Then GD and EH bisect each other at O. [The diagonals of a □ bisect each other.] (§ 111) But by hyp., G is the middle point of OA, and H of OB. OD, and BH = OH = OE. .. AG = OG = That is, AD and BE intersect at a point O which lies two-thirds the way from A to D, and from B to E. In like manner, AD and CF intersect at a point which lies two-thirds the way from A to D, and from C to F. Hence, AD, BE, and CF intersect at the common point O, which lies two-thirds the way from each vertex to the middle point of the opposite side. LOCI. 141. Def. If a series of points, all of which satisfy a certain condition, lie in a certain line, and every point in this line satisfies the given condition, the line is said to be the locus of the points. For example, every point which satisfies the condition of being equally distant from the extremities of a straight line, lies in the perpendicular erected at the middle point of the line (§ 42). Also, every point in the perpendicular erected at the middle point of a line satisfies the condition of being equally distant from the extremities of the line (§ 41). Hence, the perpendicular erected at the middle point of a straight line is the Locus of points which are equally distant from the extremities of the line. Again, every point which satisfies the condition of being within an angle, and equally distant from its sides, lies in the bisector of the angle (§ 102). Also, every point in the bisector of an angle satisfies the condition of being equally distant from its sides (§ 101). Hence, the bisector of an angle is the locus of points which are within the angle, and equally distant from its sides. EXERCISES. 51. Two straight lines are parallel if any two points of either are equally distant from the other. (Prove by Reductio ad Absurdum.) 52. What is the locus of points at a given distance from a given straight line? (Ex. 51.) 53. What is the locus of points equally distant from a pair of intersecting straight lines? 54. What is the locus of points equally distant from a pair of parallel straight lines? 55. The bisectors of the interior angles of a trapezoid form a quadrilateral, two of whose angles are right angles. (Ex. 46.) 56. If the angles at the base of a trapezoid are equal, the non-parallel sides are also equal. (Given A = 4D; to prove AB = CD. Draw BE || CD.) B H A 的 D B A E D 57. If the non-parallel sides of a trapezoid are equal, the angles which they make with the bases are equal. (Fig. of Ex. 56. Given AB = CD; to prove ▲A=LD, and also LABCLC. Draw BE || CD.) 58. The perpendiculars from the extremities of the base of an isosceles triangle to the opposite sides are equal. A E B 59. If the perpendiculars from the extremities of the base of a triangle to the opposite sides are equal, the triangle is isosceles. (Converse of Ex. 58. Prove ▲ ACD = ▲ BCE.) 60. The angle between the bisectors of the equal angles of an isosceles triangle is equal to the exterior angle at the base of the triangle. (4 ADB = 180° — (▲ BAD + ▲ ABD).) 61. If a line joining two parallels be bisected, any line drawn through the point of bisection and included between the parallels will be bisected at the point. (To prove that GH is bisected at 0.) E 62. If through a point midway between two parallels two transversals be drawn, they intercept equal portions of the parallels. (Draw OK Then at L. AB, and produce KO to meet CD 63. If perpendiculars BE and DF be drawn from vertices B and D of parallelogram ABCD to the diagonal AC, prove BE = DF. (§ 70.) 64. The lines joining the middle points of the sides of a triangle divide it into four equal triangles. (§ 130.) B 65. If from any point in the base of an isosceles triangle parallels to the equal sides be drawn, the perimeter of the parallelogram formed is equal to the sum of the equal sides of the triangle. (§ 96.) 66. The bisector of the exterior angle at the vertex of an isosceles triangle is parallel to the base. (§ 85, 1.) 67. The medians drawn from the extremities of the base of an isosceles triangle are equal. 68. If from the vertex of one of the equal angles of an isosceles triangle a perpendicular be drawn to the opposite side, it makes with the base an angle equal to one-half the vertical angle of the triangle. (To prove BAD = } ≤ C.) 69. If the exterior angles at the vertices A and B of triangle ABC are bisected by lines which meet at D, prove 70. The diagonals of a rhombus bisect its angles. (Fig. of Prop. XLIV.) 71. If from any point in the bisector of an angle a parallel to one of the sides be drawn, the bisector, the parallel, and the remaining side form an isosceles triangle. 72. If the bisectors of the equal angles of an isosceles triangle meet the equal sides at D and E, prove DE parallel to the base of the triangle. (Prove ▲ CED isosceles.) 73. If at any point D in one of the equal sides AB of isosceles triangle ABC, DE be drawn perpendicular to base BC meeting CA produced at E, prove triangle ADE isosceles. 74. From C, one of the extremities of the base BC of isosceles triangle ABC, a line is drawn meeting BA produced at D, making AD = AB. Prove CD perpendicular to BC. (§ 84.) (▲ ACD is isosceles.) 75. If the non-parallel sides of a trapezoid are equal, its diagonals are also equal. (Ex. 57.) 76. If ADC is a re-entrant angle of quadrilateral ABCD, prove that angle ADC, exterior to the figure, is equal to the sum of interior angles A, B, and C. (§ 128.) B B 77. If a diagonal of a quadrilateral bisects two of its angles, it is perpendicular to the other diagonal. A (Prove ACL DB, by § 43.) 78. In a quadrilateral ABCD, angles ABD and CAD are equal to ACD and BDA, respectively; prove BC parallel to AD. B (Prove AB = CD; then prove BE = CF.) E 79. State and prove the converse of Prop. XLIV. (§ 41, I.) B D F 80. State and prove the converse of Ex. 66, p. 67. (§ 96.) 81. The bisectors of the exterior angles at two vertices of a triangle, and the bisector of the interior angle at the third vertex meet at a common point. (Prove as in § 134.) 82. ABCD is a trapezoid whose parallel sides AD and BC are perpendicular to CD. If E is the middle point of AB, prove EC=ED. (§ 41, I.) (Draw EF || AD.) 83. The middle point of the hypotenuse of a right triangle is equally distant from the vertices of the triangle. (To prove AD=BD=CD. Draw DE || BC.) 84. The bisectors of the angles of a rectangle B form a square. (By Ex. 50, EFGH is a rectangle. Now prove AF BH and AE = BE.) = 85. If D is the middle point of side BC of triangle ABC, and BE and CF are perpendiculars from B and C to AD, produced if necessary, prove BE = CF. 86. The angle at the vertex of isosceles triangle ABC is equal to twice the sum of the equal angles B and C. If CD be drawn perpendicular to BC, meeting BA produced at D, prove triangle ACD equilateral. (Prove each of ▲ ACD equal to 60°.) A A B 87. If angle B of triangle ABC is greater than angle C, and BD be drawn to AC making AD = AB, prove ZADB = 1 (B + C), and ≤ CBD = } (B — C). (Fig. of Prop. XXXII.) 88. How many sides are there in the polygon the sum of whose interior angles exceeds the sum of its exterior angles by 540° ? |