46. I. † 31. I. 36. 1. 43. I. this. $ tb.s. PROPOSITION VI. THEOREM. 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 Line, together with the Square of half the firft Line, is equal to the Square of [the Line compounded of half the Line and the added Line, taken as one Line. L ET the Right Line A B be bifected in the Point C, and BD added directly thereto. I fay, the Rectangle under A D, and BD, together with the Square of BC, is equal to the Square of CD. For, defcribe CEFD, the Square of C D, and join DE; draw + BHG thro' B, parallel to CE, or DF, and K L M thro' H, parallel to A D, or E F, as alfo A K thro' A, parallel to CL, or DM. * Then because AC is equal to CB, the Rectangle AL fhall be equal to the Rectangle CH; but CH is equal to H F. Therefore A L will be equal to HF; and adding C M, which is common to both, then the whole Rectangle A M is equal to the Gnomon NXO. But AM is that Rectangle which is Cor. 4. ofcontained under A D, and DB; for DM is + equal to DB; therefore the Gnomon NXO is equal to the Rectangle under AD, and D B. Add LG, Cor. 4. of which is common, viz. t the Square of C B; and then the Rectangle under AD, DB, together with the Square of BC, is equal to the Gnomon NXO with L G. But the Gnomon NX 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 CD. 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 whale and added Line (taken as one Line) and the added added Line, together with the Square of half the firft Line, is equal to the Square of [the Line compounded of] half the Line and the added Line, taken as one Line; which was to be demonftrated. PROPOSITION VII. If a Right Line be any bow cut, the Square of LET the Right Line A B be any how cut in the Point C. I fay, the Squares of A B, and BC, together, are equal to double the Rectangle contained under A B, and B C, together with the Square, made of A C. For let the Square of AB be described, viz. * 46. z. ADE B, and conftruct the Figure. Then, because the Rectangle A G is + equal to the † 4;. *. Rectangle GE; if CF, 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, CE, make up the Gnomon K L M, and the Square CF. Therefore the Gnomon K L M, together with the Square CF, fhall be double to the Rectangle A F. But double the Rectangle under A B, and BC, is double the Rectangle AF; for BF is equal to B C. Therefore the Gnomon KLM, I Cor. 4and the Square C F, are equal to double the Rectangle contained under A B, and BC. And if HF, which is common, being the Square of AC, be added to both, then the Gnomon K L M, and the Squares CF, HF, A Figure is faid to be conftruct.d, when Lines drawn in a Paralilogram, 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 confiructed, when two Right Lines, parallel to the Sides, make four Parallelograms about the Diameter, and four Complements. E 3 are Hyp. † 34. I. 36. 1. 43.1. are equal to double the Rectangle contained under A B, and B C, together with the Square of A C. But the Gnomon KLM, together with the Squares C F, HF, are equal to ADEB, and CF, viz. the Squares of AB, BC. Therefore the Squares of A B, and BC, are together equal to double the Rectangle contained under A B, and B C, together with the Square of A C. Therefore, if a Right Line be any how cut, the Square of the whole Line, together with the Square of one of the Segments, is equal to double the Rectangle contained under the whole Line, and the faid Segment, together with the Square, made of the other Segment; which was to be demonftrated. PROPOSITION VIII. THEOREM. If a Right Line be any how cut into two Parts, four Times the Rectangle, contained under the whole Line, and one of the Parts, together with the Square of the other Part, is equal to the Square of [the Line compounded of] the whole Line and the firft Part, taken as one Line. LET the Right Line A B be cut any how in C. I fay, four Times the Rectangle contained under A B, and BC, together with the Square of AC, is equalto the Square of A B, and B C, taken as one Line. For, let the Right Line A B be produced to D, fo that BD be equal to BC; defcribe the Square A EFD, on AD, and conftruct the double Figure. Now, fince CB is equal to BD, and alfo to + G K, and BD is equal to KN; therefore G K fhall be likewife equal to KN: By the fame Reafoning, PR is equal to RO. And fince C B is equal to BD, and G K to K N, the Rectangle CK will be equal to the Rectangle B N, and the Rectangle G R to the Rectangle R N. But CK is equal to RN; for they are the Complements of the Parallelogram CO. Therefore BN is equal to G R, and the four Squares BN, CK, GR, RN, are equal to each other; and fo they are together quadruple CK. Again, becaufe CB is equal to BD, and B D to B K, that is, equal to CG; and the faid C B is equal alfo to G K, that is, to GP; therefore C G fhall be equal to GP. But PR is equal to RO; therefore the Rectangle AG fhall be equal to the Rectangle MP, and the Rectangle PL equal to RF. But MP is equal to PL; for they are the Complements of the Parallelogram M L. Wherefore AG is equal alfo to R F. Therefore the four Parallelograms AG, MP, PL, RF, are equal to each other, and accordingly they are together quadruple of A G. But it has been proved, that the four Squares CK, BN, GR, RN, are quadruple of CK. Therefore the four Rectangles, and the four Squares, making up the Gnomon STY, are together quadruple of AK; and because A K is a Rectangle contained under AB, and BC, for BK is equal to BC; therefore four Times the Rectangle under A B and B C, will be quadruple of A K. But the Gnomon STY has been proved to be quadruple of A K. And fo four Times the Rectangle contained under A B, and BC, is equal to the Gnomon S T Y. And if XH, being equal to + the Square of A C, which ist Cor. 4. f common, be added to both; then four Times the this. Rectangle contained under A B, and BC, together with the Square of AC, is equal to the Gnomon STY, and the Square XH. But the Gnomon STY and X H make A E FD, the whole Square of AD. Therefore four Times the Rectangle contained under A B, and B C, together with the Square of A C, is equal to the Square of A D, that is, of A B and BC taken as one Line. Wherefore, if a Right Line be any how cut into two Parts, four Times the Rectangle contained under the whole Line, and one of the Parts, together with the Square of the other Part, is equal to the Square of [the Line compounded of] the whole Line and the firft Part, taken as one Line; which was to be demonftrated. II. I. † 31. 1. 5. I. **Cor. 3. 32. I. PROPOSITION IX. THEOREM. If a Right Line be any how cut into two equal, and two unequal Parts; then the Squares of the unequal Parts, together, are double to the Square of the half Line, and the Square of the intermediate Part. LE ET any Right Line A B be cut unequally in D, and equally in C. I fay, the Squares of A D, DB, together, are double to the Squares of A C and CD together. For, let CE be drawn from the Point C at Right Angles to A B, which make equal to A C, or CB; and join EA, E B. Alfo through D let + DF be t drawn parallel to C E, and F G through F parallel to A B, and draw A F. Now, because AC is equal to CE, the Angle EAC will be equal to the Angle A E C; and fince the Angle at C is a Right one, the other Angles, A EC, E A C, together, fhall make one Right Angle, and are equal to each other: And fo AEC, EA C, are each equal to half a Right Angle. For the fame Reasons are alfo C E B, E BC, each of them half a Right Angle. Therefore the whole Angle A E B is a Right Angle. And fince the Angle GEF is half a † 29. 1. Right one, and EGF is a Right Angle; for it is + equal to the inward and oppofite Angle ECB; the other Angle E F G will be alfo equal to half a Right one. Therefore the Angle G E F is equal to the Angle EFG. And fo the Side EG is equal to the Side GF. Again, because the Angle at B is half a Right one, and FDB is a Right one, because equal to the inward and oppofite Angle ECB, the other Angle BFD will be half a Right Angle. Therefore the Angle at B is equal to the Angle BFD; and fo the Side D F is equal to the Side DB. And because AC is equal to CE, the Square of A C will be equal to the Square of CE. Therefore the Squares of A C, and C E, together, are double to the Square of A C; +47.1, but the Square of E A is † equal to the Squares of 6.1. AC, |