the polygons have to one another, the antecedents being ABE, EBC, ECD, and the consequents FGL, LGH, LHK; and the polygon ABCDE shall have to the polygon FGHKL, the duplicate ratio of that which the side AB has to the homologous side FG. Because the triangle ABE is similar to the triangle FGL, B M F с 9. ABE has to FGL, the duplicate ratio of that which the side BE has to the side GL (VI. 19) ; for the same reason, the triangle BEC has to GLH the duplicate ratio of that which BE has to GL; therefore 10. As the triangle ABE is to the triangle FGL, so is the triangle BEC to the triangle GLH (V. 11). Again, because the triangle EBC is similar to the triangle LGII, 11. EBC has to LGH, the duplicate ratio of that which the side EC has to the side LH; for the same reason, the triangle ECD has to the triangle LHK, the duplicate ratio of that which EC has to LH; therefore 12. As the triangle EBC is to the triangle LGH, so is the triangle ECD to the triangle LHK (V. 11) ; but it has been proved, that the triangle EBC is likewise to the triangle LGH, as the triangle ABE to the triangle FGL; therefore 13. As the triangle ABE to the triangle FGL, so is the triangle EBC to the triangle LGH, and the triangle ECD to the triangle LHK ; and therefore, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents (V. 12); that is, 14. As the triangle ABE to the triangle FGL, so is the polygon ABCDE to the polygon FGHKL; but the triangle ABE has to the triangle FGL, the duplicate ratio of that which the side AB has to the homologous side FG (VI. 19); therefore also 15. The polygon ABCDE has to the polygon FGHKL, the duplicate ratio of that which AB has to the homologous side FG. Wherefore, similar polygons, &c. Q.E.D. Cor. 1. In like manner it may be proved, that similar four-sided figures, or of any number of sides, are one to another in the duplicate ratio of their homologous sides; and it has already been proved in triangles (VI. 19); therefore, universally, similar rectilineal figures are to one another in the duplicate ratio of their homologous sides. Cor. 2. And if to AB, FG, two of the homologous sides, a third proportional M be taken (VI. 11), 1. AB has to M the duplicate ratio of that which AB has to FG (V. Def. 10); but the four-sided figure or polygon upon AB, has to the four-sided figure or polygon upon FG likewise the duplicate ratio of that which AB has to FG (VI. 20, Cor. 1); therefore 2. As AB is to M, so is the figure upon AB to the figure upon FG (V. 11); which was also proved in triangles (VI. 19, Cor.); therefore 3. Universally, it is manifest, that if three straight lines be proportionals, as the first is to the third, so is any rectilineal figure upon the first, to a similar and similarly described rectilineal figure upon the second. PROPOSITION 21.-Theorem. Rectilineal figures which are similar to the same rectilineal figure, are also similar to one another. A Let each of the rectilineal figures A, B be similar to the rectilineal figure C. Then the figure A shall be similar to the figure B. Demonstration. Because A is similar to C, B Again, because B is similar to C, 1. They are equiangular, and also have their sides about the equal angles proportional (VI. Def. 1). 2. They are equiangular, and have their sides about the equal angles proportionals (VI. Def. 1). therefore the figures A, B are each of them equiangular to C, and have the sides about the equal angles of each of them and of C proportionals. Wherefore therefore 3. The rectilineal figures A and B are equiangular (I. Ax. 1), and have their sides about the equal angles proportionals (V. 11); M 4. A is similar to B.(VI. Def. 1). PROPOSITION 22.-Theorem. If four straight lines be proportionals, the similar rectilineal figures similarly described upon them shall also be proportionals; and conversely, if the similar rectilineal figures similarly described upon four straight lines be proportionals, those straight lines shall be proportionals. Therefore, rectilineal figures, &c. Q.E.D. Let the four straight lines AB, CD, EF, GH be proportionals, viz., AB to CD, as EF to GH; and upon AB, CD let the similar rectilineal figures KAB, LCD be similarly described; and upon EF, GH the similar rectilineal figures MF, NH, in like manner. Then the rectilineal figure KAB shall be to LCD, as MF to NH. F E R Construction. To AB, CD take a third proportional X (VI. 11); and to EF, GH a third proportional O. Demonstration. Because AB is to CD as EF to GH, therefore CD is to X, as GH to O (V. 11); wherefore, ex æquali, 1. As AB to X, so is EF to O (V. 22) ; but as AB to X, so is the rectilineal figure KAB to the rectilineal figure LCD, and as EF to O, so is the rectilineal figure MF to the rectilineal figure NH (VI. 20, Cor. 2); therefore 2. As KAB to LCD, so is MF to NH (V. 11). Next. Let the rectilineal figure KAB be to LCD, as MF to NH; Then the straight line AB shall be to CD, as EF to GH. Construction. Make as AB to CD, so EF to PR (VI. 12), and upon PR describe the rectilineal figure SR similar and similarly situated to either of the figures MF, NH (VI. 18). Demonstration. Then, because as AB to CD, so is EF to PR, and that upon AB, CD are described the similar and similarly situated rectilineals KAB, LCD, and upon EF, PR, in like manner, the similar rectilineals MF, SR; therefore 1. KAB is to LCD, as MF to SR; but by the hypothesis, KAB is to LCD, as MF to NH; fore the rectilineal figure MF having the same ratio to two NH, SR, therefore 2. NH and SR are equal to one another (V. 9). But they are also similar, and similarly situated; therefore 3. GH is equal to PR; 4. AB is to CD, as EF to GH (V. 7). and because as AB to CD, so is EF to PR, and that PR is equal to GH; therefore If therefore, four straight lines, &c. Q.E.D. PROPOSITION 23.-Theorem. Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides. and thereeach of the Let AC, CF be equiangular parallelograms, having the angle BCD equal to the angle ECG. A Then the ratio of the parallelogram AC to the parallelogram CF, shall be the same with the ratio which is compounded of the ratios of their sides. B M F E Construction. Let BC, CG be placed in a straight line; therefore DC and CE are also in a straight line (I. 14); and complete the parallelogram DG; and taking any straight line K, make as BC to CG, so K to L (VI. 12); and as DC to CE, so make L to M (VI. 12). A D A B F E Demonstration. Because the ratios of K to L, and L to M, are the same with the ratios of the sides, viz., of BC to CG, and DC to CE; but the ratio of K to M is that which is said to be compounded of the ratios of K to L, and L to M (V. Def. A); therefore 1. K has to M the ratio compounded of the ratios of the sides; and because as BC to CG, so is the parallelogram AC to the parallelogram CH (VI. 1); but as BC to CG, so is K to L; therefore 2. K is to L, as the parallelogram AC to the parallelogram CH (V. 11). Again, because as DC to CE, so is the parallelogram CH to the parallelogram CF; but as DC to CE, so is L to M; wherefore 3. L is to M, as the parallelogram CH to the parallelogram CF (V. 11); therefore since it has been proved, that as K to L, so is the parallelogram AC to the parallelogram CH; and as L to M, so is the parallelogram CH to the parallelogram CF; ex æquali, 4. K is to M, as the parallelogram AC to the parallelogram CF (V. 22); but K has to M the ratio which is compounded of the ratios of the sides therefore also 5. The parallelogram AC has to the parallelogram CF, the ratio which is compounded of the ratios of the sides. Wherefore, equiangular parallelograms, &c. Q.E.D. PROPOSITION 24.-Theorem. Parallelograms about the diameter of any parallelogram, are similar to the whole, and to one another. |