Similarly, by placing B' at B, we prove that Cor. 1. Denoting the sides of ABC by a, b, c, and those of i.e., the perimeters of similar triangles are proportional to any pair of homologous sides. 198°. Theorem.--Two triangles which have their sides pro have the and Proof.—On A'C' let the ▲A'DC' be constructed so as to LDA'C' = LA ¿DC'A'=LC. 199°. Theorem.-If two triangles have two sides in each proportional and the included angles equal, the triangles are да B C B' Proof.—Place A' on A, and let A'C' lie along AB, and A'B' lie along AC, so that C' falls at D and B' at E. The triangles AED and A'B'C' are congruent and therefore and (194°) (177°) LAED=LB, and LADE=LC, (106°, Cor. 3) ^ABC~^AED ~ ^\A'B'C'. q.e.d. 200°. Theorem.-If two triangles have two sides in each proportional, and an angle opposite a homologous side in each equal: I. If the angle is opposite the longer of the two sides the triangles are similar. 2. If the angle is opposite the shorter of the two sides the triangles may or may not be B Proof.-Place A' at A and let B' fall at D, and A'C' along AC. Draw DE to BC. Then And since B'C' > A'B', the AA'B'C' = ▲ADE and they are therefore similar. (65°, 1) 2. If BC <AB, B'C' <A'B', and the triangles may or may not be similar. Proof. Since AD=A'B', and DE=B'C', and B'C' <A'B', .. the triangles A'B'C' and ADE may or may not be congruent (65°, 2), and therefore may or may not be similar. .. the triangles ABC and A'B'C' may or may not be similar. Cor. Evidently, if in addition to the conditions of the theorem, the angles C and C' are both less, equal to, or greater than a right angle the triangles are similar. Also, if the triangles are right-angled they are similar. 201°. The conditions of similarity of triangles may be classified as follows: 1. Three angles respectively equal. (Def. of similarity.) 2. Three sides proportional. 3. Two sides proportional and the included angles equal. 4. Two sides proportional and the angles opposite the longer of the homologous sides in each equal. If in 4 the equal angles are opposite the shorter sides in each the triangles are not necessarily similar unless some other condition is satisfied. By comparing this article with 66° we notice that there is a manifest relation between the conditions of congruence and those of similarity. Thus, if in 2, 3, and 4 of this article the words “proportional" and "homologous" be changed to "equal," the statements become equivalent to 1, 2, and 5 of Art. 66°. The difference between congruence and similarity is the nonnecessity of equality of areas in the latter case. When two triangles, or other figures, are similar, they are copies of one another, and the smaller may be brought, by a uniform stretching of all its parts, into congruence with the larger. Thus the primary idea of similarity is that every line-segment of the smaller of two similar figures is stretched to the same relative extent to form the corresponding segments of the larger figure. This means that the tensors of every pair of corresponding line-segments, one from each figure, are equal, and hence that any two or more linesegments from one figure are proportional to the corresponding segments from the second figure. Def. - Two line-segments are divided similarly when, being divided into the same number of parts, any two parts from one of the segments and the corresponding parts from the other taken in the same order are in proportion. 202°. Theorem.—A line parallel to the base of a triangle divides the sides similarly; and Conversely, a line which divides two sides of a triangle similarly is parallel to the third side. DE is to AC. Then BA and BC are divided similarly in D and E. Proof. The triangles ABC and DBE are evidently similar, B AB CB and DB EB' AD CE (195°, 1) q.e.d. and AB and CB are divided similarly in D and E. Conversely, if DE so divides BA and BC that CE: EB, DE is || to AC. AD: DB ABC and DBE having the angle B common, and the sides about that angle proportional, are similar. ¿BDE=LA, and DE is || to AC. Cor. 1. Since the triangles ABC and DBE are similar (199°) q.e.d. A B P/ B' C D D 203°. Theorem.-Two transversals to a system of parallels are divided similarly by the parallels. AA' is || to BB' is || to CC', etc. Then AD and A'D' are divided similarly. Proof.-Consider three of the ||s, AA', BB', and CC', and draw A'Q || to AD. Then AP and BQ are AB=A'P and BC=PQ. 7s, and (81°, 1) But A'QC' is a triangle and PB' is to QC'. Def.-A set of three or more lines meeting in a point is a pencil and the lines are rays. The point is the vertex or centre of the pencil. Cor. 1. Let the transversals meet in O, and let L denote any other transversal through O. Then AD, A'D', and L are all divided similarly by the parallels. But the parallels are transversals to the pencil. .. parallel transversals divide the rays of a pencil similarly. Cor. 2. Applying Cor. 1 of 202°, |