11 of this. the Triangle DBC to the Parallelogram GM; then the whole Right-lined Figure ABCD will be equal to the whole Parallelogram KFLM. Therefore the Parallelogram KFLM is made equal to the given Right-lined Figure ABCD, in an Angle FKM, equal to the given Angle E; which was to be done. Coroll. It is manifeft, from what has been faid, how to apply a Parallelogram to a given Right Line, equal to a given Right-lined Figure in a given Right-lined Angle. PROPOSITION XLVI. PROBLEM. To defcribe a Square upon a given Right Line. ET AB be the Right Line given, upon which Lit is required to defcribe a Square. Draw* AC at Right Angles to AB from the Point + 3 of this. A given therein; make † AD equal to AB, and thro' 31 of this. the Point D draw‡DE parallel to AB; also thro' B draw BE parallel to AD. * 34 of this. Then ADEB is a Parallelogram; and fo AB* is equal to DE, and AD to BE. But BA is equal to AD. Therefore the four Sides BA, AD, DE, EB, are equal to each other. And fo the Parallelogram ADEB is equilateral: I fay it is likewife equiangular, For because the Right Line AD falls upon the Parallels AB, DE, the An† 29 of this gles BAD, ADE, are + equal to two Right Angles. But BAD is a Right Angle: Wherefore ADE is also a Right Angle; but the oppofite Sides and oppo34 of this. fite Angles of Parallelograms are equal. Therefore each of the oppofite Angles ABE, BED, are Right Angles; and confequently ADBE is a Rectangle: But it has been proved to be equilateral. Therefore it is neceffarily a Square, and is described upon the Right Line AB; which was to be done. Coroll. Hence every Parallelogram that has one Right PRO PROPOSITION XLVII. THEOREM. In any Right-angled Triangle, the Square defcribed upon the Side fubtending the Right Angle, is equal to both the Squares defcribed upon the Sides containing the Right Angle. ET ABC be a Right-angled Triangle, having the Right Angle BAC. I fay the Square defcribed upon the Right Line BC, is equal to both the Squares defcribed upon the Sides BA, AC. For defcribe* upon BC the Square BDEC, and * 46 of this on BA, AC, the Squares GB, HC, and thro' the Point A draw A L parallel to BD, or CE; and let AD, F C, be joined. Then because the Angles BAC, BAG, †are Right † Def. 30. ones, two Right Lines AG, AC, at the given Point A, in the Right Line B A, being on contrary Sides thereof, make the adjacent Angles equal to two Right Angles. Therefore CA, AG, make ‡ one ftraight 14 of this. Line; by the fame Reafon AB, AH, make one straight Line. And fince the Angle DBC is equal to the Angle FBA, for each of them is a Right one, add ABC, which is common, and the whole Angle DBA is equal to the whole Angle FBC. fince the two Sides AB, BD, are equal to the two Sides FB, BC, each to each, and the Angle DBA equal to the Angle FBC; the Bafe AD will be†† 4 of this. equal to the Bafe F C, and the Triangle ABD equal to the Triangle FBC: But the Parallelogram BL * And Ax. 2. is double to the Triangle ABD; for they have the ‡ 41 of thi fame Base D B, and are between the fame Parallels BD, AL. The Square GB is alfo double to the Triangle FBC; for they have the fame Base F B, and are in the fame Parallels F B, GC. But Things that are the Doubles of equal Things are equal to* Ax. 6. each other. Therefore the Parallelogram BL is equal to the Square GB. After the fame Manner, AE, BK, being joined, we prove, that the Parallelogram CL is equal to the Square HC. Therefore the whole Square DBEC is equal to the two Squares GB, HC, But the Square DBEC is defcribed on the Right Line 2 BC BC, and the Squares G B, HC, on BA, AC. Therefore the Square BE, defcribed 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 defcribed upon the Sides containing the Right Angle. PROPOSITION XLVIII. THEOREM. If a Square defcribed upon one Side of a Triangle be equal to the Squares defcribed upon the other two Sides of the faid Triangle, then the Angle contain ed by these two other Sides is a Right Angle. F the Square described upon the Side BC of the Triangle ABC, be equal to the Squares described upon the other two Sides of the Triangle BA, AC: I fay the Angle BAC is a Right one. For let there be drawn AD from the Point A, at Right Angles to AC; likewise make AD equal to BA and join DC. * Then because DA is equal to AB, the Square defcribed on D A will be equal to the Square described on AB. And adding the common Square described on AC, the Squares defcribed on DA, AC, are equal to the Squares described on BA, AC. But the Square 47 of this. defcribed on DC is equal to the Squares described 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 defcribed on DC is equal to the Square defcribed on BC; and fo the Side CD is equal to the Side CB. And because DA is equal to A B, and AC is common, the two Sides DA, AC. are equal to the two Sides BA, AC; and the Base DC is equal to the Bafe CB. Therefore the Angle of this. DAC is equal to the Angle BAC; but DAC is a Right Angle; and fo BAC will be a Right Angle alfo. If, therefore, a Square defcribed upon one Side of a Triangle be equal to the Squares defcribed 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 demonftrated. EUCLID'S EUCLID's ELEMENTS. BOOK II. I E DEFINITIONS. VERY Right-angled Parallelogram is II. In every Parallelogram, either of thofe Pa- mon. PRQ 45 * II. I. † 3. 1. 31. I. PROPOSITION I. THEOREM. If there be two Right Lines, and one of them be divided into any Number of Parts; the Rettangle comprehended under the whole, and divided Line, fhall be equal to all the Rectangles contained under the whole Line, and the feveral Segments of the divided Line. L' ET A and BC be two Right Lines, whereof BC is cut or divided any how in the Points D, E. I fay, the Rectangle contained under the Right Lines A and BC, is equal to the Rectangles contained under A and BD, A and DE, and A and E C. For let BF be drawn from the Point B, at Right Angles, to BC; and make + BG equal to A; and let + G H be drawn thro' G parallel to BC: Likewife, let there be drawn DK, EL, CH, thro' D, E, C, parallel to B G. Then the Rectangle BH is equal to the Rectangles BK, DL, EH; but the Rectangle B H, is that contained under A and BC; for it is contained under GB, BC; and GB is equal to A; and the Rectangle B K is that contained under A and BD; for it is contained under GB and BD, and G B is equal to A; and the Rectangle DL is that contained under. A and DE, becaufe DK, that is, BG, is equal to A: So likewise the Rectangle EH is that contained under A and E C. Therefore the Rectangle under A and BC, is equal to the Rectangle 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 and divided Line fhall be equal to all the Rectangles contained under the whole Line, and the feveral Segments of the divided Line; which was to be demonftrated. PRO |