given line ; join

DO, and with center O and radius DO describe a circle: produce the

chord AB to meet the circumference of this circle in F : then F is the

point required.

37. Let D be the point required in the diameter BA produced, such

that the tangent DP is half of DB. Join CP, C being the ct-nter. Then

CPD is a right-angled triangle, having the sum of the base PC and hypo-

tenuse CD double of the perpendicular PD.

38. If BE intersect DF in K (fig. Euc. iii. 37). Join FB, FE, then

by means of the triangles, BE is shewn to be bisected in K at right angles.

39. Let AB, CD be any two diameters of a circle, O the center, and

let the tangents at their extremities form the quadrilateral figure EFGII.

Join EO, OF, then EO and OF may be proved to be in the same straight

line, and similarly HO, OK.

Note. â€” This Proposition is equally true if AB, CD be any two chords

v.'hatever. It then becomes equivalent to the following proposition: â€”

The diagonals of the circumscribed and inscribed quadrilaterals, intersect

in the same point, the points of contact of the former being the angles of

the latter figure.

40. Let C be the point without the circle from which the tangents

CA, CB are drawn, and let DE be any diameter, also let AE, BD be

joined, intersecting in P, then if CP be joined and produced to meet DE

in G : CG is perpendicular to DE. Join DA, EB, and produce them to

meet in F.

Then the angles DAE, EBD being angles in a semicircle, are right

ambles ; or DB, EA are drawn perpendicular to the sides of the triangle

DEF : whence the line drawn from F through P is perpendicular to the

third side DE.

41. Let the chord AB, of which P is its middle point, be produced

both ways to C, D, so that AC is equal to BD. From C, D, draw the

tangents to the circle forming the tangential quadrilateral CKDR, the

points of contact of the fides, being E, II, F, G. Let O be the center of

the circle. Join EH, GF, CO, GO, FO, DO. Then EH and GF may

be proved each parallel to CD, they are therefore parallel to one another.

"Whence is proved that both EF and DG bisect AB.

42. This is obvious from Euc. i. 29, and the note to iii. 22. p. 156.

43. From any point A in the circumference, let any chord AB and

tangent AC be drawn. Bisect the arc AB in D, and from D draw DE,

DC perpendiculars on the chord AB and tangent AC. Join AD, the

triangles ADE, ADC may be shewn to be equal.

44. Let A, B, be the given points. Join AB, and upon it describe a

segment of a circle which shall contain an angle equal to the given angle.

If the circle cut the given line, there will be two points; if it only touch

the line, there will be one ; and if it neither cut nor touch the line, the

problem is impossible.

I

JN BOOK III. 329

45. It may be shewn that the point required is determined by a per-

pendicular drawn from the center of the circle on the <;ivcn line.

46. Let two lines AP, BP be drawn from the given points A, B,

making equal angles with the tangent to the circle at the point of contact

P, take any other point Q in the convex circumference, and join QA,

QB : then by Prob. 4, p. 71, and Euc. i. 21.

47. Let C be the center of the circle, and E the point of contact of

DF with the circle. Join DC, CE, CF.

48. liCt the tangents at E, F meet in a point K. Produce HE, RF

to meet the diameter AB produced in S, T. Then RST is a triangle,

and the quadrilateral RFOE maybe circumscribed by a circle, and RPO

may be proved to be one of the diagonals,

49. Let C be the middle point of the chord of contact : produce AC,

BC to meet the circumference in B', A', and join AA', BB'.

50. Let A be the given point, and B the given point in the given line

CD. At B draw BE at right angles to CD, join AB and bisect it in F,

and from F draw FE perpendicular to AB and meeting BE in E. E is

the center of the required circle.

51. Let O be the center of the given circle. Draw OA perpendicular

to the given straight line ; at O in OA make the angle AOP equal to the

given angle, produce PO to meet the circumference again in Q. Then P,

Q are two points from which tangents may be drawn fulfilling the re-

quired condition.

52. Let C be the center of the given circle, B the given point in the

circumference, and A the other given point through which the required

circle is to be made to pass. Join CB, the center of the circle is a point

in CB produced. The center itself may be found in three ways.

53. Euc. III. 11 suggests the construction.

54. Let AB, AC be the two given lines which meet at A, and let D

be the given point. Bisect the angle BAC by AE, the center of the circle

is in AE. Through D draw DF perpendicular to AE, and produce DF

to G, making FG equal to FD. Then DG is a chord of the circle, and

the circle which passes through D and touches AB, will also pass through

G and touch AC. ^

55. As the center is given, the line joining this point and center of

the given circle, is perpendicular to that diameter, through the extremi-

ties of which the required circle is to pass.

56. Let AB be the given line and D the given point in it, through

which the circle is required to pass, and AC the line which the circle is

to touch. From D draw DE perpendicular to AB and meeting AC in C.

Suppose O a point in AD to be the centre of the required circle. Draw

OE perpendicular to AC, and join OC, then it may be shown that CO

bisects the angle ACD.

57. Let the given circle be described. Draw a line through the

center and intersection of the two lines. Next draw a chord pei-pendi-

cular to this line, cutting off a segment containing the given angle. The

circle described passing through one extremity of the chord and touch-

ing one of the straight lines, shall also pass through the other extremity

Of the chord and touch the other line.

58. The line drawn through the point of intersection of the two

circles parallel to the line which joins their centers, may be shewn to be

double of the line which joins their centers, and greater than any other

straight line drawn through tlie same point and terminated by the cir-

cumferences. The greatest line therefore depends on the distance be-

tween the centers ot the two circles.

330 GEOMETRICAL EXERCISES, &C.

59. Apply Euc. m. 27. i. 6.

60. Let two unequal circles cut one another, and let the line ABC

drawn throuj^h B, one of the points of intersection, be the line required,

such that AB is equal to BC. Join O, O' the centers of the circles, and

draw OP, O'P' perpendiculars on ABC, then PB is equal to BP' ; through

O' draw O'D parallel to PP' ; then ODO' is a right-angled triangle, and

a semicircle described on 00' as a diameter will pass through the point

D. Hence the synthesis. If the line ABC be supposed to move round

the point B and its extremities A, C to be in the extremities of the two

circles, it is manifest that ABC admits of a maximum.

61. Suppose the thing done, then it will appear that the line joining

the points of intersection of the two circles is bisected at right angles by

the line joining the centers of the circles. Since the radii are known,

the centers of the two circles may be determined,

62. Let the circles intersect in A, B ; and let CAD, EBF be any

parallels passing through A, B and intercepted by the circles. Join

CE, AB, DF. Then the figure CEFD may be proved to be a parallelo-

gram. Whence CAD is equal to EBF.

63. Complete the circle whose segment is ADB ; AHB being the

other part. Then since the angle ACB is constant, being in a given

segment, the sum of the arcs DE and AHB is constant. But AHB is

given, hence ED is also given and therefore constant.

64. From A suppose ACD drawn, so that when BD, BC are joined,

AD and DB shall together be double of AC and CB together. Then

the angles ACD, ADB are supplementary, and hence the angles BCD,

BDC are equal, and the triangle BCD is isosceles. Also the angles

BCD, BDC are given, hence the triangle BDC is given in species.

Again AD + DB = 2.AC + 2.BC, or CD = AC + BC.

Whence, make the triangle bdc having its angles at c?, c equal to that

in the segment BDA ; and make ca â€” cd â€” ch, and join ah. At A make

the angle BAD equal to bad^ and AD is the line required.

65. The line drawn from the point of intersection of the two lines

to the center of the given circle may be shewn to be constant, and the

center of the given circle is a fixed point.

66. This is at once obvious from Euc. iii. 36.

67. This follows directly from Euc. iii. 36.

68. Each of the lines CE, DF may be proved parallel to the common

chord AB.

69. By constructing the figure and joining AC and AD, by Euc.

III. 27, it may be proved that the line BC falls on BD.

70. By constructing the figure and applying Euc i. 8, 4, the truth

is manifest.

71. The bisecting line is a common chord to the two circles ; join the

other extremities of the chord and the diameter in each ch'cle, and the

angles in the two segments may be proved to be equal.

72. Apply Euc. m. 27 ; i. 32, 6.

73. Draw a common tangent at C the point of contact of the circles,

and prove AC and CB to be in the same straight line.

74. Let A, B, be the centers, and C the point of contact of the two

circles ; D, E the points of contact of the circles with the common tangent

DE, and CF a tangent common to the two circles at C, meeting DF in E.

Join DC, CE. Then DF, FC, FE may be shewn to be equal, and FC

to be at right angles to AB.

75. The line must be drawn to the extremities of the diameters which

are on opposite sides of the line joining the centers.

ON BOOK III. 331

76. The sum of the distances of the center of the third circle from

the centers of the two given circles, is equal to the sum of the radii of

the given circles, which is constant.

77. Let the circles touch at C either externally or internally, and

their diameters AC, BC through the point of contact will either coincide

or be in the same straight line. CDE any line through C will cut off

similar segments from the two circles. For joining AD, BE, the angles

in the segments DAC, EBC are proved to be equal.

The remaining segments are also similar, since they contain angles

which are supplementary to the angles DAC, EBC.

78. Let the line which joins the centers of the two circles be pro-

duced to meet the circumferences, and let the extremities of this line

and any other line from the point of contact be joined. From the center

of the larger circle draw perpendiculars on the sides of the right- angled

triangle inscribed within it.

79. Li general, the locus of a point in the circumference of a circle

which rolls within the circumference of another, is a curve called the

Hypocycloid ; but to this there is one exception, in which the radius of

one of the circles is double that of the other: in this case, the locus is

a straight line, as may be easily shewn from the figure.

80. Let A, B be the centers of the circles. Draw AB cutting the

circumferences in C, D. On AB take CE, DF each equal to the radius

of the required circle : the two circles described with centers A, B, and

radii AE, BF, respectively, will cut one another, and the point of inter-

section will be the center of the required circle.

81. Apply Euc. III. 31.

82. Apply Euc. iii. 21.

83. (1) When the tangent is on the same side of the two circles.

Join C, C their centers, and on CC describe a semicircle. With center

C and radius equal to the difference of the radii of the two circles, describe

another circle cutting the semicircle in D : join DC and produce it to

meet the circumference of the given circle in B. Through C draw CA

parallel to DB and join BA ; this line touches the two circles.

(2) When the tangent is on the alternate sides. Having joined C,

C ; on CC describe a semicircle ; with center C, and radius equal to the

sum of the radii of the two circles describe another circle cutting the

semicircle in D, join CD cutting the circumference in A, through C

draw CB parallel to CA and join AB.

84. The possibility is obvious. The point of bisection of the segment

intercepted between the convex circumferences will be the center of one

of the circles : and the center of a second circle will be found to be the

point of intersection of two circles described from the centers of the

given circles with their radii increased by the radius of the second circle.

The line passing through the centers of these two circles will be the locus

of the centers of all the circles which touch the two given circles.

85. At any points P, R in the circumferences of the circles, whose

centers are A, B, draw PQ, RS, tangents equal to the given lines, and

join AQ, BS. These being made the sides of a triangle of which AB

is the base, the vertex of the triangle is the point required.

86. In each circle draw a chord of the given length, describe circles

concentric with the given circles touching these chords, and then draw

a straight line touching these circles.

87. Within one of the circles draw a chord cutting off a segment

equal to the given segment, and describe a concentric circle touching

the chord : then draw a straight line touching this latter circle and the

other given circle.

8S2 GEOMETRICAL EXERCISES, &C.

88. The tangent may intersect the line joining the centers, or the line

produced. Prove that the angle in the segment of one circle is equal to

the angle in the corresponding segment of the other circle.

89. Join the centers A, B ; at C the point of contact draw a tangent,

and at A draw AF cutting the tangent in F, and making with OF an

angle equal to one-fourth of the given angle. From F draw tangents

to the circles.

90. Let C be the center of the given circle, and D the given point in

the given line AB. At D draw any line DE at right angles to AB, then

the center of the circle required is in the line AE. Through C draw a

diameter FG parallel to DE, the circle described passing through the

points E, F, G will be the circle required.

91. Apply Euc. III. 18.

92. Let A, B, be the two given points, and C the center of the given

circle. Join AC, and at C draw the diameter DCE perpendicular to AC,

and through the points A, D, E describe a circle, and produce AC to

meet the circumference in F. Bisect AF in G, and AB in H, and draw

GK, HK, perpendiculars to AF, AB respectively and intersecting in K.

Then K is the center of the circle which passes through the points A, B,

and bisects the circumference of the circle whose center is C.

93. Let D be the given point and EF the given straight line. (fig.

Euc. III. 32.) Draw DB to make the angle DBF equal to that contained

in the alternate segment. Draw BA at right angles to EF, and DA at

right angles to DB and meeting BA in A. Then AB is the diameter of

the circle.

94. Let A, B be the given points, and CD the given line. From E

the middle of the line AB, draw EM perpendicular to AB, meeting CD

in M, and draw MA. In EM take any point F ; draw FII to make the

given angle with CD ; and draw FG equal to FH, and meeting MA

produced in G. Through A draw AP parallel to FG, and CPK parallel

to FH. Then P is the center, and C the third defining point of the

circle required : and AP mav be proved equal to CP by means of the

triangles GMF, AMP; and HMF, CMP, Euc. vi. 2. Also CPK the

diameter makes with CD the angle KCD equal to FHD, that is, to the

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