For describe the Square CDE B upon BC; 46. 1. produce ED to F; and letA F be drawn + through A, † 31. 1. parallel to CD or BE. Then the Rectangle A E fhall be equal to the two Rectangles AD, CE: And the Rectangle A E is that contained under A B and B C; for it is contained under A B and B E, whereof B E is equal to BC: And the Rectangle AD is that contained under A C and C B, fince DC is equal to CB: And D B is a Square described upon B C. Wherefore the Rectangle under AB and B C is equal to the Rectangle under A Cand CB, together with the Square described upon BC. Therefore, if a Right Line be any how cut, the ReƐtangle contained under the whole Line, and one of its Parts, is equal to the Rectangle contained under the two Parts, together with the Square of the first-mentioned Part; which was to be demonftrated. PROPOSITION IV. If a Right Line be any how cut, the Square which L ET the Right Line A B be any how cut in C. I fay, the Square made on A B is equal to the Squares of A C, C B, together with twice the Rectangle contained under A C, C B. For defcribe the Square ADEB upon A B, join * 46. 1. BD, and thro' C draw + CG F parallel to A D or BE;t 31. 1. and alfo thro' G draw HK parallel to AD or D E. Then, because C F is parallel to A D, and B D falls upon them, the outward Angle BGC fhall be + equal I 29. 1. to the inward and oppofite Angle A D B; but the Angle ADB is equal to the Angle A B D, fince the 5.1. Side B A is equal to the Side AD. Wherefore the Angle C G B is equal to the Angle G BC; and fo the Side BC equal to + the Side CG; but likewise † 6. 1. the Side C Bis is equal to the Side G K, and the Side † 34. 1. CG to BK. Therefore G K is equal to K B, and E CGKB 29. I. + 34.8. 43. Z. CG KB is equilateral. I fay, it is alfo Right-angled; for, because CG is parallel to B K, and C B falls on them, the Angles KBC, GCB *, are equal to two Right Angles. But KBC is a Right Angle. Wherefore G C B alfo is à Right Angle, and the oppofite Angles CGK, GKB, fhall be Right Angles. Therefore C G K B is a Rectangle. But it has been proved to be equilateral. Therefore, C G K B is a Square defcribed upon BC. For the fame Reafon HF is alfo a Square made upon H G, and (becaufe HG is equal to A C +) it is equal to the Square of A C. Wherefore H F and CK are the Squares of A C and C B. And, because the Rectangle A G* is equal to the Rectangle GE, and A G is that which is contained under AC and CB; for G C is equal to CB; therefore GE fhall be equal to the Rectangle under AC and C B. Wherefore the Rectangles A G, and GE, are equal to twice the Rectangle contained under A C, and CB; and HF and C K, are the Squares of A C, CB. Therefore the four Figures HF, CK, AG, GE, are equal to the Squares of A C and C B, with twice the Rectangle contained under A C and CB. But HF, CK, AG, G E, make up the whole Square of A B, viz. A DE B. Therefore the Square of AB is equal to the Squares of A C and CB, together with twice the Rectangle contained under A C and C B. Wherefore, if a Right Line be any how cut, the Square which is made on the whole Line will be equal to the Squares made on the Segments thereof, together with twice the Rectangle contained under the Segment; which was to be demonftrated. Coroll. Hence it is manifeft, that the Parallelograms which ftand about the Diameter of a Square, are likewife Squares. PRO If a Right Line be cut into two equal Parts, and into two unequal ones; the Rectangle under the unequal Parts, together with the Square that is made of the intermediate Distance, is equal to the Square made of half the Line. LET any Right Line A B be cut into two equal Parts in C, and into two unequal Parts in D. I fay, the Rectangle contained under AD, and D B, together with the Square of CD, is equal to the Square of B C. For+ defcribe CEF B, the Square of BC; draw t 46. 1. B E, and through D draw * DH G, parallel to CE, or B F; and through H draw KLO, parallel to * 31. 1. CB, or EF; and A K through A, parallel to C L, or BO. this. Now the Complement C H ist equal to the Com- 43. %, plement HF. Add DO, which is common to both of them, and the whole CO is equal to the whole DF: But CO is equal to A L, because AC is equal to CB+; therefore A L is equal to DF; and adding CH, which is common, the whole A H fhall be equal to FD, D) L, together. But A H is the Rectangle contained under AD, and D B; for DH is equal to Cor. 4. of D B, and FD, DL, is the Gnomon M N X; therefore M N X is equal to the Rectangle contained under A D, and DB; and if L G, being common, and equal to the Square of CD; be added, then the Gnomon MN X, and L G, are equal to the Rectangle contained under A D, and D B, together with the Square of CD; but the Gnomon, MNX, and LG, make up the whole Square C E F B, viz. the Square of C B. Therefore the Rectangle under A D, and D B, together with the Square of C D, is equal to the Square of C B. Wherefore, if a Right Line be cut into two equal Parts, and into two unequal ones; the Rectangle under the unequal Parts, together with the Square that is. made of the intermediate Distance, is equal to the Square made of half the Line; which was to be demonftrated. E 2 PRO #46. I. † 31.1. 36. I. 43. 1, PROPOSITION VI. If a Right Line be divided into two equal Parts, LET the Right Line A B be bifected in the Point For, defcribe CEFD, the Square of CD, and join DE; draw + BHG thro' B, parallel to C E, or DF, and K L M thro' H, parallel to AD, or EF, as alfo A K thro' A, parallel to C EL, or D M. * Then, because AC is equal to C B, the Rectangle AL fhall be equal to the Rectangle CH; but CH is equal to HF. Therefore A L will be equal to HF; and adding CM, which is common to both, then the whole Rectangle A M is equal to the Gnomon NXO. But A M is that Rectangle which is † Cor. 4. of contained under A D, and D B; for DM is † equal to DB; therefore the Gnomon NXO is equal to the Rectangle under A D and D B. Add LG, which is common, viz. the Square of C B; and then 1 Cor. 4. of the Rectangle under A D, D'B, together with the this zbis. Square of B C, is equal to the Gnomon NXO with LG. But the Gnomon N X O, and L G, together, make up the Figure CEFD, that is, the Square of CD. Therefore the Rectangle under AD, and D B, together with the Square of BC, is equal to the Square of C D. Therefore, if a Right Line be divided into two equal Parts, and another Right Line be added directly to the fame, the Rectangle contained under [the Line compounded of] the whole and added Line [taken as one Line] and the added added Line, together with the Square of half the first Line, PROPOSITION VII. If a Right Line be any bow cut, the Square of the LET the Right Line A B be any how cut in the Point C. I fay the Squares of A B, and B C, together are equal to double the Rectangle contained under A B, and BC, together with the Square made of A C. For let the Square of A B be ADE B, and conftruct & the Figure. described, viz. * 46. 1. +43. Then, because the Rectangle A G is † equal to the Rectangle G E; if C F, which is common, be added to both, the whole Rectangle A F fhall be equal to the whole Rectangle CE; and fo the Rectangles A F, CE, taken together, are double to the Rectangle A F; but A F, C E, make up the Gnomon K L M, and the Square C F. Therefore the Gnomon K L M, together with the Square C F, fhall be double to the Rectangle AF. But double the Rectangle under A B, and B C, is double the Rectangle A F; for BF is equal to B C. Therefore the Gnomen K L M, 1 Cor. 4à and the Square CF, are equal to double the Rectangle contained under A B, and B C. And if HF, which is common, being the Square of A C, be added to both, then the Gnomon KLM, and the Squares CF, HF, A Figure is faid to be confiructed, when lines drawn in a Parallelogram, parallel to the Sides thereof, cut the Diameter in one Point, and make two Parallelograms about the Diameter, and two Complements. So likewife a double Figure is faid to be confiruted, when two Right Lines, parallel to the Sides, make four Farallelograms about the Diameter, and four Complements. E 3 are |