18. Corollary III. If two straight lines MN, M'N', are intersected by any number of parallels AA', BB', CC', etc., the corresponding segments of the two lines are proportional. For, let the two lines meet in O; then, by If MN and M'N' were parallel, this proportion would still hold, since we should then have AB = A'B', BC= B'C', etc. C D DI BD etc. N B'D' N' PROPOSITION II.-THEOREM. 19. Conversely, if a straight line divides two sides of a triangle proportionally, it is parallel to the third side. Let DE divide the sides AB, AC, of the triangle ABC, proportionally; then, DE is parallel to BC. For, if DE is not parallel to BC, let some other line DE', drawn through D, be parallel to BC. Then, by the preceding theorem, E' D E whence it follows that A.E' = AE, which is impossible unless DE' coincides with DE. Therefore, DE is parallel to BC. 20. Scholium. The converse of (18) is not generally true. PROPOSITION III.-THEOREM. 21. In any triangle, the bisector of an angle, or the bisector of its exterior angle, divides the opposite side, internally or externally, into segments which are proportional to the adjacent sides. D For, through B draw BE parallel to DA, meeting CA produced in E. The angle ABE BAD (I. 49), and the (I. 51); and, by hypothesis, the angle BAD = = angle ABE AEB, and AE AB (I. 90). or = B = E' angle AEB CAD CAD; therefore, the Now, in the triangle CEB, AD being parallel to EB, we have (17), that is, the side BC is divided by AD internally into segments proportional to the adjacent sides AB and AC. 2d. Let AD' bisect the exterior angle BAE; then, D'B: D'CAB: AC. For, draw BE' parallel to D'A; then, ABE' is an isosceles triangle, and AE' = AB. In the triangle CAD', we have (17), or D'B: D'CAE': AC, D'B: D'CAB: AC; that is, the side BC is divided by AD' externally into segments proportional to the adjacent sides AB and AC. 22. Scholium. When a point is taken on a given finite line, or on the line produced, the distances of the point from the extremities of the line are called the segments, internal or external, of the line. The given line is the sum of two internal segments, or the difference of two external segments. 23. Corollary. If a straight line, drawn from the vertex of any angle of a triangle to the opposite side, divides that side internally in the ratio of the other two sides, it is the bisector of the angle; if it divides the opposite side externally in that ratio, it is the bisector of the exterior angle. (To be proved). L SIMILAR POLYGONS. 24. Definitions. Two polygons are similar, when they are mutually equiangular and have their homologous sides proportional. In similar polygons, any points, angles or lines, similarly situated in each, are called homologous. The ratio of a side of one polygon to its homologous side in the other is called the ratio of similitude of the polygons. mar 3 PROPOSITION IV.-THEOREM. 25. Two triangles are similar, when they are mutually equiangular. Let ABC, A'B'C', be mutually equiangular triangles, in which A A', B = B', C C'; then, = = these triangles are similar. For, place the angle A' upon its equal angle A, and let B' fall at b and C'at c. Since the angle Abc is equal to B, be is parallel to BC (I. 55), and we have (15), or B AB: Ab AC: Ac, AB: A'B' = AC: A'C'. In the same manner, it is proved that AB: A'B' BC: B'C'; Therefore, the homologous sides are proportional, and the triangles are similar (24). 26. Corollary. Two triangles are similar when two angles of the one are respectively equal to two angles of the other (I. 73). 27. Scholium I. The homologous sides lie opposite to equal angles. 28. Scholium II. The ratio of similitude (24) of the two similar triangles, is any one of the equal ratios in the continued proportion [1]. 29. Scholium III. In two similar triangles, any two homologous lines are in the ratio of similitude of the triangles. For example, the perpendiculars AD, A'D', drawn from the homologous vertices A, A', to the opposite sides, are homologous lines of the two triangles; and the right triangles ABD, A'B'D', being similar (25), we have 44 D' C In like manner, if the lines AD, A'D', were drawn from A, A', to the middle points of the opposite sides, or to two points which divide the opposite sides in the same ratio in each triangle, these lines would still be to each other in the ratio of similitude of the two triangles. PROPOSITION V.-THEOREM. 30. Two triangles are similar, when their homologous sides are proportional. In the triangles ABC, A'B'C', let Comparing this with the given proportion [1], we see that the first ratio is the same in both; hence the second and third ratios in each are equal respectively, and, the numerators being the same, the denominators are equal; that is, A'C' Ac, and B'C' b c. Therefore, the triangles A'B'C' and Abc are equal (I. 80); and since Abc is similar to ABC, A'B'C' is also similar to ABC. = 31. Scholium. In order to establish the similarity of two polygons according to the definition (24), it is necessary, in general, to show that they fulfill two conditions: 1st, they must be mutually equiangular, and 2d, their homologous sides must be proportional. In the case of triangles, however, either of these conditions involves the other; and to establish the similarity of two triangles it will be sufficient to show, either that they are mutually equiangular, or that their homologous sides are proportional. PROPOSITION VI.-THEOREM. 32. Two triangles are similar, when an angle of the one is equal to an angle of the other, and the sides including these angles are proporportional. In the triangles ABC, A'B'C', let да B B' at b, and C' at c. Then, by the hy AB AC Therefore, be is parallel to BC (19), and the triangle Abc is similar to ABC (25). But Abc is equal to A'B'C'; therefore, A'B'C' is also similar to ABC. PROPOSITION VII.-THEOREM. 33. Two triangles are similar, when they have their sides parallel each to each, or perpendicular each to each. Let ABC, abc have their sides parallel each to each, or perpendicular each to each; then, these triangles are similar. For, when the sides of two angles B A //A |