Book I. a II. I. b 3. I. C 31. I. d 34. I. e 29. I. a 46. I. PROP. XLVI. P R O B. To defcribe a fquare upon a given straight line. Let AB be the given straight line; it is required to describe a fquare upon AB. a E From the point A draw AC at right angles to AB; and make 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 equal to DE, and AD to BE: But BA is equal to AD; therefore the four ftraight lines BA, AD, DE, EB are equal to one another, and the parallelogram ADEBD is equilateral, likewife all its angles are right angles; because the ftraight line AD meeting the parallels AB, DE, the angles BAD, ADE, are equal to two right angles; but BAD is a right angle; therefore alfo ADE is a right angle; but the oppofite A angles of parallelograms are equal d; B therefore each of the oppofite angles ABE, BED is a right angle; wherefore the figure ADEB is rectangular, and it has been demonftrated that it is equilateral; it is therefore a square, and it is described upon the given straight line AB: Which was to be done. COR. Hence every parallelogram that has one right angle has all its angles right angles. N any right angled triangle, the fquare which is defcribed upon the fide fubtending the right angle, is. equal to the fquares defcribed upon the fides which contain the right angle. Let ABC be a right angled triangle having the right angle BAC; the fquare defcribed upon the fide BC is equal to the fquares defcribed upon BA, AC On BC defcribe the fquare BDEC, and on BA, AC the fquares GB, GB, HC; and thro' A draw b AL parallel to BD or CE, and join AD, FC; then, because each of the angles BAC, BAG is a right angle, the two ftraight 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 fame ftraight lined with AG; for the fame reafon, AB and AH are in the fame ftraight 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, and the whole angle DBA is Book. I. b 31. 1. G 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 bafe AD is equal to the bafe FC, and the triangle ABD to the triangle f 4.1. FBC: Now the parallelogram BL is double of the triangle g 41. I, ABD, because they are upon the fame bafe BD, and between the fame parallels, BD, AL; and the fquare GB is double of the triangle FBC, because these alfo are upon the fame base 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 fquare GB; And in the fame manner, by joining AE, BK, it is demonftrated that the parallelogram CL is equal to the fquare HC: Therefore the whole fquare BDEC is equal to the two fquares GB, HC; and the fquare BDEC is defcribed upon the ftraight line BC, and the fquares GB, HC upon BA, AC: Wherefore the fquare upon the fide BC is equal to the fquares upon the fides BA, AC. Therefore in any right angled triangle, &c. Q. E. D. PROP. XLVIII. THE OR. [F the fquare defcribed upon one of the fides of a triangle, be equal to the fquares defcribed upon the other two fides of it; the angle contained by these two fides is a right angle. Book I. a II. I. b 47.1. € 8. 1. If the fquare described upon BC, one of the fides of the triangle ABC, be equal to the fquares upon the other fides BA, AC; the angle BAC is a right angle. a D 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 fquare of DA is equal to the fquare of AB: To each of these add the fquare of AC; therefore the fquares of DA, AC, are equal to the fquares of BA, AC; But the square of DC is equal to the fquares of DA, AC, because DAC is a right angle; and the fquare of BC, by hypothefis, is equal to the fquares of BA, AC; therefore the fquare of DC is equal to the fquare of BC; and therefore alfo B the fide DC is equal to the fide BC. And A 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 alfo BAC is a right angle. Therefore, if the fquare, &c. Q. E. D. THE E VERY right angled parallelogram is faid to be contained by any two of the ftraight lines which contain one of the right angles. II. In every parallelogram, any of the parallelograms about a diameter, together with the two complements, is called A E D K ÁGK, or EHC which are B G the parallelograms which make the gnomon.' PROP. I. THEOR. IF there be two ftraight lines, one of which is divided into any number of parts; the rectangle contained by the two ftraight lines, is equal to the rectangles contained by the undivided line, and the feveral parts of the divided line. Book II. B DEC Let A and BC be two ftraight lines; and let BC be divided into any parts in the points D, E; the rectangle contained by the ftraight lines A, BC is equal to the rectangle contained by A, BD, together with that contained by A, DE, and that contained by Á, EC. DKF 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 c 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 feveral 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. a 46. 1. b 31.1. PROP. II. THEOR. IF a ftraight line be divided into any two parts, the rectangles contained by the whole and each of the parts, are together equal to the fquare of the whole N. B. To avoid repeating the word contained too frequently, the rectangle contained by two straight lines AB, AC is fometimes fimply called the rectangle AB, AC. |