SECTION II.-Relations between the Sides and Angles of a Triangle. 9. The bisector of the angle at the vertex of an isosceles triangle bisects the base and is perpendicular to the base. Let AD bisect the angle B AC. Then the triangles A B D, ACD have AB = AC, ABD = < ACD, and BAD = 4 CAD. Therefore the triangles are equal (Theorem VI.) and BD = CD and LADB 4 ADC. = B D A C That is to say, A D bisects B C and the adjacent angles at D are equal. Hence each of them is a right angle. 10. The straight line which is drawn from the vertex of an isosceles triangle so as to bisect the base will also bisect the angle at the vertex and be perpendicular to the base. Here the triangles A D B, A DC have two sides and the included angle of the one respectively equal to two sides and the included angle of the other—that is, A B AC, BD = C D, and ▲ B = C. Therefore ▲ BAD = 4 CAD and ADB = 4 ADC, and consequently AD is perpendicular to B C. 11. The perpendicular drawn to the base of an isosceles triangle from its vertex will bisect the base as well as the angle at the vertex. Let A B C be the isosceles triangle and AD the perpendicular from the vertex A to the base; then 4 BAD 4 CAD. In Example 10 it is proved that the straight line which joins A with the middle point of B C is perpendicular to B C; and in Example 8 it is proved that only one perpendicular can be drawn from A to B C. Hence the perpendicular from A to B C is also the line from A which bisects the base. Consequently BD CD and 4 BAD= CAD (Example 10). 12. The perpendicular to the base of an isosceles triangle drawn from the middle point of the base will pass through the vertex and bisect the angle at the vertex. Let AD be the line joining the point A with the middle point, D, of the base, BC, of the isosceles triangle ABC. Then it may be proved, as in Example 10, that AD is perpendicular to BC and bisects BAC. But from the point D in B C it is only possible to draw one perpendicular to BC. Therefore A D, the bisector of the angle B A C, is the perpendicular at the middle point, D, of B C. Consequently the perpendicular to BC at the middle point is the bisector of ▲ BAC. 13. When two isosceles triangles stand on the same base, the straight line which passes through their vertices will bisect the base and be perpendicular to it. Let A B C and A D C be isosceles triangles, and let E be the middle point of the base AC. Join B, E; then, by Example 10, B E is perpendicular to A C at the point E. E Similarly, if we join D, E, the line D E will be perpendicular to AC at the point E. But since there can be but one perpendicular to AC at the point E, this perpendicular passes through both B and D SECTION III.-Relations between the Sides of a Triangle. 14. If from a point within a triangle two straight lines be drawn to the extremities of the base, these straight lines shall be together less than the two sides of the triangle. Let A B C be the triangle and D the point within it; then AD + DB will be less in the side B C. B than A C + C B. But & ADB will be greater than AC B. Continue A D to the point E Now AC+CE > AE. Add E B; ... AC+ CB > AEE B. Add A D; ... AE + EB > AD + DB. But we have shown that AC+ CB > AE + EB; .. ACCB > AD + D B. 15. Two villages, A and B, are connected by straight roads with two others, C and D. The roads AD and C B intersect, but A C and D B do not. Prove that the roads which intersect are together longer than those which do not. Let AD and CB inter sect in O. .. AO+OC > AC; DO + OB > D B. By addition, ..AD+BC>AC+D B. B 16. The line which joins the vertex to the middle point of the base of a triangle is less than half the sum of the two sides. Let A B C be the triangle and A M the line joining A with the middle point, M, of the base B C. Then we have to prove that A M the half of AB+ AC. B Continue A M to D and make M D = join D C. In the triangles A M B, D M C, M A M A, and CM = BM, A M = D M, and ▲ AMB= 4 DM C. Therefore the triangles are equal in all respects, and A B = D C ; .'. BA + A C = DC + AC. But DC + AC is greater than A D. Consequently BA + CA is greater than AD; that is, greater than twice A M. |