b. 31. I. 1 GB, HC; and thro' A draw AL paralkl to BD or CE, and join Book I. AD, FC. then because each of the angles BAC, BAG is a right m angle, the two straight lines G AC, AG upon the opposite C. 39. Def. sides of AB, make with it at H the point A the adjacent angles equal to two right angles ; therefore CA is in the same K straight line with AG. for d. 14. 1. the iame reason, AB and AH B D 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. 1. and the triangle ABD to the triangle FBC. now the parallelogram BL is double 8 of the triangle ABD, because they are upon the 8. 41. 1. fame base BD, and between the fame parallels BD, AL; and the square GB is double of the triangle FBC, because these also are upon the same base FB, and between the fame parallels FB, GC. but the doubles of equals are equal h to one another. therefore the h. 6. As. parallelogram BL is equal to 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. Q. E. D. PROP. XLVIII. THEOR. IF triangle, be equal to the squares described upon the other two sides of it ; the angle contained by these two sides is a right angle. a. II. I. Book I. If the square described upon BC one of the sides of the trianmgle ABC be cqual 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 D qual to the squares of BA, AC. but the DA, AC, because DAC is a right angle; AC; and the base DC is equal to the base BC; therefore the anC. &. I. gle DAC is equal o 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. E 1. by any two of the straight lines which contain one of the « Thus the pa In every parallelogram, any of the parallelograms about a diame ter, together with the two complements, is called a A. E Gnomon. rallelogram HG together with the complements AF, FC is the gnomon, which F H Η K is inore briefly expressed ' by the letters AGK, or B EHC which are at the op G C posite angles of the parallelograms which make the gnomon.' PROP. I. THEOR. F there be two straight lines, one of which is divid ed 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 seyeral parts of the divided line. 2. II. 1. 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 B В D E C С the rectangle 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 G at right angles to BC, and make K L H b. 3. 1. BG equal 6 to A; and thro' GFI A 6. 31. 1. draw · GH parallel to BC; and thro’ D, E, C draw • DK, EL, CH parallel to BG. then the rectangle BH is equal to the rectangles BK, DL, EH ; and BH is contained by A, BC, for it is contained by GB, BC, and GB is equal to A ; and BK is contained by A, BD, for it is contained by GB, BD, of which GB is equal to A; and DL is contained d. 34. 1. by A, DE, because DK, that is d BG, is equal to A ; and in like manner the rectangle EH 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. IF 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 A C B Upon AB describe a the square ADEB, •N. B. To avoid repeating the word Contained too frequently, the rectangle contained by two Araight lines AB, AC is fometimes simply called the rectangle AB, AC. 1. 46. 1. 1.31. 1. is the rectangle contained by BA, AC; for it is contained by Book II. DA, AC, of which AD is equal to AB; and CE is contained by AB, BC, for BE 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. PRO P. II. THEOR. IF 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. 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 a the square A C B a. 46. si CDEB, and produce ED to F, and thro’ A draw 6 AF parallel to CD b. 31. lo or BE. then the rectangle AE is equal to the rectangles AD, CE; and AE is the rectangle contained by AB, BC, for it is contained by AB, BE, of which BE is equal to BC; and AD is contained by F D E AC, CB, for CD is equal to CB; and DB is the square of BC. therefore the rectangle AB, BC is equal to the rectangle AC, CB together with the square of BC. If therefore a straight line, &ce Q. E. D. PROP. IV. THEOR. 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. |