Right Line B C, and the Squares GB, HC, on BA, AC. Therefore the Square BE, described on the Side B. C, is equal to the Squares described on the Sides BA, AC. Wherefore, in any Right-angled Triangle, the Square defcribed upon the Side, fubtending the Right Angle, is equal to both the Squares described upon the Sides, containing the Right Angle; which was to be demonftrated. PROPOSITION XLVIII. THEOREM. If a Square described upon one side of a Triangle, IF the Square, defcribed upon the Side BC of the For, let there be drawn A D from the Point A, at Then, because D A is equal to A B, the Square described on DA will be equal to the Square described on AB. And adding the common Square described on AC, the Squares described on DA, AC, are equal to the Squares described on B A, A C. But the Square described on DC is * equal to the Squares described * 47 of this. on DA, AC; for DAC is a Right Angle: But the Square on BC is put equal to the Squares on BA, AC. Therefore the Square described on DC is equal to the Square described on BC; and so the Side CD is equal to the Side CB. And because D A is equal to A B, and A C is common, the two Sides DA, AC, are equal to the two Sides BA, AC; and the Base DC is equal to the Base CB. Therefore the Angle DAC is I equal to the Angle B AC; but DAC ist & of this. a Right Angle; and so BAC will be a Right Angle alfo. If, therefore, a Square described upon one Side of a Triangle, be equal to the Squares, described upon the other two Sides of the faid Triangle, then the Angle, contained by these two other Sides, is a Right Angle; which was to be demonstrated. : EUCLID's EUCLI D's ELEMENTS. BOOK II. DEFINITIONS. I. EVERY Right-angled Parallelogram is faid to be contained under two Right Lines, comprehending a Right Angle. II. In every Parallelogram, either of those Pa rallelograms, that are about the Diameter, together with the Complements, is called a Gnos mon.. PRO PROPOSITION I. THEOREM. If there be two Right Lines, and one of them be divided into any Number of Parts; the Rectangle comprehended under the whole Line and the divided Line, shall be equal to all the Rectangles contained under the whole Line, and the Several Segments of the divided Line. L ET A and BC be two Right Lines, whereof D, E. I say, the Rectangle contained under the Right Lines A and BC, is equal to the Rectangles contained under A and B D, A and DE, and A and E C. For, let * B F be drawn from the Point B, at Right* 11. 1, Angles, to BC; and make + BG equal to A; and † 3. 1. let ‡ GH be drawn thro' G parallel to BC: Like-131. wife let I 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 BK is that contained under A and BD; for it is contained under GB and BD; and GB is equal to A; and the Rectangle DL is that contained under A and DE, because DK, that is, BG, is equal to A: So likewise the Rectangle EH is that contained under A and EC. Therefore the Rectangle under A and BC is equal to the Rectangles under A and BD, A and DE, and A and EC. 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, shall be equal to all the Rectangles contained under the whole Line, and the several Segments of the divided Line; which was to be demonftrated. 46. 1. PROPOSITION II. THEOREM. If a Right Line be any bow divided, the Rectangles contained under the whole Line, and each of the Segments, or Parts, are equal to the Square of the whole Line. LET the Right Line A B be any how divided in the Point C. I fay, the Rectangle contained under AB and BC, together with that contained under A B, and A C, is equal to the Square made on A B. For let the Square ADEB be described * on AB, 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 AB; and AF is the Rectangle contained under BA and AC; for it is contained under DA and AC, whereof A D is equal to AB; and the Rectangle CE is contained under A B and BC, since 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 how cut, the Rectangle contained under the whole Line, and one of its Parts, is equal to the Rectangle contained un der the two Parts, together with the Square of the firft-mentioned Part. LET the Right Line A B 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 : For describe * the Square CDEB upon BC;* 46. 1. produce ED to F; and let AF be drawn + thro' A,† 31. 1. parallel to CD or BE. Then the Rectangle A E shall 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 BE, 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 A B and BC is equal to the Rectangle under A C and CB, together with the Square described upon BC. 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. 1 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. ET the L For * describe the Square A DE B 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 AB or DE. 5.1. Then, because CF is parallel to A D, and BD falls upon them, the outward Angle BGC shall be ‡ equal ‡ 29. 1. to the inward and opposite Angle ADB; but the Angle AD Bis * equal to the Angle A BD, 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 † 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 KB, and 4 E CGKB |