Proof. DE is parallel and equal to JAC. (147) FG is parallel and equal to LAC. (147) .. DE is parallel and equal to FG. (Axs. 10 and 1) :: ZEDO= LOGF, Ex. 228. Demonstrate the proposition by proving the equality of A DOF and AEOG. (Draw OB.) Ex. 229. Prove the proposition by showing that DFGE is a parallelogram. Ex. 230. Prove the proposition by means of 1(b) of analysis (three methods). PROPOSITION XLIV. THEOREM 159. The three altitudes of a triangle (produced if necessary) meet in a common point. To prove Hyp. In A ABC ADI BC, BEI AC, CHL AB. O is common to AD, BE, CH. Proof. Draw A'B' | AB through C; B'C" || BC through A; AC" || AC through B. ADI BC (Hyp.), .. ADI B'C". (88) But ABCB' and ACBC" are parallelograms. (Con.) .:. AB' = BC, and AC" = BC. (134) .. AB' = AC" (Ax. 1) . . AD is the perpendicular-bisector of C'B'. [The rest of the proof is left to the student.] Ex. 231. In what kind of a triangle is the point of intersection of the altitudes of a triangle within the triangle ? without ? in one side ? Ex. 232. In what kind of a triangle do altitudes and perpendicularbisectors coincide ? F PROPOSITION XLV. THEOREM 160. The medians of a triangle meet in a common point. To prove Нур. . AE, BF, CD, are the medians of A ABC. AE, BF, CD, meet in a point. CO= 2 (DO). (158) Q.E.D. Ex. 233. If the two medians of a triangle are equal, the triangle is isosceles. B Ex. 234. Given the vertical angle B of an isosceles triangle, required ZDAC, formed by the base and the altitude upon one arm, in terms of B. 20+2 CAB = 2 rt. És – ZB, (Why?) but ZCEL CAB. (Why?) B 2 B В 2 2 Therefore the angle formed by the base and the altitude upon an arm of an isosceles triangle equals one-half the vertical angle. 161. REMARK. Geometrical propositions may be demonstrated by algebraical computation. Although a direct geometrical proof is always preferable, the above method is very useful, especially for establishing relations between angles. The propositions to be used are mainly the proposition of the sum of the angles of a polygon (and triangle), and the proposition of the exterior angles of a polygon (and triangle). If the algebraical solution should be too difficult for the beginner, numerical examples ought to be solved at first. А Ex. 238. If AD is the altitude and AE the bisector of the angle BAC of the triangle ABC, prove Z DAE = } (ZB-ZC). Ex. 239. The sum of three angles of a quadrilateral, diminished by the fourth exterior angle, is equal to a straight angle. Ex. 240. The bisector of two exterior angles of a triangle include an angle equal to one-half the third exterior angle. (Hint. — Express every angle in terms of Z A and Z B.) Ex. 241. The bisectors of the angles of a quadrilateral form a quadrilateral the sum of whose opposite angles is equal to two right angles. * Ex. 242. Prove Exs. 237 and 238 by a purely geometrical method. MISCELLANEOUS EXERCISES Ex. 243. If the bisectors of two adjacent angles are perpendicular to each other, the angles are supplementary. Ex. 244. The bisectors of vertical angles are in a straight line. Ex. 245. Perpendiculars drawn from a point within an angle, upon the sides, include an angle which is the supplement of the given angle. Ex. 246. If the vertical angles of two isosceles triangles are supplementary, the base angles are complementary. Ex. 247. If the base angles of two isosceles triangles are complementary, the vertical angles are supplementary. Ex. 248. The exterior angle at the base of an isosceles triangle is equal to a right angle increased by one-half the vertical angle. |