Book I. PRO P. XLVI. PROB. To describe a square upon a given straight line. a 11.1. b 3. I. C 31. I. d 34. I. D Let AB be the given straight line ; it is required to describe a square upon AB. From the point A draw a AC at right angles to AB; and make b 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 equals to DE, and AD to BE : But BA is equal to AD; therefore the four straight lines BA, AD, DE, EB are equal to one another, and the parallelogram ADEB E is equilateral, likewife all its angles are right angles; becaufe the straight line AD meeting the parallels AB, DE, the angles BAD, ADE, are equale to two right angles; but BAD is a right angle ; therefore also ADE is a right angle; but the opposite B angles of parallelograms are equal d; therefore each of the opposite angles ABE, BED is a right angle; wherefore the figure ADEB is rectangular, and it has been demonstrated 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. e 29. I. N any right angled triangle, the square which is de. scribed upon the side fubtending the right angle, is. equal to the squares described upon the sides which con. tain the right angle. Let ABC be a right angled triangle having the right angle GB, a 46. I C 30. def. L GB, HC; and thro' A draw 6 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 G b 31. 1. straight lines AC, AG upon the opposite fides of AB, H make with it at the point A the adjacent angles equal to two right angles; therefore K CA is in the same straight lined with AG; for the same d 14.1. B reason, AB and AH are 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, D 'E and the whole angle DBA is equal to the whole FBC; and because the two sides AB, ED e 2. Ax. are equal to the two FB, BC, cach to each, and the angle DBA equal to the angle FBC; therefore the base AD is equalf to the base FC, and the triangle ABD to the triangle f 4. 1. FBC: Now the parallelogram BL is double 5 of the triangle g 41.1, ABD, because they are upon the same base BD, and between the same 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 same 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 fam2 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 Yquare BDEC is described upon the itraight 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. F the square described upon one of the sides of a tri 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. D If Book 1. a II. I. b 47.1. If the square described upon BC, one of the sides of the tri. angle ABC, be equal to the Squares upon the other sides BA, AC; the angle BAC is a right angle. Fron, the point A draw • AD at right angles to AC, and D to the squares of DA, AC, because DAC € 8. I. E I. 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 E two complements, is called A a Gnomon. • Thus the pa• rallelogram HG, together with the complements AF, FC, is the gno F F mon, which is more brief. K ly expreffed by the letters AGK, or EHĆ which are B at the opposite angles of • the parallelograms which make the gnomon.' G 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. D 2 Let II.1. b 3. 1, Book II. Let A and BC be two straight lines, and let BC be divided minto any parts in the points D, E, the rectangle contained by the straight lines A, BC is equal B В. D E C From the point B draw a BF at right angles to BC, and make BG equal to A ; and through K L H C 31. 1. G draw GH parallel to BC ; contained by GB, BD, of which GB is equal to A ; and DL is d 31. 1. contained by A, DE, because DK, that is, 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. IF 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. CB Let the straight line AB be divided into A Upon AB describe the square A DEB, a 46. 1. b 31.s. F E and • N. B. To avoid repeating the word contained too frequently, the rectangle contained by two straight lincs AB, AC is lowetimes simply called the rectangle AB, AC. |