PROPOSITION X. THEOREM. 329. Straight lines drawn through the same point intercept proportional segments upon two parallels. Let the two parallels AE and A'E' cut the straight Lines OA, OB, OC, OD, and OE. Proof. Since A'E' is to AE, the pairs of A OAB and OA'B', OBC and OB'C', etc., are mutually equiangular and similar, AB OB and A'B' OB' BC OB (homologous sides of similar ▲ are proportional). REMARK. A condensed form of writing the above is § 319 Ax. 1 Q. E. D. where a parenthesis about a ratio signifies that this ratio is used to prove the equality of the ratios immediately preceding and following it. PROPOSITION XI. THEOREM. 330. CONVERSELY: If three or more non-parallel straight lines intercept proportional segments upon two parallels, they pass through a common point. Let AB, CD, EF, cut the parallels AE and BF so that AC: BD CE : DF. To prove that AB, CD, EF prolonged meet in a point. Proof. Prolong AB and CD until they meet in O. Join OE. If we designate by F the point where OE cuts BF, we shall have by $ 329, But by hypothesis AC: BD CE: DF'. AC: BDCE: DF These proportions have the first three terms equal, each to each; therefore the fourth terms are equal; that is, DFDF .. F' coincides with F .. EF prolonged passes through O. .. AB, CD, and EF prolonged meet in the point O. Q. E. D. 331. If two polygons are composed of the same number of triangles, similar each to each, and similarly placed, the polygons are similar. In the two polygons ABCDE and A'B'C'D'E', let the triangles AEB, BEC, CED be similar respectively to the triangles A'E'B', B'E'C', C'E'D'. In like manner we may prove BCD = Z B'C'D', etc. Hence the two polygons are mutually equiangular. (the homologous sides of similar are proportional). Hence the homologous sides of the polygons are proportional. Therefore the polygons are similar, § 319 (having their homologous equal, and their homologous sides proportional). Q. E D. PROPOSITION XIII. THEOREM. 332. If two polygons are similar, they are composed of the same number of triangles, similar each to each, and similarly placed. Let the polygons ABCDE and A'B'C'D'E' be similar. From two homologous vertices, as E and E', draw diagonals EB, EC, and E'B', E'C'. To prove A EAB, EBC, ECD similar respectively to ▲ E'A'B', E'B'C', E'C'D'. Proof. In the ▲ EAB and E'A'B', (being homologous sides of similar polygons). (having an of the one equal to an ▲ of the other, and the including sides § 326 proportional). (having an of the one equal to an ▲ of the other, and the including sides proportional). In like manner we may prove ▲ ECD and E'C'D' similar. PROPOSITION XIV. THEOREM. Q. E. D. 333. The perimeters of two similar polygons have the same ratio as any two homologous sides. Let the two similar polygons be ABCDE and A'B'C'D'E', and let P and P' represent their perimeters. AB: A'B' BC: B'C' CD: C'D', etc., = = (the homologous sides of similar polygons are proportional). = § 319 .. AB+ BC, etc. : A'B' + B'C', etc. AB: A'B', § 303 (in a series of equal ratios the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent). That is, P: P AB: A'B'. Q. E D. |