PROPOSITION I. THEOREM. If there be two Right Lines, and one of them be divided into any Number of Parts; the Reangle comprehended under the whole Line and the divided Line, fhall be equal to all the Redangles contained under the whole Line, and the Several Segments of the divided Line. L ET A and BC be two Right Lines, whereof BC is cut or divided any how in the Points D, E. I fay, the Rectangle contained under the Right Lines A and BC, is equal to the Rectangles contained under A and B D, A and D E, and A and E C. For, let* BF be drawn from the Point B, at Right* 11. 1, Angles, to BC; and make + B G equal to A; and † 3.1. let GH be drawn thro' G parallel to BC: Like- † 31. wife let there he drawn DK, EL, CH, through D, E, C, parallel to B G. Then the Rectangle B H, is equal to the Rectangles BK, DL, and EH, but the Rectangle BH is that contained under A and BC; for it is contained under GB, BC; and G B is equal to A; and the Rectangle B K is that contained under A and BD; for it is contained under G B and B D; and GB is equal to A; and the Rectangle D L is that contained under A and D E, because D K, that is, B G, is equal to A: So likewife the Rectangle E H is that contained under A and E C. Therefore the Rectangle under A and BC is equal to the Rectangles under A and B D, A and D E, and A and E C. Therefore, if there be two Right Lines given,. and one of them be divided into any Number of Parts, the Rectangle comprehended under the whole Line and the divided Line, fhall be equal to all the Rectangles contained under the whole Line, and the feveral Segments of the divided Line; which was to be demonftrated. PRO 46. i. PROPOSITION II. THEOREM. If a Right Line be any bow divided, the Rettan- LET the Right Line A B be any how divided in the For let the Square A DEB be described * on AB, and thro' C let C F be drawn parallel to A D or BE. Therefore AE is equal to the Rectangles AF and CE. But AE is a Square defcribed upon AB; and AF is the Rectangle contained under BA and AC; for it is contained under DA and A C, whereof A D is equal to AB; and the Rectangle CE is contained under A B and B C, fince B E is equal to A B. Wherefore the Rectangle under A B and A C, together with the Rectangle under AB and BC, is equal to the Square of A B. Therefore, if a Right Line be any how divided, the Rectangles contained under the whole Line, and each of the Segments, or Parts, are equal to the Square of the whole Line; which was to be demonftrated. PROPOSITION III. THEOREM. If a Right Line be any bow cut, the Rectangle 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 firft-mentioned Part. LE ET the Right Line A B be any how cut in the Point C. I fay, the Rectangle under A B and BC is equal to the Rectangle under AC and BC, together with the Square defcribed on B C. For For defcribe the Square CDEB upon BC; 46. 1. produce ED to F; and let A F be drawn + thro' A,† 31. 1. parallel to CD or BE. t Then the Rectangle A E fhall be equal to the two Rectangles A D, CE: And the Rectangle A E is that contained under A B and BC; for it is contained under A B and B E, whereof BE is equal to BC: And the Rectangle AD is that contained under AC and CB, fince DC is equal to CB: And DB is a Square defcribed upon BC. Wherefore the Rectangle under AB and BC is equal to the Rectangle under A C and CB, together with the Square defcribed upon B C. Therefore, if a Right Line be any how cut, the Rectangle 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. THEOREM. If a Right Line be any how cut, the Square which LET the Right Line A B be any how cut in C. For* defcribe the Square A DEB upon A B, join * 46. 1. BD, and thro' C draw + CGF parallel to AD or BE;† 31.1. and also thro' G draw HK parallel to A B or D E. Then, because CF is parallel to A D, and B D falls upon them, the outward Angle BGC fhall be equal † 29. 1, to the inward and oppofite Angle A DB; but the Angle A D B is equal to the Angle A B D, fince the * 5. 1. Side B A is equal to the Side A D. Wherefore the Angle CGB is equal to the Angle G BC; and fo the Side B C equal † to the Side CG; but likewise † 6. 1. the Side C B is equal to the Side GK, and the Side 1 34.1. CG to BK. Therefore GK is equal to K B, and E CGKB + 29. T. 43.1. CGK B is equilateral. I fay, it is alfo Right-angled; for, because C G is parallel to B K, and C B falls on them, the Angles K BC, GCB †, are equal to two Right Angles. But KBC is a Right Angle. Wherefore GCB alfo is a Right Angle, and the oppofite Angles CG K, G K B, shall be Right Angles. Therefore CGK B is a Rectangle. But it has been proved to be equilateral. Therefore CGKB is a Square defcribed upon B C. For the fame Reafon H F is alfo a Square made upon HG, and (becaufe HG is 34. 1. equal to AC+) it is equal to the Square of A C. Wherefore HF and CK are the Squares of A C and CB. And, because the Rectangle AG is equal to the Rectangle GE, and AG is that which is contained under AC and CB; for GC is equal to CB; therefore GE fhall be equal to the Rectangle-under A C, 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 CK, are the Squares of AC, CB. Therefore the four Figures HF, CK, AG, GE, are equal to the Squares of AC and C B, with twice the Rectangle contained under AC and CB. But HF, CK, AG, GE, make up the whole Square of A B, viz. A DE B. Therefore the Square of A B is equal to the Squares of A C and C B, together with twice the Rectangle contained under AC and CB. 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 fand about the Diameter of a Square, are likewife Squares. PRO PROPOSITION V. THEORE M. 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 DB, together with the Square of CD, is equal to the Square of BC. * For + defcribe CEF B, the Square of BC; draw † 46. 1. BE, and through D draw DHG, parallel to CE, or 31. 1. BF; and through H draw KLO, parallel to C B, or EF; and AK through A, parallel to CL, or B O.. Now the Complement C H is equal to the Com-t 43. 16 plement HF. Add D O, which is common to both of them, and the whole C O is equal to the whole DF: But CO is equal to A L, because AC is equal to CB+; therefore AL is equal to DF; and, adding † 36. 1. CH, which is common, the whole A H fhall be equal to FD, DL, together. But A H is the Rectangle contained under AD, and D B; for DH is equal to Cor. 4. of DB, and FD, D L, is the Gnomon MN X; there this. fore M N X is equal to the Rectangle contained under A D, and DB; and if LG, being common,_and equal to the Square of C D, be added, then the Gnomon M N X, and LG, are equal to the Rectangle contained under AD, and D B, together with the Square of CD; but the Gnomon, MN X, and LG, make up the whole Square CEF B, viz. the Square of C B. Therefore the Rectangle under A D, and DB, together with the Square of CD, 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. PRO E 2 |