but the angle APB is greater than the angle PQB; much more then is the angle APB greater than the angle АСВ*, Q. E. D. THEOREM 14. If two triangles have two sides of the one equal to two sides of the other, each to each, but the included angles unequal, their bases are unequal, the base of that which has the greater angle being greater than the base of the other t. • Part. En. Let ABC, DEF be two triangles, having AB=DE, AC=DF, but the included angle BAC greater than the included angle EDF; it is required to prove that the base BC is greater than the base EF. Proof. Place the point A on D, and AB along DE; then because AB = DE, (Hyp.) therefore the point B will fall on the point E; and because the angle BAC is greater than the angle EDF, the line AC will fall outside DF, as DG; (Hyp.) and BC will fall as EG. Let DH be the bisector of the angle FDG, (Ax. 4.) meeting EG in H. * Euclid, I. 21. + Euclid, 1. 24. Join FH. (Hyp.)) DH common, and the included angle FDH= the included angle GDH, (Constr.) therefore HF=HG; (Th. 5.) and therefore EH and HF together are equal to EG. But EH and HF together are greater than EF; (Th. 12.) therefore EG or BC is greater than EF. Q. E. D. THEOREM 15. If two triangles have the three sides of the one equal to the three sides of the other, each to each, then the triangles are identically equal, and of the angles those are equal which are opposite to equal sides * Part. En. Let ABC, DEF be two triangles which have AB=DE) CA= FD,) Proof. The angle BAC must be either equal to EDF, or greater than EDF, or less than EDF. But BAC is not greater than EDF; for then the base BC would be greater than the base EF. (Th. 14.) Nor is BAC less than EDF; for then the base BC would be less than the base EF: (Th. 14.) therefore the angle BAC is equal to the angle EDF. Q. E. D. * Eucl. 1. 8. Alternative Proof. If the point B were placed on E, and BC were placed along EF, then because BC= EF (Hyp.), therefore the point C would fall on F: and let the triangles BAC, EDF fall on opposite sides of EF; BA, AC falling as EG, GF, and the angle BAC as EGF. Join DG. Then because EG=ED, (Hyp.) therefore the angle EDG= the angle EGD: (Th. 6.) and because FG=FD, (Hyp.) therefore the angle FDG = the angle FGD: (Th. 6.) therefore the whole angle EDF=the whole angle EGF; EGF= BAC; (Constr.) therefore EDF= BAC; and therefore the triangles BAC, EDF are equal in all respects. (Th. 5.) Q. E. D. NOTE. The student should examine for himself the cases in which DG passes through an extremity of the base, and passes outside the base. but THEOREM 16. If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the one is greater than the base of the other, then the angle contained by the sides of that which has the greater base is greater than the angle contained by the sides of the other. Inst. Im E 34. JESE . Dries va 3.. =)E. F ZIO rungtes teenHg3 44 RUE ZEDE Ingres. I te ther. 00:13 a us, vous coco 24 dies prostit ne i ne EULE Pgdes i bla by 46% te "snigte Ara anai espects to Rec Billig flot uredu, topposite Part. En. Let the two triangles ABC, DEF have the two angles ABC=DEF, ACB=DFE, it is required to prove the triangles are equal in all respects. Proof. Let the point A be placed on the point D, and AB along DE, then because AB=DE, (Hyp.) therefore the point B will fall on the point E. And because the angle ABC is equal to the angle DEF, (Hyp.) therefore the line BC will lie along EF. And the point C will fall on F, for if it fell otherwise as G, then, since the angle ACB is equal to the angle EFD, (Hyp.) the angle EGD would be equal to the angle EFD, the exterior angle equal to the interior and opposite, which is impossible; (Th. 9.) therefore the triangles would coincide and are equal in all respects, AC being equal to DF, BC to EF, and the angle BAC to the angle EDF* Q. E. D. * Eucl. 1. 26. Part 2. |