THEOREM XV. 154. Of any two obliques between a given point and line, that which makes the greater sect from the foot of the perpendicular is the greater. PROOF. By 153, take F on the same side of the perpendicular as B; then, in ▲ AFC, because, by hypothesis, ☀ ACF is rt., :. ¥ AFC <rt. ¥, (143. Any two angles of a triangle are together less than a straight angle.) (143. Any two angles of a triangle are together less than a straight angle.) .. AF < AB. (148. If angle A is less than angle B, therefore a is less than b.) 155. Corollary. No more than two of all the sects can be equal. PROOF. For no two sects on the same side of the perpendicular can be equal. THEOREM XVI. 156. Any two sides of a triangle are together greater than the third side. PROOF. By 101, produce BC to D; making, by 133, CD = b. By 100, join AD. Similarly it may be shown that a + c>b, and that b + c>a. 157. COROLLARY. Any side of a triangle is greater than the difference between the other two sides. THEOREM XVII. 158. If from the ends of one of the sides of a triangle two sects be drawn to a point within the triangle, these will be together less than the other two sides of the triangle, but will contain a greater angle. HYPOTHESIS. D is a point within ▲ ABC. CONCLUSIONS. (I.) AB + BC>AD + DC. (II.) X ADC >B. PROOF. (I.) Join AD and CD. Produce CD to meet AB in F. Then (CB+BF)> (CF), and [DF+ FA]> [DA]. .. (156. In a triangle any two sides are together greater than the third.) CB + BA = (CB + BF) + FA> (CF) + FA (II.) Next, in ▲ AFD = CD + [DF + FA]> CD + [DA]. × ADC > × AFD. (142. Exterior angle of a triangle is greater than opposite interior angle.) And for the same reason, in ▲ CFB EXERCISES. X AFD> B, :: ¥ ADC>& B. 19. The perimeter of a triangle is greater than the sum of sects from a point within to the vertices. THEOREM XVIII. 159. If two triangles have two sides of the one equal respectively to two sides of the other, but the included angle of the first greater than the included angle of the second, then the third side of the first will be greater than the third side of the second. HYPOTHESIS. ABC and DEF are two triangles, in which PROOF. Place the triangles so that EF shall coincide with BC, and the point D fall on the same side of BC as A. By 134, bisect X ABD, and let G be the point in which the bisector meets AC. By 100, join DG. Then, in the triangles ABG and BDG, by construction, & ABG = DBG; by hypothesis, AB = BD; and BG is common. :. Δ ABG = Δ DBG, (124. Two triangles are congruent if two sides and the included angle are equal in But CG + GD > CD, (156. Any two sides of a triangle are together greater than the third.) :. CA>FD. 160. COROLLARY. If cf, a = d, and & B< E, then E> B; and so, by 159, 161. INVERSES. We have now proved, when a = d, and EXERCISES. 20. In ▲ ABC, AB < AC. D is the middle point of BC. Prove the angle ADB acute. |
