Construction. Join DF, and at the points A, B in the straight line AB, make the angle BAG equal to the angle at C, and the angle ABG equal to the angle CDF (I. 23); and at the point G, B, in the straight line GB, make the angle BGH equal to the angle DFE, and the angle GBH equal to the angle FDE (İ. 23). Then the figure ABHG shall be similar, and similarly situated, to CDEF. Demonstration. Because the angle BAG is equal to C, and the angle ABG to the angle CDF (constr.); therefore the remaining angle 1GB is equal to the remaining angle CFD (I. 32); therefore 1. The triangle FCD is equiangular to the triangle GAB; and because the angle BGH is equal to the angle DFE, and the angle GBH to the angle FDE (constr.); therefore the remaining angle GHB is equal to the remaining angle FED, and 2. The triangle FDE is equiangular to the triangle GBH. Then, because the angle AGB is equal to the angle CFD, and BGH to DFE, therefore 3. The whole angle AGH is equal to the whole angle CFE (Ax. 2); for the same reason, 4. The angle ABH is equal to the angle CDE; and the angle at A is equal to the angle at C (constr.); and the angle GHB to FED; therefore 5. The rectilineal figure ABHG is equiangular to CDEF. Likewise these figures have their sides about the equal angles proportionals; for, because the triangles GAB, FCD are equiangular, therefore 6. BA is to AG, as GB to CF (VI. 4); and because AG is to GB, as CF to FD, and as GB is to GH, so is FD to FE, by reason of the equiangular triangles BGH, DFE; therefore, ex æquali, 7. AG is to GH, as CF to FE (V. 22). P In the same manner it may be proved that AB is to BH, as CD to DE; and GH is to HB, as FE to ED (VI. 4); therefore wherefore 8. The rectilineal figures ABHG, CDEF are equiangular, and have their sides about the equal angles proportionals; 9. ABHG is similar to CDEF (VI. Def. 1). ́ Secondly. Let it be required to describe upon a given straight line AB, a rectilineal figure similar, and similarly situated, to the rectilineal figure CDKEF of five sides. Construction. Join DE, and upon the given straight line AB, describe the rectilineal figure ADHG similar, and similarly situated, to the quadrilateral figure CDEF, by the former case; and at the points B, H, in the straight line BH, make the angle HBL equal to the angle EDK, and the angle BHL equal to the angle DEK. Then the figure ABLHG is similar, and similarly situated, to CDKEF. Demonstration. Because the angle HBL is equal to the angle EDK, and the angle BHL equal to the angle DEK (constr.); therefore 1. The remaining angle at L is equal to the remaining angle at K (I. 32); and, because the figures ABHG, CDEF, are similar, 2. The angle GHB is equal to the angle FED (VI. Def. 1); and BHL is equal to DEK (constr.); therefore 3. The whole angle GHL is equal to the whole angle FEK; for the same reason, therefore 4. The angle ABC is equal to the angle CDK; 5. The five-sided figures AGHLB, CFEKD are equiangular. And, because the figures AGHB, CFED, are similar, therefore 6. GH is to HB, as FE to ED (VI. Def. 1); but as HB to BL, so is ED to EK (VI. 4); therefore, ex æquali, 7. GH is to HL, as FE to EK (V. 22) ; for the same reason, 8. AB is to BL, as CD to DK, and BL is to LH, as DK to KE, because the triangles BLH, DKE, are equiangular (VI. 4); therefore 9. The five-sided figures ABLHG, CDKEF, are equiangular, and have their sides about the equal angles proportionals. wherefore 10. ABLHG is similar to CDKEF. In the same manner, a rectilineal figure of six sides may be described upon a given straight line similar to one given, and so on. Q.E.F. PROPOSITION 19.-Theorem. Similar triangles are to one another in the duplicate ratio of their homologous sides. Let ABC, DEF be similar triangles, having the angle B equal to the angle E, and let AB be to BC, as DE to EF, so that the side BC may be homologous tc EF (V. Def. 12). Then the triangle ABC shall have to the triangle DEF the duplicate ratio of that which BC has to EF. Construction. Take BG a third proportional to BC, EF (VI. 11), so that BC may be to EF, as EF to BG, and join GA. Demonstration. Then, because as AB to BC, so DE to EF; alternately, 1. AB is to DE, as BC to EF (V. 16) ; but as BC to EF, so is EF to BG (constr.); therefore, as AB to DE, so is EF to BG (V. 11); therefore 2. The sides of the triangles ABG, DEF, which are about the equal angles, are reciprocally proportional : but triangles, which have the sides about two equal angles reciprocally proportional, are equal to one another (VI. 15); therefore 3. The triangle ABG is equal to the triangle DEF; and because as BC is to EF, so EF to BG; and that if three straight lines be proportionals, the first is said to have to the third, the duplicate ratio of that which it has to the second (V. Def. 10); therefore 4. BC has to BG the duplicate ratio of that which BC has to EF; but as BC is to BG, so is the triangle ABC to the triangle ABG (VI. 1); therefore 5. The triangle ABC has to the triangle ABG, the duplicate ratio of that which BC has to EF; but the triangle ABG is equal to the triangle DEF; therefore also 6. The triangle ABC has to the triangle DEF, the duplicate ratio of that which BC has to EF. Therefore, similar triangles, &c. Q.E.D. Cor. From this it is manifest, that if three straight lines be proporonals, as the first is to the third, so is any triangle upon the first, to a similar and similarly described triangle upon the second. PROPOSITION 20.-Theorem. Similar polygons may be divided into the same number of similar triangles, having the same ratio to one another that the polygons have and the polygons have to one another the duplicate ratio of that which their homologous sides have. Let ABCDE, FGHKL be similar polygons, and let AB be the side homologous to FG; Then the polygons ABCDE, FGHKL may be divided into the same number of similar triangles, whereof each shall have to each the same ratio which the polygons have ; and the polygon ABCDE shall have to the polygon FGHKL the duplicate ratio of that which the side AB has to the side FG. E M Construction. Join BE, EC, GL, LH. Demonstration. Because the polygon ABCDE is similar to the polygon FGHKL, therefore 1. The angle BAE is equal to the angle GFL (VI. Def. 1), and BA is to AE, as GF to FL (VI. Def. 1); therefore, because the triangles ABE, FGL have an angle in one, equal to an angle in the other, and their sides about these equal angles proportionals, 2. The triangle ABE is equiangular to the triangle FGL (VI. 6) ; and therefore similar to it (VI. 4); wherefore 3. The angle ABE is equal to the angle FGL; and, because the polygons are similar, the whole angle ABC is equal to the whole angle FGH (VI. Def. 1); therefore 4. The remaining angle EBC is equal to the remaining angle LGH (I. 32 and Ax. 3); and because the triangles ABE, FGL are similar, 5. EB is to BA, as LG to GF (VI. 4) ; and also, because the polygons are similar, AB is to BC, as FG to GH (VI. Def. 1); therefore, ex æquali, 6. EB is to BC, as LG to GH (V. 22) ; that is, the sides about the equal angles EBC, LGH are proportionals; therefore 7. The triangle EBC is equiangular to the triangle LGH, (VI. 6), and similar to it (VI. 4); for the same reason, the triangle ECD likewise is similar to the triangle LHK; therefore 8. The similar polygons ABCDE, FGHKL are divided into the same number of similar triangles. Also, these triangles shall have, each to each, the same ratio which |