c. 30. Des. С GB, HC; and thro' A draw 6 AL parallel to BD or CE, and join Book I. AD, FC. /then because each of the angles BAC, BAG is a right angle", the two straight lines b. 31.1. AC, AG upon the opposite G fides of AB, make with it at H the point A the adjacent angles ecual to two right angles; therefore CA is in the same K straight lined with AG. for d. 14. I. the fame reason, AB and AH ale in the same straight line. and because the angle DBC is equal to the angle FBA, each of them being a right angle, add to each the angle ABC, E and the whole angle DBA is equal o to the whole FBC./and because the two sides AB, BD are c. 2. Ax. equal to the two FB, BC, each to each, and the angle DBA equal to the angle FBC; therefore the base AD is equal to the base FC, f. 4. I. and the triangle ABD to the triangle FBC. now the parallelogram BL is double 8 of the triangle ABD, because they are upon the fame g. 41.1. base BD, and between the fame parallels BD, AL; and the square GB is double of the triangle FBC, because thicse also are upon the fame base FB, and between the fame parallels TB, GC. but the doubles of cquals are equal h to one another. therefore the paralle- h. 6. Ax. logram BL is equal to the square GB. and in the same manner, by joining AE, BK, it is demonstrated that the parallelogram CL is equal to the square HC. Therefore the whole square BDEC is equal to the two squares GB, HC. and the square BDEC is described upon the straight line BC, and the squares GB, HC upon BA, AC. wherefore the square upon the side BC is equal to the squares upon the sides BA, AC. Therefore in any right angled triangle, &c. R. E. D. PROP. XLVIII. THEOR. angle, be equal to the squares described upon the other two sides of it; the angle contained by these two fides is a right angle. 1. II.1. Book 1. If the square described upon BC one of he sides of the triangle mABC be equal to the squares upon the other sides BA, AC; the angle BAC is a right angle. From the point A draw * AD at right angles to AC, and make AD equal to BA, and join DC. then because DA is equal to AB, the square of D A is equal to the square D equal to the squares of BA, AC. but the 5. 47. 1. square of DC is equalb to the squares of DA, А B С fore also the side DC is equal to the side BC. and because the side DA is equal to AB, ani AC common to the two triangles DAC, BAC, the two DA, AC are equal to the two BA, AC; and the base DC is equal to the base BC; therefore the angle DAC is e6, 8.1. qual 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. Τ Η Σ DEFINITION S. I. II. together with the two com- E $ D mon. “Thus the parallelo gram HG together with F K K IF 1 PROP, I. THEOR. 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 Let 2. JI. I. Book II. Let A and BC be two straight lines; and let BC be divided into any parts in the points D, E; the rectangle contained by the straight lines A, BC is equal to the rect B D E C angle contained by A, BD; and to that contained by A, DE; and also to that contained by A, EC. From the point B draw * BF at right angles to BC, and make BG G b. 3. 1. equal 6 to A; and thro’G draw c K L H c. 31. 1. GH parallel to BC; and thro' D, F A for it is contained by GB, BD, of which GB is equal to A; and d. 31.1. DL is contained by A, DE, because DK, that is a BG, is equal to A; and in like manner the rectangle E H is contained by A, EC. therefore the rectangle contained by A, BC is equal to the several rectangles contained by A, BD, and by A, DE, and also by A, EC. Wherefore if there be two straight lines, &c. Q. E. D. 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. IT Q. 46. I, b. 31. 3. Let the straight line AB be divided into A CB Upon AB describe a the square ADEB, * N. B. To avoid repeating the word Contained too frequently, the rectangle contained by two straight lines AB, AC is sometimes simply called the rectangle AB, AC. rectangle F E EUCLID. rectangle contained by BA, AC; for it is contained by DA, AC, of Book II. which AD is equal to AB; and CE is contained by AB, BC, for BEM is equal to AB. therefore the rectangle contained by AB, AC together with the rectangle AB, BC, is equal to the square of AB. If therefore a straight line, &c. Q. E. D. IF F 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 foresaid part. . a. 46. s. b. 31. I. Let the straight line AB be divided into any two parts in the point C; the rectangle AB, BC is equal to the rectangle AC, CB together with the square of BC. Upon BC describe " the square В PRO P. IV. THE OR. IF 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, Upon |