lelograms are equal (1. 34.); therefore each of the opposite angles A BE, BED is a right angle; wherefore the figure ADE B is rectangular, and it has been demonstrated that it is equilateral; it is therefore a square (Def. 30.), 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. H PROP. XLVII. THEOR. In any right angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle. Let ABC be a right angled triangle having the right angle BAC; the upon the side bc is equal to the squares described upon BA, AC. On Bc describe (1. 46.) the square BDEC, and on BA, AC the squares GB, HC; and through a draw (1. 31.) AL parallel to BD or CE, and join AD, FC; then, because each of the angles BAC, BAG is a right angle G (Def. 30.), the two straight lines AC, Ag upon the opposite sides of AB, make with it at the point A the adja- f cent angles equal to two right angles; K therefore ca is in the same straight line (1. 14.) with AG; for the same 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 L E each the angle ABC, and the whole angle dba is equal (Ax. 2.) to the whole FBC; and because the two sides A B, BD are equal to the two FB, BC, each to each, and the angle DBA equal to the angle FBC; therefore the base Ad is equal (1. 4.) to the base Fc, and the triangle ABD to the triangle FBC: Now the parallelogram bl is double (1. 41.) of the triangle ABD, because they are upon the same basé 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 (Ax. 6.) B D to one another : Therefore the parallelogram bl is equal to the square GB: And, in the same manner, by joining A E, 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 G B, 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. D PROP. XLVIII. THEOR. If the square described upon one of the sides of a 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. If the square described upon BC, one of the sides of the triangle ABC, be equal to the squares upon the other sides BA, AC, the angle Bac is a right angle. From the point a draw (1. 11.) AD at right angles to AC, and make an equal to BA, and join Dc: Then, because DA is equal to A B, 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 (1. 47.) 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 also the side dc is equal to the side BC. And because the side 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 (1. 8.) 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. C EUCLID'S ELEMENTS OF GEOMETRY. BOOK II. DEFINITIONS. I. EVERY right angled parallelogram is said 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 diameter, together with the two A E D complements, is called a Gnomon. “ Thus the parallelogram Hg, together with the complements AF, FC, H K is the gnomon, which is more B с briefly expressed by the letters AGK, or Euc, which are at the opposite angles of the parallelograms which make the gnomon.” G PROP. I. THEOR. If there be two straight lines, one of which 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 A and Bc be two straight lines ; and let Bc be divided L into any parts in the points D, E; the rectangle contained by the straight lines A, BC is equal to the B D E С rectangle contained by A, BD, together with that contained by A, DE, and that contained by A, EC. From the point B draw (1. 11.) BF G at right angles to BC, and make BG F АН equal (1. 3.) to A; and through a draw (1.31.) GH parallel to BC; and through D, E, C, draw (1. 31.) DK, EL, cu parallel to BG; then the rectangle bh is equal to the rectangles BK, DL, EH; and be 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 (1. 34:) BG, is equal to A; and in like manner the rectangle Es 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. A AB. PROP. II. THEOR. 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. Let the straight line AB be divided into any two parts in the point c; the rectangle contained by AB, с в BC, together with the rectangle* AB, AC, shall be equal to the square of Upon A B describe (1. 46.) the square ADEB, and through c draw (1. 31.) CF parallel to AD or BE; then AE is equal to the rectangles AF, CE; and AE is the square of AB; and AF is the rectangle contained by BA, AC; for E it is contained by 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 A B. If therefore a straight line, &c. Q.E. D. * N. B. - To avoid repeating the word contained too frequently, the rectangle contained by two straight lines A B, ac is sometimes simply called the rectangle A B, AC. D F If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts is equal with the rectangle contained by the two parts, together with the square of the aforesaid part. Let the straight line AB be divided into two parts in the point c; the rectangle A B, BC is equal to the rectangle ac, CB, together with the square of BC. Upon Bc describe (1. 46.) the square CDEB, A C B and produce ED to F, and through a draw (1. 31.) AF parallel to cd or BE; then the rectangle AE is equal to the rectangles AD, CE; and a E is the rectangle contained by AB, BC, for it is contained by AB, BE, of which we is equal to BC; and AD is contained by AC, CB, for cd is equal to CB; and do is the square of Bc; therefore the rectangle A B, BC is equal to the rectangle AC, CB together with the square of BC. If therefore a straight, &c. Q.E.D. F D 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 A B be divided into any two parts in c; the square of AB is equal to the squares of AC, CB, together with twice the rectangle contained by AC, CB. Upon A B describe (1. 46.) the square ADE B, and join BD, and through c draw (1. 31.) CGF parallel to AD or BE, and through a draw uk parallel to AB or DE: And because CF is parallel to AD, and BD falls upon them, the exterior angle BGC is equal (1. 29.) to the interior and opposite angle ADB; but ADB is equal (1. 5.) to the angle ABD, because BA is equal to AD, being sides of a square ; wherefore the angle |