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proof depends on Euc. vi. 19; 20, Cor. 2. (2) Let ABC be the triangle, BC being the base. Draw AD at right angles to BA meeting the base produced in D. Bisect BC in E, and on ED describe a semicircle, from B draw BP to touch the semicircle in P. From BA cut off BF equal to BP, and from F draw FG perpendicular to BC. The line FG bisects the triangle. Then it may be proved that BFG: BAD :: BE: BD, and that BAD: BAC :: BĎ: BC; whence it follows that BFG: BAC :: BE: BC or as 1 : 2.

43. Let ABC be the given triangle which is to be divided into two parts having a given ratio, by a line parallel to BC. Describe a semicircle on AB and divide AB in D in the given ratio; at D draw DE perpendicular to AB and meeting the circumference in E; with center A and radius AE describe a circle cutting AB in F: the line drawn through F parallel to BC is the line required. In the same manner a triangle may be divided into three or more parts having any given ratio to one another by lines drawn parallel to one of the sides of the triangle. 44. Let these points be taken, one on each side, and straight lines be drawn to them; it may then be proved that these points severally bisect the sides of the triangle.

45. Let ABC be any triangle and D be the given point in BC, from which lines are to be drawn which shall divide the triangle into any number (suppose five) equal parts. Divide BC into five equal parts in E, F, G, H, and draw AE, AF, AG, AH, AD, and through E, F, G, H draw EL, FM, GN, HO parallel to AD, and join DL, DM, DN, DO; these lines divide the triangle into five equal parts.

By a similar process, a triangle may be divided into any number of parts which have a given ratio to one another.

46. Let ABC be the larger, abc the smaller triangle, it is required to draw a line DE parallel to AC cutting off the triangle DBE equal to the triangle abc. On BC take BG equal to bc, and on BG describe the triangle BGH equal to the triangle abc. Draw HK parallel to BC, join KG; then the triangle BGK is equal to the triangle abc. On BA, BC take BD to BE in the ratio of BA to BC, and such that the rectangle contained by BD, BE shall be equal to the rectangle contained by BK, BG. Join DE, then DE is parallel to AC, and the triangle BDE is equal to abc.

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

Then AB': AB. BC::AB: BC,

and AB.BC: BC2 :: AB: BC,

whence AB: AB. BC :: AB. BC: BC",

or the rectangle contained by two adjacent sides of a rectangle, is a mean proportional between their squares.

and

48. In a straight line at any point A, make Ac equal to Ad in the given ratio. At A draw AB perpendicular to cAd, and equal to a side of the given square. On cd describe a semicircle cutting AB in b; join bc, bd; from B draw BC parallel to bc, and BD parallel to bd: then AC, AD are the adjacent sides of the rectangle. For, CA is to AD as cA to Ad, Euc. vi. 2; and CA. AD AB, CBD being a right-angled triangle.

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49. From one of the given points two straight lines are to be drawn perpendicular, one to each of any two adjacent sides of the parallelogram; and from the other point, two lines perpendicular in the same manner to each of the two remaining sides. When these four lines are drawn to intersect one another, the figure so formed may be shewn to be equi. angular to the given parallelogram.

50. It is manifest that this is the general case of Prop. 4, p. 197. If the rectangle to be cut off be two-thirds of the given rectangle ABCD. Produce BC to E so that BE may be equal to a side of that square which is equal to the rectangle required to be cut off; in this case, equal to two-thirds of the rectangle ABCD. On AB take AF equal to AD or BC; bisect FB in G, and with center G and radius GE, describe a semicircle meeting AB, and AB produced, in H and K. On CB take CL equal to AH and draw HM, LM parallel to the sides, and HBLM is two-thirds of the rectangle ABCD.

51: Let ABCD be the parallelogram, and CD be cut in P and BC produced in Q. By means of the similar triangles formed, the property may be proved.

52. The intersection of the diagonals is the common vertex of two triangles which have the parallel sides of the trapezium for their bases.

53. Let AB be the given straight line, and C the center of the given circle; through C draw the diameter DCE perpendicular to AB. Place in the circle a line FG which has to AB the given ratio; bisect FG in H, join CH, and on the diameter DCE, take CK, CL each equal to CH; either of the lines drawn through K, L, and parallel to AB is the line required.

51. Let C be the center of the circle, CA, CB two radii at right angles to each other; and let DEFG be the line required which is trisected in the points E, F. Draw CG perpendicular to DH and produce it to meet the circumference in K; draw a tangent to the circle at K: draw CG, and produce CB, CG to meet the tangent in L, M, then MK may be shewn to be treble of LK.

55. The triangles ACD, BCE are similar, and CF is a mean proportional between AC and CB.

56. Let any tangent to the circle at E be terminated by AD, BC tangents at the extremity of the diameter AB. Take O the center of the circle and join OC, OD, OE; then ODC is a right-angled triangle and OE is the perpendicular from the right angle upon the hypotenuse.

57. This problem only differs from problem 59, infra, in having the given point without the given circle.

58. Let A be the given point in the circumference of the circle, Cits center. Draw the diameter ACB, and produce AB to D, taking AB to BD in the given ratio: from D draw a line to touch the circle in E, which is the point required. From A draw AF perpendicular to DE, and cutting the circle in G.

59. Let A be the given point within the circle whose center is C, and let BAD be the line required, so that BA is to AD in the given ratio. Join AC and produce it to meet the circumference in E, F. Then EF is a diameter. Draw BG, DH perpendicular on EF: then the triangles BGA, DHA are equiangular. Hence the construction.

60. Through E one extremity of the chord EF, let a line be drawn parallel to one diameter, and intersecting the other. Then the three angles of the two triangles may be shewn to be respectively equal to one another.

61. Let AB be that diameter of the given-circle which when produced is perpendicular to the given line CD, and let it meet that line in C; and let P be the given point: it is required to find D in CD, so that DB may be equal to the tangent DF. Make BC: CQ:: CQ : CA, and join PQ; bisect PQ in E, and draw ED perpendicular to PQ meeting CĎ in D; then D is the point required. Let O be the center of the circle, draw the tangent DF; and join OF, OD, QD, PD. Then QD may be shewn

to be equal to DF and to DP. When P coincides with Q, any point D in CD fulfils the conditions of the problem; that is, there are innumerable solutions.

62. It may be proved that the vertices of the two triangles which are similar in the same segment of a circle, are in the extremities of a chord parallel to the chord of the given segment.

63. For let the circle be described about the triangle EAC, then by the converse to Euc. 1. 32; the truth of the proposition is manifest. 64. Let the figure be constructed, and the similarity of the two triangles will be at once obvious from Euc III. 32.; Euc. 1. 29.

65. In the arc AB (fig. Euc. iv. 2) let any point K be taken, and from K let KL, KM, KN be drawn perpendicular to AB, AC, BC respectively, produced if necessary, also let LM, LN be joined, then MLN may be shewn to be a straight line. Draw AK, BK, CK, and by Euc. III, 31, 22, 21; Euc. I. 14.

66. Let AB a chord in a circle be bisected in C, and DE, FG two chords drawn through C; also let their extremities DG, FE be joined intersecting CB in H, and AC in K; then AK is equal to HB. Through H draw MHL parallel to EF meeting FG in M, and DE produced in L. Then by means of the equiangular triangles, HC may be proved to be equal to CK, and hence AK is equal to HB.

67. Let A, B be the two given points, and let P be a point in the locus so that PA, PB being joined, PA is to PB in the given ratio. Join AB and divide it in C in the given ratio, and join PC. Then PC bisects the angle APB. Euc. vi. 3. Again, in AB produced, take AD to AB in the given ratio, join PD and produce AP to E, then PD bisects the angle BPE. Euc. vI. A. Whence CPD is a right angle, and the point P lies in the circumference of a circle whose diameter is CD.

68. Let ABC be a triangle, and let the line AD bisecting the vertical angle A be divided in E, so that BC: BA+AC:: AE: ED. By Euc. VI. 3, may be deduced BC: BA+AC:: AC: AD. Whence may be proved that CE bisects the angle ACD, and by Euc. iv. 4, that E is the center of the inscribed circle.

69. By means of Euc. Iv. 4, and Euc. vI. C. this theorem may be shewn to be true.

70. Divide the given base BC in D, so that BD may be to DC in the ratio of the sides. At B, D draw BB', DD' perpendicular to BC and equal to BD, DC respectively. Join B'D' and produce it to meet BC produced in O. With center O and radius OD, describe a circle. From A any point in the circumference join AB, AC, AO. Prove that AB is to AC as BD to DC. Or thus. If ABC be one of the triangles. Divide the base BC in D so that BA is to AC as BD to DC. Produce BC and take DO to OC as BA to AC: then O is the center of the circle.

71. Let ABC be any triangle, and from A, B let the perpendiculars AD, BE on the opposite sides intersect in P: and let AF, BG drawn to F, G the bisections of the opposite sides, intersect in Q. Also let FR, GR be drawn perpendicular to BC, AC, and meet in R: then R is the center of the circumscribed circle. Join PQ, QR; these are in the same line.

Join FG, and by the equiangular triangles, GRF, APB, AP is proved double of FR. And AQ is double of QF, and the alternate angles PAQ, QFR are equal. Hence the triangles APQ, RFQ are equiangular.

72. Let C, C' be the centers of the two circles, and let CC' the line joining the centers intersect the common tangent PP' in T. Let the

line joining the centers cut the circles in Q, Q', and let PQ, P'Q' be joined; then PQ is parallel to PQ. Join CP, C'P', and then the angle QPT may be proved to be equal to the alternate angle Q'P'T.

73. Let ABC be the triangle, and BC its base; let the circles AFB, AFC be described intersecting the base in the point F, and their diameters AD, AE, be drawn; then DA: AE :: BA: AC. For join DB, DF, EF, EC, the triangles DAB, EAC may be proved to be similar.

74. If the extremities of the diameters of the two circles be joined by two straight lines, these lines may be proved to intersect at the point of contact of the two circles; and the two right-angled triangles thus formed may be shewn to be similar by Euc. III. 34.

75. This follows directly from the similar triangles.

76. Let the figure be constructed as in Theorem 4, p. 162, the triangle EAD being right-angled at A, and let the circle inscribed in the triangle ADE touch AD, AE, DE in the points K, L, M respectively. Then AK is equal to AL, each being equal to the radius of the inscribed circle. Also AB is equal to GC, and AB is half the perimeter of the triangle AED.

Also if GA be joined, the triangle ADE is obviously equal to the difference of AGDE and the triangle GDE, and this difference may be proved equal to the rectangle contained by the radii of the other two circles.

77. From the centers of the two circles let straight lines be drawn to the extremities of the sides which are opposite to the right angles in each triangle, and to the points where the circles touch these sides. Euc. vI. 4.

78. Let A, B be the two given points, and C a point in the circumference of the given circle. Let a circle be described through the points A, B, C and cutting the circle in another point D. Join CD, AB, and produce them to meet in E. Let EF be drawn touching the given circle in F; the circle described through the points A, B, F, will be the circle required. Joining AD and CB, by Euc. III. 21, the triangles CEB, AED are equiangular, and by Euc. vI. 4, 16, III. 36, 37, the given circle and the required circle each touch the line EF in the same point, and therefore touch one another. When does this solution fail?

Various cases will arise according to the relative position of the two points and the circle.

79. Let A be the given point, BC the given straight line, and D the center of the given circle. Through D draw CD perpendicular to BC, meeting the circumference in E, F. Join AF, and take FG to the diameter FE, as FC is to FA. The circle described passing through the two points A, G and touching the line BC in B is the circle required. Let H be the center of this circle; join HB, and BF cutting the circumference of the given circle in K, and join EK. Then the triangles FBC, FKE being equiangular, by Euc. vI. 4, 16, and the construction, K is proved to be a point in the circumference of the circle passing through the points A, G, B. And if DK, KH be joined, DKH may be proved to be a straight line:- the straight line which joins the centers of the two circles, and passes through a common point in their circumferences.

80. Let A be the given point, B, C the centers of the two given circles. Let a line drawn through B, C meet the circumferences of the circles in G, F; E, D, respectively. In GD produced, take the point H, so that BH is to CH as the radius of the circle whose center

is B to the radius of the circle whose center is C. Join AH, and take KH to DH as GH to AH. Through A, K describe a circle ALK touching the circle whose center is B, in L. Then M may be proved to be a point in the circumference of the circle whose center is C. For by joining HL and producing it to meet the circumference of the circle whose center is B in N; and joining BN, BL, and drawing CO parallel to BL, and CM parallel to BN, the line HN is proved to cut the circumference of the circle whose center is B in M, O; and CO, CM are radii. By joining GL, DM, M may be proved to be a point in the circumference of the circle ALK. And by producing BL, CM to meet in P, P is proved to be the center of ALK, and BP joining the centers of the two circles passes through L the point of contact. Hence also is shewn that PMC passes through M, the point where the circles whose centers are P and C touch each other.

NOTE. If the given point be in the circumference of one of the circles, the construction may be more simply effected thus:

Let A be in the circumference of the circle whose center is B. Join BA, and in AB produced, if necessary, take AD equal to the radius of the circle whose center is C; join DC, and at C make the angle DCE equal to the angle CDE, the point E determined by the intersection of DA produced and CE, is the center of the circle.

81. Let AB, AC be the given lines and P the given point. Then if O be the center of the required circle touching AB, AC, in R, S, the line AO will bisect the given angle BAC. Let the tangent from P meet the circle in Q, and draw OQ, OS, OP, AP. Then there are given AP and the angle OAP. Also since OQP is a right angle, we have OP-QO2 =OPOS=PQ a given magnitude. Moreover the right-angled triangle AOS is given in species, or OS to OA is a given ratio. Whence in the triangle AOP there is given, the angle AOP, the side AP, and the excess of OP above the square of a line having a given ratio to OA, to determine OA. Whence the construction is obvious.

82. Let the two given lines AB, BD meet in B, and let C be the center of the given circle, and let the required circle touch the line AB, and have its center in BD. Draw CFE perpendicular to HB intersecting the circumference of the given circle in F, and produce CE, making EF equal to the radius CF. Through G draw GK parallel to AB, and meeting DB in K. Join CK, and through B, draw BL parallel to KC, meeting the circumference of the circle whose center is C in L; join CL and produce CL to meet BD in O. Then O is the center of the circle required. Draw OM perpendicular to AB, and produce EC to meet BD in N. Then by the similar triangles, OL may be proved

equal to OM.

83. (1) In every right-angled triangle when its three sides are in Arithmetical progression, they may be shewn to be as the numbers 5, 4, 3. On the given line AC describe a triangle having its sides AC, AD, DC in this proportion, bisect the angles at A, C by AE, CE meeting in E, and through E draw EF, EG parallel to AD, DC meeting in F and G.

(2) Let AC be the sum of the sides of the triangle, fig. Euc. v. 13. Upon AC describe a triangle ADC whose sides shall be in continued proportion. Bisect the angles at A and C by two lines meeting in E. From E draw EF, EG parallel to DA, DC respectively.

84. Describe a circle with any radius, and draw within it the straight line MN cutting off a segment containing an angle equal to the given angle, Euc. 111. 34. Divide MN in the given ratio in P, and at P draw PA perpendicular to MN and meeting the circumference in A. Join

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