EH by A, EC ; for DK, EL, CH, are each equal to BG”, that Book II. is, equal to A; but the rectangle BH is equal to the rectangles m BK, DL, EH; and BH is contained by A, BC; therefore the e 34. I. rectangle by A, and BC, is equal to the rectangles by A, BD; A, DE, and A, EC. Whereiore, &c. PRO P. II. THE O R. IE.a right line be any, how, cut, the rečtangles, contained by the whole line, and each of the segments, are equal to the square of the whole line. Let the right line AB be any how cut in C, the rectangles contained by AB, BC, and AB, AC, together, are equal to the square of ÁB. Upon AB describe the square ADEB *; thro' a 46. 1. C draw CF parallel to ADb, or BE; then, because AD is e- b 31. 1. į qual to AB, the rectangle under AD, AC, is equal to the rec tangle under AB, AC; and the rectangle under EB, BC, is equal to the rectangle under AB, BC; but the rectangle under AD, AC, that is, the rectangle AF, together with the rectangle under EB, BC, that is, CE, are equal to the square of AB; that is, the square AE. Wherefore, &c. PRO P. III. THEO R. [Fa right line be any how cut, the rectangle under the whole is two parts, together with the Square of the first mentioned part. Let the right line AB be any how cut in C, the rectangle under AB, BC, is equal to the rectangle under AC, CB, together with the square of BC. Upon BC describe the square BCDE"; produce ED to F, and draw AF parallel to CD or 1 46. 1, BE; the rectangle under AB, BC, that is, AE, is equal to the rectangles AC, CD; that is, the rectangle under AC, CB, and the square of CB. Wherefore, &c. PRO P. IV. TH E O R. Fa right line be any how cut, the square of the whole line is er under these parts. Let a 5. I. b 29 1. BOOK II. Let the right line AB be any low cut in C, the square of w AB is equal to the squares of AC, CB, and twice the rectangle under AC, CB For, upon AB describe the square ADEB; through C draw Cr parallel to AD, or BE; draw DB, cutting CF in G; through which point draw HGK parallel to AB, or DE. (Then the figure is faid to be constructed.) Now, because AB is equal to All the angle ADB is equal to ABD“; and the angle CGB to ADBb; therefore the angle CBG is equal to CGBC; therecix. 1. 1. fore CG is equal to CB • ; therefore CGKB is equilateral °; but the angjes BCG, CBK, are equal to two right anglesb, and CBK is a right anglef; therefore BCG is likewise a right angle; therefore all the angles are right ones o ; and CGKB is & Def. 30, a square 5. For the same reason HF is a square. But the rec tangles AG, GE, are equal"; and AG is the rectangle under h 43. I. AČ, CB; for CG is equal to CB; but the squares HF, CK, with the rectangles AG, GE, make up the square of AB: Therefore the square of AB is equal to the squares of AC, CB, and twice the rectangle under AC, CB. Wherefore, &c. Cor. Hence every parallelogram about the diameter of a square is a square, do I. e 34. 1. f 46, 1. I. PRO P. V. THE O R. IES parts, the rectangle under the two unequal parts, toget ber with the square of the intermediate part, are equal to the square of half the line. to EC. Let the right line AB be cut equally in C, and unequally in D, the rectangle under AD, DB, together with the square of CD, are equal to the square of CB. For, upon CB describe the square CEFB; construct the fi gure, and produce OL to K; and through A draw AK parallel a 36. . The parallelograms AL, CO, are equal a, and CH is equal b 43. 1. to HF); add DO to both ; then CO is equal to DF¢; therefore & Ax: 1:1. AL is likewise equal to DFd; add CH to both ; then the rec tangle AH is equal to the gnomon LDF; add LG, that is, the Cor. 4. fquare of CD, to both; then the rectangle AH, that is, the rectangle under AD, DB, with LG, that is, the square of CD, are equal to the gnomon LDF, and LG; that is, equal the {quareof CB. Wherefore, &c. |