Upon AB defcribe a the fquare ADEB, and join BD, and through C drawb 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 BGC is equal to the interior and oppofite angle ADB; but ADB is equal to the angle ABD, becaufe BA is equal to AD, being fides of a fquare; wherefore the angle CGB A is equal to the angle GBC; and there- с B G K FE is a right angle; wherefore GCB is a right angle; and therefore alfo the angles f CGK, GKB oppofite to thefe are right angles, and CGKB is rectangular: But it is alfo equilateral, as was demonstrated; wherefore it is a fquare, and it is upon the fide CB: For the fame reafon HF alfo is a fquare, and it is upon the fide HG which is equal to AC: Therefore HF, CK are the fquares of AC, CB; and becaule the complement AG is equal to the complement GE, and that AG is the rectangle contained by AC, CB, for GC is equal to CB; therefore GE is also 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, CB, and to twice the rectangle AC, CB: But HF, CK, 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 ftraight line, &c. Q. E. D. COR. From the demonftration, it is manifeft, that the parallelograms about the diameter of a fquare are likewite fquares. PROP Book II. PROP. V. THE OR. 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 the fquare CEFB, join BE, and through a 46. f. D draw DHG parallel to CE or BF; and through H draw b 31. 1. KLM parallel to CB or EF; and alfo through A draw AK parallel to CL or BM: And because the complement CH is e с qual to the complement HF, to each of these add DM; € 43. I. therefore the whole CM is equal to the whole DF; and the whole AH is equal to DF and CH: But AH is the rectangle contained by AD, DB, for DH is equal e 2. to DB; and DF together with CH is the gnomon CMG; Cor. 4. therefore the gnomon CMG is equal to the rectangle AD, DB: To each of thefe add LG, which is equal to the fquare of CD; therefore the gnomon CMG, together with LG, is equal to the rectangle AD, DB, together with the 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. From this propofition it is manifeft, that the difference of the fquares of two unequal lines AC, CD, is equal to the rectangle contained by their fum and difference. Book II, a 46. I. b 31. I. [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 of the line bifected, is equal to the fquare of the ftraight line which is made up of the half and the part produced. Let the ftraight line AB be bifected in C, and produced to the point D; the rectangle AD, DB, together with the square of CB, is equal to the fquare of CD. Upon CD defcribe the fquare CEFD, join DE, and through B draw 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 becaufe AC A is equal to CB, the rectangle c 36. I. AL is equal to CH; but d 43. I. CH is equal to HF; there- HF: To each of these add CM; therefore the whole с BD AD, DB, for DM is equal e Cor. 4. 2. 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 rectangle AD, DB, together with the fquare of CB, is equal to the gnomon CMG and the figure LG: But the gnomon CMG and LG make up the whole figure CEFD, which is the fquare of CD; therefore the rectangle AD, DB, together with the fquare of CB, is equal to the fquare of CD. Wherefore, if a straight line, &c. Q. E. D. IF PROP. VII. THEOR. Fa ftraight line be divided into any two parts, the fquares of the whole line, and of one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the fquare of the other part. Let the ftraight line AB be divided into any two parts in the he point C; the fquares of AB, BC are equal to twice the Book II. rectangle AB, BC together with the fquare of AC. Upon AB defcribe the fquare ADEB, and conftruct the a 46. I. figure as in the preceding propofitions: And because AG is B G K equal to GE, add to each of them CK; the whole AK is b 43. I. therefore equal to the whole CE; therefore AK, CE are double of A AK: But AK, CE are the gnomon AKF together with the fquare CK; therefore the gnomon AKF, toge- H ther with the fquare CK, is double of AK: But twice the rectangle AB, BC is double of AK, for BK is equal to BC: Therefore the gnomon AKF, together with the fquare CK, is equal to twice the rectangle D AB, BC: To each of these equals c Cor. F E add HF, which is equal to the fquare of AC; therefore the gnomon AKF, together with the fquares CK, HF, is equal to twice the rectangle AB, BC and the fquare of AC: But the gnomon AKF, together with the fquares CK, HF, make up the whole figure ADEB and CK, which are the fquares of AB and BC: Therefore the fquares of AB and BC are equal to twice the rectangle AB, BC, together with the fquare of AC. Wherefore, if a ftraight line, &c. Q. E. D. It a ftraight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the fquare of the other part, is equal to the fquare of the ftraight line which is inade up of the whole and that part. Let the straight line AB be divided into any two parts in the point C; four times the rectangle AB, BC, together with the Iquare of AC, is equal to the fquare of the ftraight line made up of AB and BC together. Produce AB to D, fo that BD be equal to CB, and upon AD defcribe the fquare AEFD; and conftruct two figures fuch as in the preceding. Because CB is equal to BD, and that CB is equal to GK, and BD to KN; therefore GK is a 34. I. equal b36. I. € 43.1. Book IL equal to KN: For the fame reafon, PR is equal to RO; and because CB is equal to BD, and GK to KN, the rectangle CK is equal to BN, and GR to RN: But CK is equal to RN, because they are the complements of the parallelogram CO; therefore alfo BN is equal to GR; and the four rectangles BN, CK, GR, RN are therefore equal to one another, and fo are quadruple of one of them CK: Again, because CB is equal to BD, and that BD is Cor. 4. 2. equal to BK, that is, to CG; e CB D GK N HL and CB equal to GK, that is, to A GP; therefore CG is equal to GP: And because CG is equal to M GP, and PR to RO, the rectangle AG is equal to MP, and PL to X RF: But MP is equal to PL, because they are the complements of the parallelogram ML; wherefore AG is equal alfo to RF: Therefore the four rectangles E AG, MP, PL, RF are equal to one another, and fo are quadruple of one of them AG. And it was demonftrated, that the four CK, BN, GR, RN are quadruple of CK: Therefore the eight rectangles which contain the gnomon AOH, are quadruple of AK: And becaufe AK is the re tangle contain ed by AB, BC, for BK is equal to BC, four times the rectangle AB, BC is quadruple of AK: But the gnomon AOH was demonftrated to be quadruple of AK; therefore four times the rectangle AB, BC is equal to the gnomon AOH. Cor. 4. 2. To each of thefe add XH, which is equal to the fquare of AC: Therefore four times the rectangle AB, BC, together with the fquare of AC, is equal to the gnomon AOH and the fquare XH: But the gnomon AOH and XH make up the fi gure AEFD which is the fquare of AD: Therefore four times the rectangle AB, BC. together with the fquare of AC, is equal to the fquare of AD, that is, of AB and BC added together in one ftraight line. Wherefore, if a ftraight line, &c. Q. E. D. PROP. |