line bisecting the angle opposite the base of an isosceles triangle bisects the base at right angles and also bisects the triangle; also the line drawn from the vertex perpendicular to the base of an isosceles triangle bisects the base, the vertical angle, and the triangle. And, conversely, the perpendicular bisecting the base of an isosceles triangle bisects the angle opposite, and also the triangle. 44. Cor. 2. An equilateral triangle is equiangular. THEOREM XI. 45. If two angles of a triangle are equal, the sides opposite are also equal. In the triangle ABC let the angle A equal the angle C; then AB is equal to BC. Bisect the angle ABC by the line BD. Now by hypothesis the angle B D A is equal to the angle C', and by construction the angle A B D is equal to the angle DBC; therefore (35) the angle ADB is equal to the angle BDC; and the two triangles A BD, DBC, having the side BD common and the angles including BD respectively equal, are equal (41) in all respects; therefore AB BC. 46. Corollary. An equiangular triangle is equilateral. THEOREM XII. 47. The greater side of a triangle is opposite the greater angle;" and, conversely, the greater angle is opposite the greater side. In the triangle ABC let B be greater than C; then the side AC is greater than A B. At the point B make the angle CBD equal to the angle C; then (45) B A C D Conversely. Let A C>AB; then the angle ABC > C. For if the angle ABC is not greater than the angle C, it must be either equal to it or less. It cannot be equal, because then the side A B A C (45), which is contrary to the hypothesis. It cannot be less, because then, by the former part of this theorem, AC < A B, which is contrary to the hypothesis. Hence, the angle ABC > C. THEOREM XIII. 48. Two triangles mutually equilateral are equal in all respects. Let the triangle ABC have. A B, BC, CA respectively equal to AD, DC, CA of the triangle ADC; then ABC is equal in all respects to A D C. B D C A D Place the triangle ADC so that the base AC will coincide with its equal AC, but so that the vertex D will be on the side of AC, opposite to B. Join BD. Since by hypothesis ABAD, ABD is an isosceles triangle; and the angle ABD ADB (42); also, since BC CD, BCD is an isosceles triangle; and the angle DBCCDB; therefore the whole angle ABC ADC; therefore the triangles ABC and AD C, having two sides and the included angle of the one equal to two sides and the included angle of the other, are equal (40). 49. Scholium. In equal triangles the equal angles are opposite the equal sides. THEOREM XIV. 50. Two right triangles having the hypothenuse and a side of the one respectively equal to the hypothenuse and a side of the other are equal in all respects. Let A B C have the hypothenuse A B and the side BC equal to the hypothenuse BD and the side BC of BDC; then are the two triangles equal in all A respects. B C Place the triangle BDC so that the side BC will coincide with its equal B C, then CD will be in the same straight line with AC (10). An isosceles triangle A B D is thus formed, and BC being perpendicular to the base divides the triangle into the two equal triangles ABC and BDC (43). THEOREM XV. 51. If from a point without a straight line a perpendicular and oblique lines be drawn to this line, 1st. The perpendicular is shorter than any oblique liné. 2d. Any two oblique lines equally distant from the perpendicu lar are equal. 3d. Of two oblique lines the more remote is the greater. gle A D E being a right angle is greater than the angle A E D; therefore A D A E (47). triangles ADE and ADC, having two sides A D, DE, and the included angle B ADE respectively equal to the two E A sides AD, DC, and the included angle A D C, are equal (40), and AE is equal to A C. 3d. If D BDE; then, as A D E is a right angle, A E D is acute; hence A E B is obtuse, and must therefore be greater than ABE (36); hence A B>A E (47). 52. Corollary. Two equal oblique lines are equally distant from the perpendicular. THEOREM XVI. 53. If at the middle of a straight line a perpendicular is drawn, 1st. Any point in the perpendicular is equally distant from the extremities of the line. 2d. Any point without the perpendicular is unequally distant ́from the same extremities. Let CD be the perpendicular at the middle of the line AB; then 1st. Let D be any point in the perpendicular; draw DA and D B. Since C A = C B, DA DB (51). = 2d. Let E be any point without the perpendicular; draw EA and EB, and from the point D, where E A cuts D C, draw D B. The an gle ABE > ABD=BAD; hence, in the triangle A E B. since the angle ABE > BAE, EA > EB (47). QUADRILATERALS. DEFINITIONS. 54. A Trapezium is a quadrilateral which has no two of its sides parallel; as A B C D. 55. A Trapezoid is a quadrilateral which has only two of its sides parallel; as EFGH. 56. A Parallelogram is a quadrilateral whose opposite sides. are parallel; as IJ KL, or MNOP, or QRST, or U V W X. 57. A Rectangle is a right-angled parallelogram; as IJ KL. 58. A Square is an equilateral rectangle; as M NO P. 59. A Rhomboid is an oblique-angled par allelogram; as QRST. 60. A Rhombus is an equilateral rhomboid; as U V W X. X W 61. A Diagonal is a line joining the vertices of two angles. not adjacent; as DB. |