PROPOSITION II. WE TH E O R E M. contained under the whole Line, and each of the ET the Right Line AB be any how divided in L the Point C. I say, the Rectangle contained under AB, BC, together with that contained under AB and AC, is equal to the Square made on AB. For let the Square ADEB be described * on A B, * 46. To and thro' C let CF be drawn parallel to AD or BE. Therefore AE is equal to the Rectangles AF and CE. But A E is a Square described upon A B; and AF is the Rectangle contained under B A, AC; for it is contained under D A and AC, whereof AD is equal to AB; and the Rectangle CE is contained under AB, BC, fince BE is equal to AB. Wherefore the Rectangle under AB and AC, together with the Rectangle under AB and BC, is equal to the Square of A B. Therefore, if a Right Line be any bor divided, the Rectangles contained under the whole Line, and each of the Segments, or Parts, are equal to the Square of the whole Line. PROPOSITION III. THE OR EM. contained under the whole Line, and one of its ET the Right Line AB be any how cut in the Point C. I say, the Rectangle under A B and BC is equal to the Rectangle under AC and BC, together with the Square described on BC. For 46. So For describe * the Square CDEB upon BC 1 31. 1. produce ED to F; and let AF be drawn + thro A, parallel to CD or BE. Then the Rectangle A E shall be equal to the two Rectangles AD, CE: And the Rectangle A E is that contained under AB and BC; for it is contained under A B and BE, whereof B E 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 described upon B C. Wherefore the Rectangle under A B and BC is equal to the Rectangle under AC and CB, together with the Square described upon BC. Therefore if a Right Line be any how cut, the Rectanö gle 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 demonstrated. PROPOSITION IV. THE OR EM. is made on the wbole Line will be equal to the * 46. 1. # 31. 1. 1.29. I. Squares of AC, CB, together, with twice the Rectangle contained under AC, CB. For* describe the Square ADEB upon AB, join BD, and thro' C draw | CGF parallel to AD or BE; and also thro' G draw HK parallel to AB or DE. Then because CF is parallel to AD, and BD falls upon them, the outward Angle BGC shall be † equal to the inward and opposite Angle ADB, but the Angle ADB is * equal to the Angle ABD, since the Side B A is equal to the Side AD. Wherefore the Angle CGB is equal to the Angle GBC; and fo the Side BC equal to the Side CG; but likewise the Side CB is I equal to the Side GK, and the Side CG to BK. Therefore GK is equal to KB, and CGKB is 5. I. † 6. 1. 34. li is equilateral. I say, it is also Right-angled; for becaufe CG is parallel to BK, and CB falls on them, the Angles KBC, GCB, are I equal to two Right Angles. But KBC is a Right Angle. Wherefore GBC also is a Right Angle, and the opposite Angles GCB, CGK, GKB, shall be Right Angles. Therefore CGKB is a Rectangle . But it has been proved to be equilateral. Therefore CGKB is a Square defcribed upon BC. BC. For the same Reason HF is allo a Square made upon HG, that is equal to the Square of AC. Wherefore HF and CK are the Squares of AC and CB. And because the Rectangle AG is equal to the Rectangle GE, and AG is that which * 43 1 is contained under AC and CB, for GC is equal to CB: GE shall be equal to the Rectangle under AC, and CB. Wherefore the Rectangles AG, GE, are equal to twice the Rectangle contained under AC, CB; and HF, CK, are the Squares of AC, CB. Therefore the four Figures HF, CK, AG, GE, are equal to the Squares of AC and CB, with twice the Rectangle contained under AC and CB. But HF, CK, AG, GE, make up the whole Square of AB, viz. ADEB. Therefore the Square of AB is equal to the Squares of AC, CB, together with twice the Rectangle contained under AČ, 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 Segments; which was to be demonstrated, Coroll. Hence it is manifeft, that the Parallelograms which stand about the Diameter of a Square, ask PRO PROPOSITION V. 31. 1. THEOREM into two unequal ones ; the Rectangle under the the Square made of half the Line. L Parts in C, and into two unequal Parts in D." I fay the Rectangle contained under AD, DB, together with the Square of CD, is equal to the Square of BC. 46. I. For + describe CEFB, the Square of BC, draw BE, and thro’ D draw* DHG, parallel to CE, or EF, and AK thro’ A, parallel to CL, or BO. | 43. 1. Now the Complement CH is f equal to the Com 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 AL, because AC is equal to CB; therefore AL is equal to DF, and adding CH, which is common, the whole AH shall be equal to FD, DL, together. But AH is the Rect• Cor. 4. angle contained under AD, DB; for DH is * equal of this. to DB, and FD, DL, is the Gnomon MNX; therefore MNX is equal to the Rectangle contained under AD, DB, and if LG, being common, and equal to the Square of CD be added ; then the Gnomon MNX, and LG, are equal to the Rectangle contained under AD, DB, together with the Square of CD; but the Gnomon MNX, and LG, make up the whole Square CEF B, viz. the Square of CB. Therefore the Rectangle under AD, DB, together with the Square of CD, is equal to the Square of CB. 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 demonstrated, PRO * PROPOSITION VI. THE ORE M. another Right Line be added direztly to the fame, ET the Right Line AB be bisected in the Point C, and BD added directly thereto. I say the Rectangle under AD, and DB, together with the Square of BC, is equal to the Square of CD. For describe * CÈFD, the Square of CD, and 46: 1. join DE; draw + BHG thro' B, parallel to CE, † 31. 1. er DF, and KLM thro' H, parallel to AD, or EF, as also AK thro' A, parallel to CL; or DM. Then because AC is equal to CB, the Rectangle AL shall be * equal to the Rectangle CH, but CH* 36. I. is I equal to HF. Therefore AL will be equal to 1 43. 1. HF; and adding CM, which is common to both, then the whole Rectangle AM, is equal to the Gnomon NXO. But A M is that Rectangle which is contained under AD, DB, for D M is * equal to * Cor: 4. DB ; therefore the Gnomon NXO is equal to of this. the Rectangle under AD, and DB. And adding LG, which is common, viz. + the Square of CB; + Cor. 4and then the Rectangle under AD, DB, together of this with the Square of B C, is equal to the Gnomon NXO with LG.' But the Gnomon NXO, and LG, together, make up the Figure ĆEFD, that is the Square of CD. Therefore the Rectangle under AD, and DB, 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 dire&tly to the same, the Rectangle contained under the Line, compounded of the whole and added Line, (taken as one Line,) and the E added |