Book VI. PROP. XXI. THEOR. RECTILINE Al figures which are similar to the same rectilineal figure, are also similar to one another. 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 also have their fides about the equal angles proportionalsa. Again, because Ba, 1. Def. 6. is finiar to C, they are cquiangular, and have their fides about the equal angles proportio A B nals a therefore the fi. gures 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 the rectilineal figures A and B are equiangular", and have their sides about the equal b. 1. Ax. 1. angles proportionalso. Therefore A is similar a to B. Q. E, D, c. II. S. F four straight lines be proportionals, the similar recti lineal figures similarly described upon them shall also be proportionals. and if the similar rectilineal figures fimilarly described upon four straight lines be proportionals, those straight lines shall be proportionals. 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. the rectili-. neal figure KAB is to LCD, as MF to NH. To AB, CD take a third proportional * X; and to EF, GH a a. 11. 6. third proportional O. and because AB is to CD, as EF to GH, b. II. S. therefore CD is b to X, as GH to 0; wherefore ex aequali', as AB to Book VI. to X, so EF to O. but as AB to X, so is d the rectilineal KAB to the rectilineal LCD, and as EF to O, so is d the rectilineal MF to the d. 2. Cor. rectilineal NH. therefore as KAB to LCD, so b is MF to NH. 20.6. And if the rectilineal KAB be to LCD, as MF to NH; the b. 11. S. straight line AB is to CD, as EF to GH. Make as AB to CD, fo EF to PR, and upon PR describe f the f. 18. 6. rectilineal figure SR similar and similarly situated to either of the fi, K 1 e. 12. 6. e F G X B C D S O H P B gures MF, NH. then because as AB to CD, fo 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 fie milar rcctilineals MF, SR; KAB is to LCD, as MF to SR; but, by the Hypothesis, KAB is to LCD, as MF to NH; and therefore the rectilineal MF having the same ratio to each of the two NH, SR, these are equal 3 to one another. they are also similar, and similarly situated; therefore GH is equal to PR. and because as AB to CD, so is EF to PR, and that PR is equal to GH; AB is to CD, as EF to GH. If therefore four straight lines, &c. Q. E. D. See N. ANGULAR parallelograms have to one another the ratio which is compounded of the ratios of their sides. E 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. Let Let BC, CG be placed in a straight line, therefore DC and CE are Book VI. álso in a straight line ; and complete the parallelogram DG, and, taking any straight line K, make b as BC to CG, so K to L; and as a. 14. s. DC to CE, so make o L to M. therefore the ratios of K to L, and b. 12. 6. i to M are the same with the ratios of the sides, viz, of BC to CG, and DC 10 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. wherefore alsoc. A. Def. s. K has to M, the ratio compounded A D H of the ratios of the sides. and because as BC to CG, so is the parallelogram AC to the parallelogram CHd; but as BC to CG, so B C d. f. 6. is K to L; therefore K ise to L, as the parallelogram AC to the patallelogram CH. again, because as DC to CE, so is the parallelogram CH to the parallelogram CF; but KLM E as DC to CE, so is L to M; wherefore L is e to M, as the parallelogram CH to the parallelogram CF. 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 the patallelogram CH to the parallelogram CF; ex aequalif, K is to M, as f. 22. 5. the parallelogram AC to the parallelogram CF. but K has to M the ratio which is compounded of the ratios of the sides; therefore also the parallelogram AC has to the parallelogram CF the ratio which is compounded of the ratios of the fides. Wherefore equiangular parallelograms, &c. Q. E. D. e. 11. S. PROP. XXIV. THEOR. lelogram, are similar to the whole, and to one ano ther. Let ABCD be à parallelogram, of which the diameter is AC; and EG, HK the parallelograms about the diameter. the parallelograms EG, HK are similar both to the whole parallelogram ABCD, and to one another. Because DC, GF are parallels, the argle ADC is equal to the a. 29. 1, angle AGF. for the same reason, because BC, EF are parallels, the angle c. 4. 6. Book VI. angle ABC is equal to the angle AEF. and each of the angles MBCD, EFG is equal to the opposite angle DAB 6, and therefore are b. 34. 1. equal to one another; wherefore the parallelograms ABCD, AEFG are equiangular. and because the angle ABC is equal to the angle A E B and because the opposite sides of paralle F lograms are equal to one another 6, AB G H d. 7.5. is d to AD, as AE to AG; and DC to K С equal angles are proportionals; and they e. 1. Def. 6. are therefore similar to one another for the same reason, the pa. rallelogram ABCD is similar to the parallelogram FHCK. Wherefore each of the parallelograms GE, KH is similar to DB. but recti lineal figures which are similar to the same rectilineal figure, are also 1. 21. 0. similar to one anotherf, therefore the parallelogram GE is similar to KH. Wherefore the parallelograms, &c. Q. E. D. Scc N. To describe a rectilineal figure which shall be fimilar to one, and equal to another given rectilineal figure. Let ABC be the given rectilineal figure, to which the figure to be described is required to be similar, and D that to which it must be equal. It is required to describe a rectilineal figure similar to ABC and equal to D. a. Cor.43.1. Upon the straight line BC describe a the parallelogram BE equal to the figure ABC; also upon CE describe the parallelogram CM S 29.1. equal to D, and having the angle FCE equal to the angle CBL. b. 214.1. therefore BC and CF are in a straight line 6, as also LE and EM. c. 13. 6. between BC and CF find a mean proportional GH, and upon GH d. 18.6. described the rectilineal figure KGH similar and similarly situated to the figure ABC. and because BC is to GH, as GH to CF, and if three straight lines be proportionals, as the first is to the third, e. 2. Cor. so is the figure upon the first to the similar and similarly described figure 20. 6. figure upon the second; therefore as BC to CF, so is the rectilineal Book VI. figure ABC to KGH. but as BC to CF, fo is f the parallelogramm BE to the parallelogram EF. therefore as the rectilineal figure ABC f. 1.6. is to KGH, so is the parallelogram BE to the parallelogram EF 8. g. 11. 5. and the rectilineal figure ABC is equal to the parallelogram BE; H E M therefore the rectilineal figure KGH is equal h to the parallelogram h. 14. 5. h. 14.'s. EF. but EF is equal to the figure D, wherefore also KGH is equal to D; and it is similar to ABC. Therefore the rectilineal figure KGH has been described similar to the figure ABC, and equal to D. Which was to be done. PROP. XXVI. THEOR. and be similarly situated; they are about the same diameter. Let the parallelograms ABCD, AEFG be similar and similarly situated, and have the angle DAB common. ABCD and AEFG are about the fame diameter. For if not, let, if possible, the А. G parallelogram BD have its diameter AHC in a different straight line from K H AF the diameter of the parallelo E F gram EG, and let GF meet AHC in H; and thro' H draw HK parallel to AD or BC. therefore the paral- B lelograms ABCD, AKHG being about the same diameter, they are fimilar to one another. where- a. 24 6. fore as DA to AB, so is 6 GA to AK. but because ABCD and b. s. Def. 6. AEFG are similar parallelograms, as DA is to AB so is GA to AE. there. |