Book I. PROP. XLVI. PROB. To describe a square upon a given straight line. Let AB be the given straight line; it is required to describe a square upon AB. a II. 1. b 3. 1. с 31. 1. d 34. 1. e 29. I. a 46. 1. From the point A drawa AC at right angles to AB; and makeb AD equal to AB, and through the point D draw DE parallel to AB, and through B draw BE parallel to AD; therefore ADEB is a parallelogram; whence AB is equald to DE, and E AD to BE: But BA is equal to AD; C B Cor. Hence every parallelogram that has one right angle has all its angles right angles. N any right angled triangle, the square which is described upon the side subtending the right angle, is. equal to the squares described upon the fides which con. tain the right angle. Let ABC be a right angled triangle having the right angle BAC; the square described upon the fide BC is equal to the squares described upon BA, AC. On BC defcribe the square BDEC, and on BA, AC the squares GB, HC; and thro' A drawb AL parallel to BD or CE, and Book. I. join AD, FC; then, because each of the angles BAC, BAG is a right angle, the two straight lines AC, AG upon the oppofite fides of AB, make with it at the point A the adjacent angles equal to two right angles; therefore CA is in the same straight lined with AG; for the fame reason, AB and AH are in the fame 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, G and the whole angle DBA is D L E equal to the whole FBC; and because the two fides AB, BD e 2. Ax. are equal to the two FB, BC, each to each, and the angle DBA equal to the angle FBC; therefore the base AD is e qual f to the base FC, and the triangle ABD to the triangle f 4.1. FBC: Now the parallelogram BL is double & of the triangle g 41. 1. ABD, because they are upon the fame bafe BD, and between the fame parallels, BD, AL; and the square GB is double of the triangle FBC, because these also are upon the fame bafe FB. and between the fame parallels FB, GC: But the doubles of equals are equal to one another: Therefore the parallelo- h 6. Ax. gram BL is equal to the square GB: And in the fame 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 defcribed upon the straight line BC, and the squares GB, HC upon BA, AC: Wherefore the square upon the fide BC is equal to the squares upon the fides BA, AC. Therefore in any right angled triangle, &c. Q. E. D. PROP. XLVIII. THEOR. F the square described upon one of the fides of a triLangle, be equal to the squares described upon the other two fides of it; the angle contained by these two fides is a right angle. Book I. 4 If the square described upon BC, one of the fides of the triangle ABC, be equal to the squares upon the other fides BA, AC; the angle BAC is a right angle. a Ir. I. b 47. 1. € 8. 1. Fron. 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 DA is equal to the square of AB: To each of these add the square of AC; therefore the squares of DA, AC, are equal to the squares of BA, AC; But the square of DC is equal A b to the squares of DA, AC, because DAC is a right angle; and the square of BC, by hypothesis, is equal to the squares of BA, AC; therefore the square of DC is equal to the square of BC; and therefore alfo B 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, 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 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. D C THE E DEFINITIONS. I. VERY right angled parallelogram is faid to be contained by any two of the straight lines which contain one of the right angles. II. In every parallelogram, any of the parallelograms about a dia meter, together with the two complements, is called A a Gnomon. Thus the pa 'rallelogram HG, toge 'ther with the comple *ments AF, FC, is thegno F 'mon, which is more brief- H K PROP. I. THEOR. IF 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. a 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 ΙΙ. Ι. b 3. 1, C 31. 1. d 34. 1. a 46. 1. b 31.1. the straight lines A, BC is equal DEC KLH A From the point B draw BF at right angles to BC, and makeG BG equal to A; and through G draw GH parallel to BC; and through D, E, C draw DKF 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 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 alfo by A, EC. Wherefore, if there be two straight lines, &c. Q. E. D. PROP. II. THEOR. IF a fstraight 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. * N. B. To avoid repeating the word contained too frequently, the rectangle contained by two straight lincs AB, AC is fometimes simply called the rectangle AB, AC. |