c 8. 1. Book I. therefore also the fide DC is equal to the side BC. And because the fide DA is equal to AB, and AC common to the two triangles DAC, BAC, and the base DC likewise equal to the base BC, the angle DAC is equal to the angle BAC: But DAC is a right angle; therefore also BAC is a right angle. Therefore, if the square, &c. Q. E. D. 1 ELE Book II. ELEMENTS OF GEOMETRY. جليم E BOOK II. DEFINITIONS. 1. VERY right angled parallelogram, or rectangle, is faid to be contained by any two of the straight lines which are about one of the right angles. "Thus, the right angled parallelogram AC is called the "rectangle contained by AD and DC, or by AD and AB, "&c. For the sake of brevity, instead of the rectangle con"tained by AD and DC, we shall fimply say the rectangle "AD.DC, placing a point between the two fides of the rect"angle. Also, instead of the square of a line, for instance of N. "AD, we may sometimes in what follows write AD2." "The fign + placed between the names of two magnitudes, "fignifies that those magnitudes are to be added together; and "the fign - placed between them, fignifies that the latter is to "be taken away from the former." "The fign fignifies, that the things between which it is " placed are equal to one another." Book II. II. In every parallelogram, any of the parallelograms about a diameter, together with the 6 AGK, or EHC.' H E D F K G C I PROP. I. THEOR. F there be two straight lines, one of which is divided into any number of parts; the rectangle contained by the two straight lines, is equal to the rectangles contained by the undivided line, and the several parts of the divided line. Let A and BC be two straight lines; and let BC be divided into any parts in the points D, E; the rectangle A.BC is equal to the several rectangles A.BD, A.DE, A.EC. From the point B draw a BF at right angles to BC, and make BG equal to A; and through G draw c GH parallel to BC; and through D, E, C, draw c DK, EL, CH parallel to BG; then BH, BK, DL and EH are rectangles, and BH=BK+DL+ G F EH. KLH A But BHBG.BC=A.BC, because BG=A: Also BK BG.BD A.BD, because BG=A; and DL=DK.DE 34. 1. A.DE, because DK=BG=A. In like manner, EH= A.EC. Therefore A.BC=A.BD+A.DE+ A.EC; that is, the rectangle A.BC is equal to the several rectangles A.BD, A.BD, A.DE, A.EC. Therefore, if there be two straight Book II. lines, &c. Q. E. D. I PROP. II. THEOR. F a straight line be divided into any two parts, the rectangles contained by the whole and each of the parts, are together equal to the square of the whole line. Let the straight line AB be divided into any two parts in the point 'C; the rectangle AB.BC, together with the rectangle AB.AC, is equal to the square of AB; or AB.AC+AB.BC=AB2 On AB describe a the square ADEB, and through C draw CF b parallel to AD or BE; then AF+ CEAE. But AF = AD.AC= AB.AC, because AD = AB; CE C B 1 D =BE.BC=AB.BC; and AE=AB2. Therefore AB.AC+AB.BC=AB2. Therefore, if a straight line, &c. Q. E. D. TE a 46. 1. b 31. 1. F PROP. III. THEOR. a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the forefaid part. Let the straight line AB be divided into two parts in the point C; the rectangle AB.BC is equal to the rectangle AC.BC, together with BC2. 52 Book II. a 46. 1. 31.1. Upon BC describe a the A because BE=BC. AB.BC=AC.CB+BC2. Therefore, if a straight line, &c. Q. E. D. a 46. 1. b 31. 1. IF PROP. IV. THEOR. a straight line be divided into any two parts, the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts. Let the straight line AB be divided into any two parts in C; the square of AB is equal to the squares of AC, CB, and to twice the rectangle contained by AC, CB, that is, ABAC2+CB2+2AC.CB. Upon AB describe a the square ADEB, and join BD, and through C draw b CGF parallel to AD or BE, and through G draw HK parallel to AB or DE. And because CF is pa A C 29. 1. d 5.1. e 6. I rallel to AD, and BD falls upon e f 34. 1. to the fide CG: but CB is equal f D F E alfo to GK, and CG to BK; where fore the figure CGKB is equilateral. It is likewise rectangular; for the angle CBK being a right angle, the other gCor.46.1. angles of the parallelogram CGKB are also right angles 8. Wherefore |