are equal to the double Rectangle contained under A B, and B C, together with the Square of A C. But the Gnomon KLM, together with the Squares CF, HF, are equal to A DEB, and CF, viz. the Squares of A B, 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 said Segment, together with the Square, made of the other Segment ; which was to be demonftrated. PRPOPSITION VIII. THEOREM. four Times the Rectangle, contained under the Line and i be first part, taken as one Line. say, four Times the Rectangle contained under AB, and B C, together with the Square of AC, is equal to the Square of A B, and B C, taken as one Line. For, let the Right Line A B be produced to D, so that B D be equal to BC; describe the Square A EFD, on A D, and construct the double Figure. Now, since C B is * equal to BD, and also to t, GK, and B D is equal to K N; therefore G K shall be likewise equal to KN: By the fame Reasoning, PR is equal to RO. And fince C B is equal to BD, and G K to K N, the Rectangle CK will I be equal to the Rectangle B N, and the Rectangle G R to the Rectangle RN. But C K is equal to RN; for they are the Complements of the 'Parallelogram CO. Therefore B'N is equal to GR, and the four Squares BN, CK, GR, RN, are equal to each other; and so they are together quadruple CK. Again, because C B is equal to BD, and B D to B K, that is, equal to to CG; and the said C B'is equal also to G K, that is, to GP; therefore CG shall be equal to G P. But P R is equal to RO; therefore the Rectangle AG (hall be equal to the Rectangle MP, and the Rectangle P L equal to RF. But MP is equal to PL; for they are the Complements of the Parallelogram ML. Wherefore A G is equal also to RF. Therefore the four Parallelograms AG, MP, PL, RF, are equal to each other, and accordingly they are together quadruple of AG. 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 A B, and B C, for B K is equal to BC; therefore four times the Rectangle under A B, and BC; will be quadruple of A K. But the Gnomon STY has been proved to be quadruple of A K. And lo four Times the Rectangle contained under A B and B C, is equal to the Gnomon S TY. And if X H, being equal to + the Square of A C, which is + Cor. 4. of common to be added to boih; then four Times the ibis. Rectangle contained under A B, and B C, together with the Square of A C, is equal to the Gnomon STY, and the Square X H. But the Gnomon STY and X H make A EFD, the whole Square of AD. Therefore four Times the Rectangle contained under A B and BC, 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 first part, taken as one Line, which was to be de. monstrated. PROPOSITION IX. THEOREM. and two unequal Parts; then the Squares of L ET any Right Line A B be cut unequally in D, and equally in C. I fay, the Squares of A D DB, together, arc double to the Squares of A C and C D together. For, let * CE be drawn from the Point C at Right Angles to A B, which make equal to AC, or CB; and join E A, E B. Also through D let + D F be 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 f equal to the Angle E AC; and fince the Angle at C is a Right one, the other Angles, AEC, EAC, together, Tall * make one Right Angle, and are equal to each other. And fo A EC, EA C, are each equal to half a Right Angle. For the fame Reasons are also CEB, EBČ, each of them half a Right Angle. Therefore the whole Angle A E B is a Right Angle. And since the Angle GEF is half a Right one, and EGF is a Right Angle; for it is t equal to the inward and opposite Angle ECB; the other Angle E F G will be also equal to half. a Right Therefore the Angle G E F is equal to the Angle E F G. And so the Side E G is I equal to the Side GF. Again, because the Angle at B is half a Right one, and F D B is a Right one, because equal to the inward and opposite Angle E CB, the other Angle BF D will be half a Right Angle. Therefore the Angle at B is equal to the Angle BFD; and fa the Side D F is equal to the Side D B. And because A C is equal to C E, the Square of A C will be equal to the Square of CE. Therefore the Squares of AC, and CE together, are double to the Square of AC; but the Square of E A is f equal to the Squares of AC, one. A C, and CE, together, since ACE is a Right Angle. Therefore the Square of E A is double to the Square of AC. Again, because E G is equal to GF, and the Square of E G is equal to the Square of G F; therefore the Squares of EG, GF, together, are double to the Square of GF. But the Square of EF is + equal to the Squares of EG, GF. Therefore + 47. s. the Square of EF is double the Square of GF: But GF is equal to CD; and so the Square of EF is double to the Square of C D. But the Square of A E is likewise double to the Square of AC. Wherefore the Squares of A E and E F, are double to the Squares of A C, and CD. But the Square of A F is f equal to the Squares of AE and EF;, because the Angle A E F is a Right Angle, and consequently the Square of A F is double to the Squares of A C and CD. But the Squares of A D, and D F, are equal to the Square of A F: For the Angle at D is a Right Angle. Therefore the Squares of A D, and D F, together, shall be double to the Squares of AC, and CD, together. But DF is equal to D B. Therefore the Squares of A D, and D B, together, will be double to the Squares of A C, and CD, together. Wherefore, 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 Squares of the half Line, and the Squares of the intermediate Part; which was to be demonstrated. PROPOSITION X. THEOREM. to it be dire£tly added another ; the Square made is compounded of ] obe balf, and the added. Line. LET. the Right Line A B be bisected in C, and any strait Line Ď D added directly thereto. I say, the Squares of A D, and D B, together, are double to the Squares of A C, and CD, together, For, For, draw * CE from the Point C at Right Angles to AB, which make equal to AC, or CB; and draw + 311. AE, EB; likewise through E let E F be drawn pa rallel to AD, and through D, DF+ parallel to CE. Then, because the Right Line E F falls upon the 1 29. 1. Parallels EC, FD, the Angles CEF, EFD, are I equal to two Right Angles. Therefore the Angles FEB, E FD, are together less than two Right Angles. But Right Lines making, with a third Line, Angles together less than two Right Angles, being in* Ax. 12. finitely produced, will meet*. Wherefore E B, É D, produced, will meet towards B D. Now let them be produced, and meet each other in the Point G, and let A G be drawn.' And then, because A C is equal to CE, the Angle † 5.1.1. CAEC will be equal to the Angle E A C ti But the a Right one. Therefore A E B is a Right Angle. I 15. 1. And lince EBC is half a Right Angle, DB G will I also be half a Right Angle, fince it is vertical to E B C. · But B D G is a Right Angle allo; for it is * 29.1. * equal to the alternate Angle DCE. Therefore the remaining Angle D G B is half a Right Angle, and so + 6.1. equal to D BG. Wherefore the Side B D is + equal to the Side D G. Again, because EGF 'is half a Right Angle, and the Angle at F is a Right Angle, 34. I. for it is equal * to the opposite Angle at C; the remaining Angle FEG will be also half a Right one, and is equal to the Angle EGF; and so the Side GF + 6.1. is + equal to the Side E F. And since C E is equal to CA, and the Square of E C equal to the Square of CA; therefore the Squares of EC, CA, together, are double to the Square of CA. But the Square of 147.1. EA is I equal to the Squares of EC and CA. Wherefore the Square of E A is double to the Square of A C. Again, because G F is equal to F E, the Square of GF also is equal to the Square of FE. Wherefore the Squares of GF and F E are double to the Square of FE. But the Square of E G is equal to the Squares of G F, F E. Therefore the Square of E G is double to the Square of EF: But E F is equal to CD. Wherefore the Square of E G shall be double |