IF F there be two ftraight lines, one of which is divided into any number of parts; the rectangle contained by the two straight lines, is equal to the rectangles contained by the undivided line, and the feveral parts of the divided line. Let A and BC be two ftraight lines; and let BC be divided into any parts in the points D, E; the rectangle contained by the ftraight lines A, BC is equal to the rectangle contained by A, BD, toge- B ther with that contained by From the point B draw a F DE C K L H A 31. 1. through G draw c GH parallel to BC; and through D, E, C, draw c DK, EL, CH parallel to BG; then the rectangle BH is equal to the rectangles BK, DL, EH, and BH is contained by A, BC, for it is contained by GB, BC, and GB is equal to A; and BK is contained by A, BD, for it is contained by GB, BD, of which GB is equal to A; and DL is contained by A, DE, d34. 1. because DK, that is, d BG, is equal to A; and in like man ner the rectangle EH is contained by A, EC: Therefore the rectangle contained by A, BC is equal to the feveral rectangles contained by A, BD, and by A, DE, and alfo by A, EC. Therefore, if there be two ftraight lines, &c. Q. E. D. IF Fa ftraight line be divided into any two parts, the rectangles contained by the whole and each of the parts, are together equal to the square of the whole line. Let C B Let the ftraight line AB be divided into any two parts in Book II. the point C; the rectangle contained by AB, BC, together with the rectangle AB, AC, fhall be equal to the fquare of AB. * D F E Upon AB defcribe a the fquare ADEB, and through C draw b ̄CF, parallel to AD or BE: then AE is equal to the rectangles AF, CE; and AE is the fquare of AB; and AF is the rectangle contained by BA, AC; for it is contained by DA, AC, of which AD is equal to AB; and CE is contained by AB, BC, for BE is equal to AB: therefore the rectangle contained by AB, AC together with the rectangle AB, BC, is equal to the fquare of AB. If therefore a straight line, &c. Q. E. D. IF Fa ftraight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the fquare of the forefaid part. Let the straight line AB be divided into two parts in the point C; the rectangle AB, BC is equal to the rectangle AC, CB, together with the fquare of BC. Upon BC defcribe a the fquare CDEB, and produce A F C D a 46. I. b 31. I. a 46. I. B b 31. r. E equal to BC; and AD is contained by AC, CB, for CD is *N. B. To avoid repeating the word contained too frequently, the Aangle contained by two ftraight lines AB, AC is fometimes fimply called the rectangle AB, AČ. 52 Book II equal to CB; and DB is the fquare of BC: therefore the rectangle AB BC is equal to the rectangle AC, CB together with the fquare of BC. If therefore a straight, &c. Q. E. D. a 46. I. I PROP. IV. THEOR. Fa ftraight line be divided into any two parts, the fquare of the whole line is equal to the fquares of the two parts, together with twice the rectangle contained by the parts. Let the ftraight line AB be divided into any two parts in C; the fquare of AB is equal to the fquares of AC, CB, and to twice the rectangle contained by AC, CB. Upon AB describe a the fquare ADEB, and join BD, and b 31. 1 through C draw b CGF parallel to AD or BE, and through G draw HK parallel to AB or DE. And because CF is parallel to AD, and BD falls upon them, the exterior angle e 29. 1. BGC is équal to the interior and oppofite angle ADB; but d5 1. ADB is equal d to the angle ABD, because BA is equal to AD, being fides of a fquare; wherefore the angle CGB is e 6. 1. equal to the angle GBC; and therefore the fide BC is equal f 34. 1. to the fide CG: but CBs equal f alfo to GK, and CG to BK; where-, fore the figure CCKB is equilateral. It is likewife rectangular; for H the angle CBK being a right angle, the other angles of the parallelo Cor. 46.1. gram CGKB are alfo right angles g. Wherefore CGKB is a fquare, and it is upon the fide CB. For the fame reafon HF alfo is a fquare, A D e B G K F E and it is upon the fide HG, which is equal to AC; therefore HF, CK are the fquarcs of AC, CB. And because the 43. 1. complement AG is equal h to the complement GE, and because AG is the rectangle contained by AC, CG, that is, by AC, CB; GE is alfo equal to the rectangle AC, CB; wherefore AG, GE are equal to twice the rectangle AC, CB. And HF, CK are the fquares of AC, CB; wherefore the four figures HF, CK, AG, GE are equal to the fquares of AC, AC, CB, and to twice the rectangle AC, CB: But HF, CK, Book II. AG, GE make up the whole figure ADEB, which is the fquare of AB: Therefore the fquare of AB is equal to the fquares of AC, CB, and twice the rectangle AC, CB. Wherefore if a straight line, &c. QE. D. COR. From the demonftration, it is manifeft that the parallelograms about the diameter of a fquare are likewife fquares. IF PROP. V. THEOR. TF a ftraight line be divided into two equal parts, and alfo into two unequal parts; the rectangle contained by the unequal parts, together with the fquare of the line between the points of fection, is equal to the fquare of half the line. Let the ftraight line AB be divided into two equal parts in the point C, and into two unequal parts at the point D; the rectangle AD, DB, together with the fquare of CD, is equal to the fquare of CB. Upon CB defcribe a the fquare CEFB, join BE, and a 46. 1. through D draw b DHG párallel to CE or BF; and through b 31. 1. H draw KLM parallel to CB or EF; and alfo through A draw AK parallel to CL or BM: And because the complement CH is equal to the complement HF, to each of thefe c 43. 1. add DM; therefore the whole CM is equal to the whole DF; but CM is equal d to AL, K because AC is equal to A & 36. 1. DB; and DF together with CH is the gnomon CMG; therefore the gnomon CMG is equal to the rectangle AD, DB. To each of thefe add LG, which is e uale to the fquare of CD; therefore the gnomon CMG, together with LG, is equal to the rectangle AD, DB, together with the E 3 fquare e Cor 4. 20 Book II. fquare of CD: But the gnomon CMG and LG make up the whole figure CEFB, which is the fquare of CB: Therefore the rectangle AD, DB, together with the fquare of CD, is equal to the fquare of CB. Wherefore, if a straight line, &c. Q. E. D. a 46. I. From this propofition it is manifeft, that the difference of the squares of two unequal lines AC, CD, is equal to the rectangle contained by their fum and difference. IF PROP. VI. THE OR. [F a ftraight line be bifected, and produced to any point; the rectangle contained by the whole line thus produced, and the part of it produced, together with the fquare of half the line bisected, is equal to the fquare of the ftraight line which is made up of the half and the part produced. Let the straight line AB be bifected in C, and produced to the point D; the rectangle AD, DB, together with the fquare of CB, is equal to the fquare of CD. Upon CD describe a the fquare CEFD, join DE, and b31. 1. through B draw b BHG parallel to CE or DF, and through H draw KLM parallel to AD or EF, and alfo through A draw AK parallel to CL or DM: and because AC is equal to CB, the rectangle AL A c36. . is equal c to CH; but & 43. I. CH is equal d to HF; therefore allo AL is equal K add CM; therefore the B C D AM is the rectangle con-. E G F Cor. 4.2. tained by AD, DB, for DM is equal e to DB: therefore the gnomon CMG is equal to the rectangle AD, DB; add to each of thefe LG, which is equal to the fquare of CB, therefore the ieangle AD, DB, together with the fquare of CB, is equal to the gnomon CMG, together with LG: But the gnomon CMG together with LG makes up the whole figure CEFD, |