straight lines be proportionals, as the first is to the third, so is any triangle described on the first to a similar and similarly described triangle on the second. PROPOSITION 20. THEOREM. と 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 are to one another in the duplicate ratio of their homologous sides. Let ABCDE, FGHKL be similar polygons, and let AB be the side homologous to the side FG: the polygons ABCDE, FGHKL may be divided into the same number of similar triangles, of which each shall have to each the same ratio which the polygons have; and the polygon ABCDE shall be to the polygon FGHKL in the duplicate ratio of AB to FG. Join BE, EC, GL, LH. Then, because the polygon ABCDE is similar to the polygon FGHKL, [Hypothesis. the angle BAE is equal to the angle GFL, and BA is to AE as GF is to FL. [VI. Definition 1. And, because the triangles ABE and FGL have one angle of the one equal to one angle of the other, and the sides E D F about these equal angles proportionals, therefore the triangle ABE is equiangular to the triangle FGL, and therefore these triangles are similar; [VI. 6. [VI. 4. therefore the angle ABE is equal to the angle FGL. But, because the polygons are similar, [Hypothesis. therefore the whole angle ABC is equal to the whole angle therefore the remaining angle EBC is equal to the remain- [Axiom 3. And, because the triangles ABE and FGL are similar, and also, because the polygons are similar, [Hypothesis. therefore AB is to BC as FG is to GH; [VI. Definition 1. therefore, ex æquali, EB is to BC as LG is to GH; [V. 22. that is, the sides about the equal angles EBC and LGH are proportionals ; therefore the triangle EBC is equiangular to the triangle LGH; and therefore these triangles are similar. [VI. 6. [VI. 4. For the same reason the triangle ECD is similar to the triangle LHK. Therefore the similar polygons ABCDE, FGHKL may be divided into the same number of similar triangles. x Also these triangles shall have, each to each the same ratio which the polygons have, the antecedents being ABE, EBC, ECD, and the consequents FGL, LGH, LÍK ; and the polygon ABCDE shall be to the polygon FGHKL in the duplicate ratio of AB to FG. For, because the triangle ABE is similar to the triangle FGL, therefore ABE is to FGL in the duplicate ratio of EB to LG. [VI. 19. For the same reason the triangle EBC is to the triangle Therefore the triangle ABE is to the triangle FGL as the [V. 11. Again, because the triangle EBC is similar to the triangle LGH, therefore EBC is to LGH in the duplicate ratio of EC to LH. [VI. 19. W For the same reason the triangle ECD is to the triangle LHK in the duplicate ratio of EC to LH. Therefore the triangle EBC is to the triangle LGH as the triangle ECD is to the triangle LHK. [V. 11. But it has been shewn that the triangle EBC is to the triangle LGH as the triangle ABE is to the triangle FGL. Therefore as the triangle ABE is to the triangle FGL, so is the triangle EBC to the triangle LGH, and the triangle ECD to the triangle LHK; [V. 11. and therefore as one of the antecedents is to its consequent so are all the antecedents to all the consequents; [V. 12. that is, as the triangle ABE is to the triangle FGL so is the polygon ABCDE to the polygon FGHKL. [VI. 19. But the triangle ABE is to the triangle FGL in the duplicate ratio of the side AB to the homologous side FG; therefore the polygon ABCDE is to the polygon FGHKL in the duplicate ratio of the side AB to the homologous 'side FG. Wherefore, similar polygons &c. Q.E.D. COROLLARY 1. In like manner it may be shewn that similar four-sided figures, or figures of any number of sides, are to one another in the duplicate ratio of their homologous sides; and it has already been shewn for triangles; therefore universally, similar rectilineal figures are to one another in the duplicate ratio of their homologous sides. COROLLARY 2. If to AB and FG, two of the homologous sides, a third proportional M be taken, [VI. 11. then AB has to M the duplicate ratio of that which AB has to FG. [V. Definition 10. But any rectilineal figure described on AB is to the similar and similarly described rectilineal figure on FG in the duplicate ratio of AB to FG, [Corollary 1. Therefore as AB is to M, so is the figure on AB to the figure on FG; [V. 11. and this was shewn before for triangles. [VI. 19, Corollary. Wherefore, universally, if three straight lines be proportionals, as the first is to the third, so is any rectilineal figure described on the first to a similar and similarly described rectilineal figure on the second. *Statement し PROPOSITION 21. THEOREM. Rectilineal figures which are similar to the same rectilineal figure, are also similar to each other. Let each of the rectilineal figures A and B be similar to the rectilineal figure C: the figure A shall be similar to the figure B. For, because A is similar to C, [Hyp. A is equiangular to C, and A and C have their sides about the equal angles proportionals. [VI. Def. 1. Again, because B is similar to C, [Hyp. A C B B is equiangular to C, and B and C have their sides about the equal angles proportionals. [VI. Definition 1. Therefore the figures A and B are each of them equiangular to C, and have the sides about the equal angles of each of them and of C proportionals. Therefore A is equiangular to B, [Axiom 1. [V. 11. and A and B have their sides about the equal angles proportionals; therefore the figure A is similar to the figure B. [VI. Def. 1. Wherefore, rectilineal figures &c. Q.E.D. If four straight lines be proportionals, the similar rectilineal figures similarly described on them shall also be proportionals; and if the similar rectilineal figures similarly described on four straight lines be proportionals, those straight lines shall be proportionals. Let the four straight lines AB, CD, EF, GH be proportionals, namely, AB to CD as EF is to GH; and on AB, CD let the similar rectilineal figures KAB, LCD be similarly described; and on EF, GH let the similar rectilineal figures MF, NH be similarly described: the figure KAB shall be to the figure LCD as the figure MF is to the figure NH. To AB and CD take a third proportional X, and to EF and GH a third proportional O. [VI. 11. Then, because AB is to CD as EF is to GH, [Hypothesis. and AB is to CD as CD is to X; and EF is to GH as GH is to 0; [Construction. [Construction. [V. 11. therefore CD is to X as GH is to 0. And AB is to CD as EF is to GH; therefore, ex æquali, AB is to X as EF is to 0. [V. 22. But as AB is to X, so is the rectilineal figure KAB to the rectilineal figure LCD ; [VI. 20, Corollary 2. and as EF is to O, so is the rectilineal figure MF to the rectilineal figure NH; [VI. 20, Corollary 2. |