Join BE, EC, GL, LH. Because the polygon ABCDE Book VI. is similar to the polygon FGHKL, the angle BAE is equal to the angle GFLa, and BA : AE : : GF : FL2; therefore a Def. 1. 6. the triangle ABE is equiangular, and similar to the triangle b 6. 6. FGL; wherefore the angle ABE is equal to the angle FGL. Because the polygons are similar, the whole angle ABC is equal to the whole angle FGHa; therefore the remaining angle EBC is equal to the remaining angle LGH. Now because the triangles ABE, FGL are similar, EB : BA :: LG : GF2; and because the polygons are similar, AB: BC:: FG: GHa; therefore, ex æqualis, EB: BC :: LG : GH; therefore the sides about the equal angles EBC, LGH are proportional; therefore the triangle EBC is equiangularb to the triangle LGH, and similar to it. For the same reason the triangle ECD is similar to the triangle LHK. Therefore the similar polygons ABC DE, FGHKL are divided into the same number of similar triangles. с 22. 5. Also, these triangles have, each to each, the same ratio which the polygons have to each other, the antecedents being ABE, EBC, ECD, and the consequents FGL, LGH, LHK; and the polygon ABCDE has 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, ABE has to FGL the duplicate ratio of that which the side e 19.6. BE has to the side GL. For the same reason the triangle BEC has to GLH the duplicate ratio of that which BE has ⚫ to GL. Therefore, as the triangle ABE is to the triangle FGL, so is the triangle BEC to the triangle GLHf. Again, f 11. 5. because the triangle EBC is similar to the triangle LGH, EBC has to LGH the duplicate ratio of that which the side f 11. 5. Book VI. 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, as the triangle EBC is to the triangle LGH, so is the triangle ECD to the triangle LHKf. But it has been proved that the triangle EBC is to the triangle LGH, as the triangle ABE 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; therefore, as the triangle ABE is 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. Therefore also 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. g 12. 5. COR. 1. In like manner it may be proved that similar four sided figures, or of any number of sides, are to one another in the duplicate ratio of their homologous sides; and it has already been proved in triangles. Therefore, universally, similar rectilineal figures are to one another in the duplicate ratio of their homologous sides. COR. 2. If to AB, FG, two of the homologous sides, a third proportional M be taken, AB has to M the duplicate h Def. 11. ratio of that which AB has to FGh. But the four sided figure, or polygon, upon AB has to the four sided figure, or polygon, upon FG the duplicate ratio of that which AB has to FG; therefore as AB is to M, so is the figure upon AB to the 5. i Cor. 19.6. figure upon FG, which was also proved in trianglest. Therefore, universally, it is manifest that if three straight lines be proportional, 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, COR. 3. Because all squares are similar figures, the ratio of any two squares to each other is the same with the duplicate ratio of their sides; and hence, also, any two similar rectilineal figures are to each other as the squares of their homo- . logous sides. PROP. XXI. THEOR. RECTILINEAL figures, which are similar to the same rectilineal figure, are also similar to one another. Book VI. Let each of the rectilineal figures A, B be similar to the rectilineal figure C; the figure A is similar to the figure B. Because A is similar to C, they are equiangular, and have their sides about the equal angles proportionala; and because a Def. 1.6. B is similar to C, they are equiangu lar, and have their sides about the e C about the equal angles of each of them, and of C, proportional. Wherefore the rectilineal figures A and B are equiangularb, b Ax. 1. 1. and have their sides about the equal angles proportionals. c 11.5. Therefore A is similar to B. Therefore, rectilineal figures, &c. Q. E. D. PROP. XXII. THEOR IF four straight lines be proportional, the similar rectilineal figures similarly described upon them will also be proportional; and if the similar rectilineal figures similarly described upon four straight lines be proportional, those straight lines will be proportional. Book VI. a 11. 6. b 11. 5. с 22. 5. d 2. Cor. 20.6. e 12. 6. f 18. 6. Let the four straight lines AB, CD, EF, GH be proportional, 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. The rectilineal figure KAB is to LCD as MF to NH. To AB, CD take a third proportionala X, and to EF, GH a third proportional O. Because AB:CD::EF: GH, ex æqualic, AB: X :: EF : 0. But AB: X :: KAB:LCDd, and EF: O:: MF:NH; therefore KAB: LCD :: MF: NHb. If the figure KAB be to the figure LCD, as the figure MF to the figure NH, AB is to CD, as EF to GH. To AB, CD, EF take a fourth proportional PRe, and upon PR describes the rectilineal figure SR, similar and similarly g 9. 5. situated to either of the figures MF, NH. Because AB is to CD, as EF to PR, and 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, KAB is to LCD, as MF to SR. But, by hypothesis, KAB is to LCD, as MF to NH; therefore the rectilineals NH, SR are equals to each other. They are also similar and similarly situated; therefore GH is equal to PR; therefore, because AB is to CD, as EF to PR, AB is to CD, as EF to GH. Therefore, if four straight lines, &c. Q. E. D. PROP. XXIII. THEOR. EQUIANGULAR parallelograms have to one another the ratio which is compounded of the ratios of their sides. Let AC, CF be equiangular parallelograms, having the angle BCD equal to the angle ECG; the ratio of the parallelogram AC to the parallelogram CF, is the same with the ratio which is compounded of the ratios of their sides. Book VI. Let BC, CG be placed in a straight line, therefore DC, CE are also in a straight line; complete the parallelogram a 14. 1. DG, take any straight line K, and makeb BC to CG, as K to b 12. 6.. L, and DC to CE, as L to M; therefore the ratios of K to L, and L to M, are the same with the ratios of the sides, 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 c Def. 10.5. 1: L to M; wherefore also K DH ed of the ratios of the sides of the parallelograms. Now BC is to CG as the parallelogram AC to the parallelogram CHd, and BC is to CG as K to L; therefore L is to M as the pa rallelogram CH to the parallelogram CF. Since it has been proved that K:L:: parallelogram AC: parallelogram CH, and L: M:: parallelogram CH: parallelogram CF, K: M:: parallelogram AC: parallelogram CFf. f 22. 5. But K has to M the ratio which is compounded of the ratios of the sides; therefore also the parallelogram AC has to the |