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 said, 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. ET AB be the Right Line given, upon which it is required to describe a Square. 11 of this. Draw* AC at Right Angles to AB from the Point + 3 of this. A given therein; maket AD equal to AB, and thro' 1 31 of this. the Point D draw IDE parallel to AB; also thro' B draw B E parallel to AD. 34 of tbis. Then A DEB 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 so the Parallelogram ADEB is equilateral : I say it is likewise 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 opposite Sides and oppo34 3f this. Site Angles of Parallelograms are equal. Therefore each of the opposite Angles A BE, BED, are Right Angles; and consequently ADBE is a Rectangle: But it has been proved to be equilateral. Therefore it is necessarily a Square, and is described upon the Right Line A B; which was to be done. Coroll. Hence every Parallelogram that has one Right Angle is a Rectangle. PRO PROPOSITION XLVII. THEOREM. upon the Side subtending the Right Angle, is e- described upon the sides containing the Right Angle. LET ET ABC be a Right-angled Triangle, having the Right Angle BAC. I say the Square desçribed upon the Right Line B C, is equal to both the Squares described upon the sides BA, AC. For describe * upon BC the Square BDEC, and 46 of this on BA, AC, the Squares GB, HC, and thro' the Point A draw AL parallel to BD, or CE; and let AD, FC, be joined. Then because the Angles BAC, BĄG, are Right f Def. 30. ones, two Right Lines AG, AC, at the given Point A, in the Right Line BA, being on contrary Sides thereof, make the adjacent Angles equal to two Right Angles. Therefore CA, AĞ, make one straight ( 14 of this. Line; by the fame Reason AB, AH, make one straight Line. And since the Angle DBC is equal to the Angle F BA, for each of them is a Right one, add ABC, which is common, and the whole Angle DBA is * equal to the whole Angle F BC. And * Ax. 2. since 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 t t 4 of this. equal to the Base F C, and the Triangle ABD equal to the Triangle FBC: But the Parallelogram BL is I double to the Triangle ABD; for they have the 41 of this fame Base D B, and are between the same Parallels BD, AL. The Square GB is also double to the Triangle FBC; for they have the fame Base FB, and are in the same Parallels FB, 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 described on the Right Line BC, 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 described upon the Side fubtending the Right Ångle, is equal to both the Squares described upon the Sides containing the Right Angle. PROPOSITION XLVIII. . I THEOREM. equal to the Squares described upon the orber two Triangle ABC, be equal to the Squares described upon the other two sides of the Triangle BA, AC: I say 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 A B. And adding the common Square described on AC, the Squares described on DA, AC, are equal to the Squares described on BA, AC. But the Square # 47 of this described on DC is * equal to the Squares described on D A, 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 DA is equal to A B, and AC is common, the two Sides DA, AC. are equal to the two Sides BA, Ą C; and the Base DC is equal to the Base CB. Therefore the Angle 18 of this. DAC is I equal to the Angle BAC; but DAC is a Right Angle; and fo BAC will be a Right Angle also. If, therefore, a Square described upon one side of a Triangle be equal to the Squares described upon the other two Šides of the faid Triangle, then the Angle contained by these two other Sides is a Right Angle; which was to be demonstrated. EUCLID's 45 E U C L I D's ELEMENTS BOOK II. DEFINITIONS. E 1. VERY Right-angled Parallelogram is said to be contained under two Right Lines, comprehending a Right Angle. rallelograms that are about the Diameter, toge, mon. X PRO PROPOSITION I. THEOREM. divided into any Number of Parts; the Rett- L 11. I. + 3. 1. 31. 1. ET A and BC be two Right Lines, whereof BC is cut or divided any how in the Points D, E. I say, the Rectangle con tained under the Right Lines A and B C, 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 + B G equal to A; and let 1 GH be drawn thro' G parallel to BC: Likewise, let I there be drawn DK, EL, CH, thro' D, E, C, parallel to BG. Then the Rectangle BH is equal to the Reétangles BK, DL, EH; but the Rectangle BH, 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, 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 B C, 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 fall be equal to all the Rectangles contained under the whole Line, and the several Segments of the divided Line ; which was to be demonstrated. PRO |