Elements of Geometry

(69), and must coincide if placed on the same side of the common base.

31. The angles ABA', ACA' can easily be shown to be the sums, or the differences, of equal angles.

32. The exterior angles ABD, ACE are equal (44), and the including sides are equal (Hyp. and Const.); hence ▲ ABD = ^ ACE; etc. 33. The interior angles adjacent to the equal exterior angles are equal (44); hence the opposite sides are equal.

34. This is a special case of Prop. VIII.

35. This is a special case of Prop. VI.

36. Since AD=AC, and BE=BC, AD+AB+BE=AC+AB+BC. 37. CAD = ≤ CBE, CA, AD, = CB, BE respectively; etc. Or show, as in Ex. 32, that CD = CE; etc.

38. AG = AB (75), GE = GB + BE = AG+2 AG = 3 AG. 39. Join AB and produce AB to C so that AC = 2 AB. With A and B as centers, and radius equal to AC, describe circumferences intersecting in X; then AX= BX = 2 AB.

40. Bisect AB in D, then DC in E, and proceed with AE as with AC in Ex. 39.

41. Since the line drawn through X and Y is perpendicular to AB at its mid point (75), it must coincide with the perpendicular drawn to AB at that point (41).

42. ZBDEL CED, they being supplements of equal angles ADE, AED.

43. After bisecting ABC by BF (81), bisect each of the angles ABF, CBF.

44. F might lie between A and DE, might coincide with A, might be in the prolongation of FA. If F coincides with A the one point cannot determine the required bisector.

45. AE, EF would be equal to AD, DF respectively, and AF be

common ; hence the

AEF, ADF would be equal, and ▲ FAE = / FAD (70).

46. The APB, APC, BQA, BQC, CRA, CRB, are equal (66); hence AP BQ, etc.; ZAPB = ZAPC, etc.; ≤ BAP = CAP, etc. 47. Proceed as in Art. 83.

48. In the first case, the perpendicular cannot fall without the triangle, since then an acute angle would be greater than a right angle

(83). In the second case the perpendicular cannot fall within the triangle, since then a right angle would be greater than an obtuse angle. 49. Proof much the same as in Art. 86.

50. Since the greatest side lies opposite the greatest angle, the angles that include the greatest side must be acute; hence by the first case of Ex. 48, the perpendicular

must fall within the triangle.

51. Since BC is the greatest side, the perpendicular to BC falls within the triangle (Ex. 48); then AB > BD and AC > DC; AB+ AC > BC.

52. Let AM be the median drawn to BC. Since BM, MA = CM, MA respectively, but AC>AB, LAMC> LAMB (91). Join any point P between A and M with B and C. Then BM, B MP=CM, MP respectively, but ▲AMC > LAMB; .. PC> PB (90).

53. At the given point make a right angle (95) and bisect it (81). 54. From the given point P, draw a perpendicular PQ to the given line (92), and at P draw PS, making half a right angle with PQ (Ex. 53). Let PS meet the given line in S; PSQ is the required angle (123). 55. CPA, CPB are the supplements of equal & APD, BPD. 56. The distance of P from A or B would have to be equal to AB. 57. Since only two equal lines can be drawn from the point to the given line, there can be only two that make equal angles with it. 58. The two oblique lines form with the given line a triangle, of which the greater side is opposite the greater angle.

59. CD may be regarded as the bisector of the st. ZADB.

60. For, if it were equally distant from the sides, it would be in the bisectors of the angle (101′′).

61. Any point unequally distant from the sides of an angle is not in the bisector of that angle. For if it were in the bisector, it would be equidistant from those sides.

62. The perpendicular forms with the sides of the angle and its bisector two triangles that are equal (63); hence the angles opposite the bisector are equal (70).

63. Any line that makes equal angles with the sides of a given angle is perpendicular to its bisector. For it forms with the sides and bisector of the angle two equal triangles (63).

64. (1) If the sides containing the angle are equal, the bisector is perpendicular to the third side at its mid point (77). (2) If the sides are unequal, the bisector cannot pass through the mid point of the third side, since then by drawing perpendiculars from that point to those sides, they could be proved equal, as the perpendiculars are equal.

65. If PF, QE be drawn, ▲ POF =AQOE (66);

whence QPF = LPQE, and PF is || to QE.

66. The three equal angles are together equal to a straight angle (120), hence each is equal to one third of a straight angle.

67. Let A be the vertical angle, B and C the equal base angles. Then ZA+B+ZC=ZA +2 ZB = 2 ZA=2 right angles; ZA is equal to a right angle.

68. With the same symbol as above for the angles, we have

ZA+ZB+2 C = ZA+2ZB = ZA+ 2 n ZA = (2 n + 1)≤ A = a

straight angle; ...

LA=

1 2n+1

straight angles.

69. If <A>ZB+ZC, 2ZA>ZA+ZB + C > 2 right angles; ... <A> a right angle. If ZA<ZB+Z C, we show similarly that A<a right angle.

E C

LEAB;

70. Let the bisectors of A and C cut CD, AB in E, F respectively. Since ZEAB = LECF (Ax. 7), and /ECF = /CFB (110), /CFB= CF is parallel to EA (112).

71. Let the diagonals AC, BD bisect each other in O. Then BC and LOAD = 2 OCB; .. AD

▲ AOD=▲ COB; AD =

is parallel to BC; ... ABCD is a parallelogram (142).

72. If A is supplementary to B, then AD is parallel to BC; and if ZB is supplementary to ≤ C, then AB is parallel to CD, and ABCD is a parallelogram.

73. If ▲ A = ZA', then also B = Z B' (113,44); hence C = C', and D=LD' (136).

74. Let the equal diagonals intersect in O. Since OA=0B=0C=OD, and OAB = ZOBA, also LOAD = 2 OCB = ≤ OBC, ▲ BAD= LABC= a right angle; etc.

75. If the diagonal AC bisects A and C, so that ▲ BAC=/ BCA, then BA = BC; etc.

76. For then the opposite angle is a right angle (136), whence the remaining angles are also right angles (113).

77. Let the diagonals AC, BD intersect in O. Since OA = OC, OB = OD (146), and ZAOB = ≤ DOC, ▲ AOB = ▲ COD; etc.

78. Let the bisectors of A and B meet in O. Since ZA+ZB=2 rt. 4, ZOAB + Z OBA: = art. 2; .. ≤ AOB = a rt. ≤ (121).

79. The figure formed will be a quadrilateral whose diagonals bisect each other, hence it is a parallelogram (Ex. 71). If AB is perpendicular to CD, the parallelogram will be equilateral.

80. Since AC = AD = CB = DB (Const.), the figure formed is equilateral; and if EA: = EC, the figure is a square (Ex. 74).

81. A square or a rhombus, according as the vertical angle is or is not a right angle.

84. If n be the number of sides of the polygon, an interior angle

85. Since

ABD = ^ CBD (140), and ▲ BOE = ^ DOF, OEAD

= OFCB (Ax. 3).

86. AABC can be superposed on its equal ▲ DCB; then as ▲ CEO will coincide with its equal ▲ BFO, OEAB will coincide with OFDC. 87. Draw BD cutting FH in O. BF=DH (149), Z OBF=ZODH, OF = OH, and OB = OD (63);

and

OFB:

ZOHD (110); ..

hence the diagonal CG will also pass through O (146).

88. If AD BC, then AD = DG (136), then Z A = ≤ DGA = Z B. 89. The CBK, DAG are easily proved equal, whence BK = AG. Now AK+ GB = 2 DC = AB + GK ; .. GK 2 DC – AB.

90. BQ BF BC (149). Also PQ (AB + EF)

= {{AB + } (AB + CD)} = } (} AB + } CD) = } (3 AB + CD).

QUESTIONS, p. 69.

91. Such a triangle is rigid because its three sides determine a triangle (69); i.e., the sides being given, the angles are also given. But any polygon of more than three sides can be deformed, that is made to change form while the length of its sides remains unchanged. This principle, so important in mechanical construction, may be illustrated by means of the wooden frame of a slate or similar contrivance.

92. No; illustrate by drawing an intercept parallel to any side of a given triangle.

93. (1) Erect a perpendicular to either side at the vertex. (2) Produce either side.

94. By a right angle.

95. (1) 54°, 45°, 0°. (2) 144°, 135°, 90°.

GEOM. KEY. -2