PROPOSITION VIII. THEOREM. 286. Two triangles which have their sides respectively perpendicular to each other are similar. In the triangles EFD and BA C, let E F, FD and E D, be perpendicular respectively to A C, BC and A B. We are to prove A EFD and BA C similar. Produce FD until it meets B C at H. Then, in the quadrilateral B E DH, (in a rt. ▲ the sum of the two acute = a rt. <); and .. LEFD = LC. LEFD + HFC = a rt. Z. ..A EFD and B A C are similar, (two are similar when two of the one are equal respectively to two of Ax. 9. Ax. 3. § 280 the other). Q. E. D. 287. SCHOLIUM. When two triangles have their sides respectively parallel or perpendicular, the parallel sides, or the perpendicular sides, are homologous. PROPOSITION IX. THEOREM. 288. Lines drawn through the vertex of a triangle divide proportionally the base and its parallel. In the triangle ABC let HL be parallel to A C, and let BS and BT be lines drawn through its vertex to the base. (two A which are mutually equiangular are similar). Ex. Show that, if three or more non-parallel straight lines divide two parallels proportionally, they pass through a common 289. If in a right triangle a perpendicular be drawn from the vertex of the right angle to the hypotenuse : I. It divides the triangle into two right triangles which are similar to the whole triangle, and also to each other. II. The perpendicular is a mean proportional between the segments of the hypotenuse. III. Each side of the right triangle is a mean proportional between the hypotenuse and its adjacent segment. IV. The squares on the two sides of the right triangle have the same ratio as the adjacent segments of the hypote nuse. V. The square on the hypotenuse has the same ratio to the square on either side as the hypotenuse has to the segment adjacent to that side. In the right triangle ABC, let BF be drawn from the vertex of the right angle B, perpendicular to the hypotenuse A C. I. We are to prove the AABF, ABC, and FBC similar. In the rt. A BAF and BA C, the acute A is common. Now as the rt. A ABF and CBF are both similar to ABC, by reason of the equality of their 4, they are similar to each other. II. We are to prove AF BF:: BF: FC. In the similar ▲ A B F and C B F, A F, the shortest side of the one, : FC, the medium side of the other. : III. We are to prove A C : AB AB: AF. A C, the longest side of the one, A C, the longest side of the one, $259 $ 259 In the proportion AC AB::AB: AF, BC2 FC ABACX AF, (the product of the extremes is equal to the product of the means). and in the proportion AC: BC :: BC: FC, Cancel the common factor A C, and we have PROPOSITION XI. THEOREM. 290. If two chords intersect each other in a circle, their segments are reciprocally proportional. Let the two chords AB and EF intersect at the (two are similar when two of the one are equal to two of the other). Whence A O, the medium side of the one, § 278 Q. E. D. |