PROPOSITION XXXIV. THEOREM. 118. Of two angles of a triangle, that is the greater which is opposite the greater side. In the triangle A B C let A B be greater than AC. We are to prove LACB> LB. Draw EC. § 112 LAEC = LACE, (being & opposite equal sides). But LAEC > LB, $ 105 (an exterior 2 of a A is greater than either opposite interior 2), Ex. If the angles ABC and ACB, at the base of an isosceles triangle, be bisected by the straight lines BD, CD, show that D B C will be an isosceles triangle. PROPOSITION XXXV. THEOREM. 119. The three bisectors of the three angles of a triangle meet in a point. B Let the two bisectors of the angles A and C meet at 0, and O B be drawn. OK, We are to prove BO bisects the < B. Draw the Is OK, O P, and O H. Iden. Cons. $ 110 (having the hypotenuse and an acute < of the one equal respectively to the hypotenuse and an acute 2 of the other). ..OP (homologous sides of equal ). Iden. Cons. § 110 (having the hypotenuse and an acute 2 of the one equal respectively to the hypotenuse and an acute Z of the other). .. OP: Ax. 1 OH, 1 OH= 0 K, and OB= 0 B, $ 109 (having the hypotenuse and a side of the one equal respectively to the hypote nuse and a side of the other), Q. E. D. PROPOSITION XXXVI. - THEOREM. 120. The three perpendiculars erected at the middle points of the three sides of a triangle meet in a point. A Let D D, E E, F F, be three perpendiculars erected at D, E, F, the middle points of A B, A C, and BC. We are to prove they meet in some point, as 0. The two Is D D' and E E' meet, otherwise they would be parallel, and A B and A C, being is to these lines from the same point A, would be in the same straight line; but this is impossible, since they are sides of a A. Then, since 0 is in D D', which is I to A B at its middle point, it is equally distant from A and B. Also, since O is in E E', I to A C at its middle point, it is equally distant from A and C. $ 59 ..O is equally distant from B and C; :: 0 is in F F' I to B C at its middle point, $ 59 (the locus of all points equally distant from the extremities of a straight line is the I erected at the middle of that line). Q. E. D. PROPOSITION XXXVII. THEOREM. 121. The three perpendiculars from the vertices of a triangle to the opposite sides meet in a point. B! In the triangle A BC, let B P, A H, CK, be the per pendiculars from the vertices to the opposite sides. We are to prove they meet in some point, as 0. ..Δ Α Β Α' § 107 (having a side and two adj. s of the one equal respectively to a side and two adj. s of the other). .:. A B AC, .:. ДС В С. A ABC, $ 107 (having a side and two adj. of the one equal respectively to a side and tuo adj. e of the other). it is I to A' C', $ 67 (a straight line which is I to one of two lls is I to the other also). But B is the middle point of A'C'; .. BP is I to A' C' at its middle point. In like manner we may prove that A H is I to A' B' at its middle point, and C KI to B'C' at its middle point. .. BP, A H, and C K are Is erected at the middle points of the sides of the A A' B'C'. ::. these Is meet in a point. (the three Is erected at the middle points of the sides of a A meet in a point). § 120 Q. E. D. |