Book I. PRO P. XLVI. PRO B. To describe a square upon a given straight line. a II. 1. a a Let AB be the given straight line; it is required to describe a square upon AB. From the point A draw * AC at right angles to AB; and b 3. I. make b AD equal to AB, and through the point D draw DE c 31. 1. parallel to AB, and through B draw BE parallel to AD; thered 34. I. fore ADEB is a parallelogram; whence AB is equal to DE, and AD to BE : But BA is equal to AD; C therefore the four straight lines BA, AD, DE, EB are equal to one ano. ther, and the parallelogram ADEB D E is equilateral, likewise all its angles are right angles; because the straight line AD meeting the parallels AB, DE, the angles BAD, ADE, are ee 29. I. quale to two right angles; but BAD is a right angle; therefore also ADE B Cor. Hence every parallelogram that has one right angle has all its angles right angles. IN 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 BAC; the square described upon the side BC is equal to the squares described upon BA, AC On BC describe a the square BDEC, and on BA, ACthe squares a 46.17 GB, b 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 up C 30. def. on the opposite sides 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 B d 14.1. reason, AB and AH are in с 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 L equal e to the whole FBC; and because the two sides AB, ED 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 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 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 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 fides BA, AC. Therefore in any right angled triangle, &c. Q. E. D. e 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. à II. I. b 47.1. b If the square described upon BC, one of the îides of the triangle ABC, be equal to the squares upon the other fides BA, AC; the angle BAC is a right angle. 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 А. 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; therefure the square of DC is equal to the square of BC ; and therefore allo 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. : THE E a 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 comple(ments AF, FC, is the gnoo mon, which is more brief. H K ly exprefied by the letters S AGK, or EHC which are B 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. Let D2 II.1. b 3. I, b с C 31. 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 B D E C BD, together with that contained by A, DE, and that contained by A, EC. From the point B draw a BF at right angles to BC, and make BG equal to A; and through K L H G draw GH parallel to BC ; and through D, E, C draw · DKF А 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 d 34. I. 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. с ; 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. N. B. To avoid repeating the word contained too frequently, the rectangle contained by two straight lincs AB, AC is sometimes simply called the rectangle AB, AC. |