Book VI. parallelogram CF the ratio which is compounded of the ratios of the sides. Wherefore, equiangular parallelograms, &c. Q. E. D. d 1. 6. f 22. 5. Otherwise. “AC: CH:: BC: CGd, and CH:CF : : DC: CE, therefore AC:CF:: BC.DC: CG.CE." a 29. 1. PROP. XXIV. THEOR. THE parallelograms about the diameter of any parallelogram are similar to the whole and to one another. Let ABCD be a parallelogram, of which the diameter is AC, and EG, HK the parallelograms about the diameter; the parallelograms EG, HK are similar to the whole parallelogram ABCD, and to each other. Because DC, GF are parallels, the angle ADC is equala to the angle AGF; and because BC, EF are parallels, the angle ABC is equal to the angle AEF; and because each of the angles BCD, EFG is equal to the opposite angle DABb, they are equal A E b 34. 1. B to one another. Wherefore the F H are equiangular. Because the an- EAF, the triangles are equiangu € 4. 6. d 7. 5. lar to each other; therefore AB is to BC as AE to EFc. But the opposite sides of parallelograms are equal to each otherb; therefore AB is to AD as AE to AGd, and DC to CB as GF to FE, and CD to DA as FG to GA. Therefore the sides of the parallelograms ABCD, AEFG about the equal angles are proportional; therefore they are similar. Book VI For the same reason the parallelogram ABCD is similar to the parallelogram FHCK. Wherefore each of the parallelo- e Def. 1. 6. grams GE, KH is similar to DB; therefore the parallelogram GE is similar to KHf. Wherefore, the parallelograms, &c. f 12. 6. Q. E. D. PROP. XXV. PROВ. TO describe a rectilineal figure which shall be similar to one, and equal to another given rectilineal figure. Let ABC and D be two given rectilineal figures; it is required to describe a rectilineal figure similar to ABC, and equal to D. Upon the straight line BC describe the parallelogram BE a Cor. 45.1. equal to the figure ABC, and upon CE describe the parallelogram CM equal to D, and having the angle FCE equal to the angle CBL; therefore BC, CF are in a straight lineb, b 529. 1. and also LE, EM. Between BC and CF find a mean pro portional GH, and upon GH described the rectilineal figure APA C B L D K F G H 14.1. с 13. 6. d 18. 6. KGH similar and similarly situated to the figure ABC. Because BC is to GH as GH to CF, BC is to CF as the figure ABC to the figure KGH. But BC is to CF as the paral- e 2. Cor. lelogram BE to the parallelogram EFf; therefore the figure 2 A 20.6. f 1. 6. g 11. 5. h 14. 5. Book VI. ABC is to the figure KGH as the parallelogram BE to the parallelogram EFg. But the rectilineal figure ABC is equal to the parallelogram BE; therefore the rectilineal figure KGH is equal to the parallelogram 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. IF two similar parallelograms have a common angle, 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 same diameter, G D H E F C For, if not, let, if possible, the meter, are similar to each others; wherefore, as DA to AB, b Def. 1. 6. so is GA to AK. But because ABCD and AEFG are similar parallelograms, as DA is to AB, so is GA to AE; therefore, as GA to AE, so GA to AK; wherefore AK is equald to AE, the less to the greater, which is impossible. Therefore ABCD and AKHG are not about the same diameter; wherefore ABCD and AEFG must be about the same diameter. Therefore, if two similar parallelograms, &c. Q. E. D. a 24. 6. с 11. 5. d 9. 5. PROP. XXVII. THEOR. OF all the rectangles contained by the segments of a given straight line the greatest is the square which is described on half the line. Book VI. Let AB be a given straight line, which is bisected in C, N. and let D be any point in it; the square on AC is greater than the AR rectangle AD.DB. CDB For the rectangle contained by AD and DB, together with the square of CD, are equal to the square of AC2; therefore a 5. 2. the square of AC is greater than the rectangle AD.DB. Therefore, &c. Q. E. D. PROP. XXVIII. PROB. TO divide a given straight line, so that the rectangle contained by its segments may be equal to a given space; but that space must not be greater than the square of half the given line. Let AB be the given straight line, and let the square upon the given straight line C be the space to which the rectangle contained by the segments of AB must be equal, and this square, by the determination, is not greater than that upon half the straight line AB. Book VI. Bisect AB in D; if the square upon AD be equal to the square upon C, the thing required is done. But if it be not equal to it, AD must be greater than C, according to the determination. Draw DE at Because AB is divided equally a 5. 2. b 47. 1. in D, and unequally in G, AG.GB+DG*=*DB2=EG2. But ED2+DG2=EG2; therefore AG.GB+DGo = ED2 + DG2; therefore if DG be taken away, AG.GB=ED2, or C2; therefore the given line AB is divided in G, so that the rectangle contained by the segments AG, GB is equal to the square upon the given straight line C. Which was to be done. PROP. XXIX. PROB. TO produce a given straight line, so that the rectangle contained by the segments between the extremities of the given line and the point to which it is produced may be equal to a given space. Let AB be the given straight line, and let the square upon the given straight line C be the space to which the rectangle under the segments of AB produced must be equal. Bisect AB in D, and draw BE at right angles to AB, and equal to C; join DE, and from the centre D, at the distance DE, describe a circle meeting AB produced in Because AB is bisected in D, and produced to G, G. |