D E DE parallel (31.1.) to AB, and through B draw BE parallel to AD; therefore ADEB is a parallelogram; whence AB is equal (34. 1.) to DE and AD to BE; but BA is equal to C AD: therefore the four straight lines BA, AD, DE, EB are equal to one another, and the parallelogram ADEB is equilateral; it is likewise rectangular; for the straight line AD meeting the parallels AB, DE, makes the angles BAD, ADE equal (29. 1.) to two right angles; but BAD is a right angle; therefore also ADE is a right angle; now the opposite angles of parallelograms are equal (34.1.); A B 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. 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 square described upon the side BC is equal to the squares described upon BA, AC. On BC describe (46. 1.) the square BDEC, and on BA, AC the squares GB, HC; and through A draw (31.1.) AL parallel to BD or CE, and join AD, FC: then, because each of the angles BAC, BAG is a right angle (25. def.), the two straight lines AC, AG fore the base AD is equal (4. 1.) to the base FC, and the triangle ABD to the triangle FBC. But the parallelogram BL is double (41. 1.) of the triangle 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 BFC, because these also are upon the same base FB, and between the same parallels FB, GC. Now the doubles of equals are equal (6. Ax.) to one another; therefore the parallelogram 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 square BDEC is de scribed 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. 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 (11. 1.) 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 D 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 (47. 1.) 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 B DA is equal to AB, and AC common to the two triangles DAC, BAC, and the base DC likewise equal to the base BC, the angle DAC is equal (8. 1.) 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. A C ELEMENTS OF GEOMETRY. BOOK II. DEFINITIONS. 1. EVERY right angled parallelogram, or rectangle, is said to be contained by any two of the straight lines which are about one of the right angles. "Thus the right angled parallelogram AC is called the rectangle "contained by AD and DC, or by AD and AB, &c. For the sake "of brevity, instead of the rectangle contained by AD and DC, we "shall simply say the rectangle AD.DC, placing a point between "the two sides of the rectangle. Also, instead of the square of a line "for instance of AD, we shall frequently in what follows write "AD." "The sign+placed between the names of two magnitudes, signi"fies that those magnitudes are to be added together, and the sign "-placed between them, signifies that the latter is to be taken away "from the former." "The sign = signifies, that the things between which it is placed "are equal to one another." 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 straightlines 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 into any parts in the points D, E; the rectangle A.BC is equal to the several rectangles A.BD, A.DE, A.EC. From the point B draw (11.1.) BF at right angles to BC, and make BG equal (3. 1.) to A; and through G draw (31.1.) GH parallel to BC; and through D, E, C, draw (31.1.) DK, EL, CH parallel to BG; then BH, BK, DL, and EH are rectangles, and BH-BK G +DL+EH. But BH=BG.BC=A.BC, because BG=A: Also BK-BG. F BD = A.BD, because BG = A; B D E C KLH A and DL = DK.DE = A. DE, because (34. 1.) DK = BG=A. In like manner, EH=A.EC. Therefore A.BC = A.BD+A.DE+ A.EC; that is, the rectangle A. BC is equal to the several rectangles A.BD, A. DE, A.EC. Therefore, if there be two straight lines, &c. Q. E. D. 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 AB, BC, together with the A rectangle AB.AC, is equal to the square of AB; or AB.AC+AB.BC = ABo. On AB describe (46.1.) the square ADEB, and through C draw CF (31.1.) parallel to AD or BE; then AF+CE=AE. But AF = AD. AC = AB.AC, because AD = AB; CE = BE. BC = = AB.BC; and AE = AB. Therefore AB.AC+AB.BC=AB. There fore, if a straight line, &c. Q. E. D. D FE PROP. III. THEOR. If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to 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 AB.BC is equal to the rectangle AC.BC, together with BC. Upon BC describe (46.1.) the A square CDEB, and produce ED to F, and through A draw (31. 1.) AF parallel to CD or BE; then AE=AD+CE. = C But AE= AB.BE= AB.BC, because BE= BC. So also AD=. AC.CD = AC.CB; and CE BC; therefore AB.BC=AC.CB + BC2. Therefore, if a straight FD line, &c. Q. E. D. PROP. IV. THEOR. B E 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 AB be divided into any two parts in C; the square of AB is equal to the squares of AC, CB, and to twice the rectangle contained by AC, CB, that is, AB=AC+CB2+2AC.CB. Upon AB describe (46. 1.) the square ADEB, and join BD, and through C draw (31. 1.) CGF parallel to AD or BE, and through G draw HK parallel to AB or DE. And because CF is parallel to AD, A and BD falls upon them, the exterior angle BGC is equal (29. 1.) to the interior and opposite angle ADB; but ADB is equal (5.1.) to the angle ABD, because BA is equal to AD, being sides of a square; wherefore the angle CGB is equal to the angle GBC; and therefore the side BC is equal (6. 1.) to the side CG: but CB is equal (34. 1.) also to GK and CG to BK; wherefore the figure CGKB is equilateral. It is likewise rectangular; for the angle CBK being a C B G H K right angle, the other angles of the parallelogram CGKB are also right angles (Cor. 46. 1.) Wherefore CGKB, is a square, and it is upon the side CB. For the same reason HF also is a square, and |