added Line, together with the Square of half the 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 demonstrated. PROPOSITION VII. THEOREM. the whole Line, together with the Square of one other Segment. L ET the Right Line A B be any how cut in the Point C. I say the Squares of AB, BC, together, are equal to double the Rectangle contained under AB, BC, together with the Square, made of AC. For let the Square of AB be * described, viz. ADE B, and construct s the Figure. + 43. 1. Then because the Rectangle AG, is f equal to the Rectangle GE. IF CF, which is common, be added to both, the whole Rectangle AF shall be equal to the whole Rectangle CE, and so the Rectangles AF, CE, are double to the Rectangle AF; but AF, CE, make up the Gnomon KLM, and the Square CF. Therefore the Gnomon KLM, together with the Square CF, shall be double to the Rectangle A F. But double the Rectangle under AB, 1 Cor. 4. BC, is double the Rectangle AF, for BF is I equal to BC. Therefore the Gnomon KLM, and the Square CF, are equal to twice the Rectangle contained under AB, BC. And if HF, which is common, being the Square of AC, be added to both; then the Gnomon KLM, and the Squares CF, HF, are equal to double the Rectangle contained § A Figure is said to be constructed, when Lines, drawn in a Paral. lelogram parallel to the Sides thereof, cut tbe Diameter in one Point, and make two Parallelograms about the Diameter, and two Complements. So likewise a double Figure is said to be constructed, when two Right Lines parallel to the Sides, make four Parallelograms about tbe Diameter, and Four Complements under under AB, BC, together with the Square of A C. But the Gnomon KLM, together with the Squares CF, HF, are equal to ADE B, and CF, viz. the Squares of AB, BC. Therefore the Squares of AB, BC, are together equal to double the Rectangle con tained under AB, BC, together with the Square of AC. Therefore, if à 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 demonstrated. PROPOSITION VIII. THEOREM. four times the Rectangle, contained under the I say four times the Rectangle contained under AB, BC, together with the Square of AC, is equal to the Square of AB, and BC taken as one Line. For let the Right Line A B be produced to D, fo that BD be equal to BC, describe the Square AE FD, on AD, and construct the double Figure. Now fince CB is * equal to BD, and also to * Hypo +GK, and BD is equal to KN: GK shall be + 34. I. likewife equal to KN; by the fame Reasoning, PR is equal to RO. And fince CB is equal to BD, and GK to KN, the Rectangle CK will I be I 36. 1. equal to the Rectangle BN, and the Rectangle GR to the Rectangle RN. But CK is * equal to RN;* 42. To for they are the Complements of the Parallelogram CO. Therefore BN is equal to GR, and the four Squares B N, KC, GR, RN, are equal to each other; and so they are together Quadruple CK. Again, because CB is equal to BD, and BD to BK, that is, equal to CG; and the faid CB is equal also E 2 also to GK, that is, to GP; therefore CG shall and the four Squares, mak- the Gnomon STY. And if XH, being equal to + Cor. 4. † the Square of AC, which is common, be added to both: Then four times the Rectangle contained of ekis. PROPOSITION IX. THEORE M. two unequal Parts; then the Squares of the un- ET any Right Line A B be cut unequally in D, and equally in C. I say the Squares of AD, DB, together, are double to the Squares of AC and CD together. For let * CE be drawn from the Point C at Right * 11. 1. Angles to AB, which make equal to AC, or CB, and join EA, EB. Also thro' D let + DF be drawn t 31. 1. parallel to CE, and F G thro' F parallel to AB, and draw AF. Now because AC is equal to CE, the Angle EAC will be equal to the Angle AEC; and since the # 5. I. Angle at C is a Right one, the other Angles AEC, EA C, together, shall * make one Right Angle, and * 3. Car. are equal to each other : And so AEC, EAC, are each 32. I. equal to half a Right Angle. For the same Reason are also CEB, EBC, each of them half Right Angles. Therefore the whole Angle AEB is a Right Angle. And since the Angle GEF is half a Right cne, and EGF is a Right Angle; for it is t equal to the † 29. I. inward and opposite Angle ECB, the other Angle EFG will be also equal to half a Right one. Therefore the Angle GEF is equal to the Angle EF G. And so the Side EG is I equal to the Side GF. Again, I 6. I. 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 so the Side DF is equal to the Side DB. And because AC is equal to CE, the Square of AC will be equal to the Square of CE. Therefore the Squares of AC, CE, together, are double to the Square of AC; but the Square of EA is 4 equal to the Squares of AC, CE, together, since 1 47. Is E 3 ACE + 47. I, ACE is a Right Angle. Therefore the Square of Therefore the Square of EF is double the Square of ; THEOREM. it be direEfly added another; the Square made on e any straight Line BD added directly thereto. For |