Book I. a 46. 1. b 31. 1. 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 the square BDEC, and on BA, AC the squares GB, HC; and through A drawb AL parallel to BD or CE, and join AD, FC; then, because each of the angles BAC, c 25. def. BAG is a right angle, d 14. 1. B G H C cause the angle DBC is equal to the angle FBA, f 4. 1. e each of them being a right angle, adding to each the e 2. Ax. angle ABC, the whole angle DBA will be equal to the whole FBC; and because the two sides AB, 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 to the base FC, and the triangle ABD to the trig 41. 1. angle FBC. But the parallelogram BL is double 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 h 6. Ax. parallels FB, GC. Now the doubles of equals are equal h to one another; therefore the parallelogram BL is equal Book I. 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, HG; 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. 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 a AD at right angles to AC, a 11. 1. and make AD equal to BA, and join DC. Then, be cause DA is equal to AB, the square D b 47. I 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, and the base DC likewise equal to the base BC, the angle DAC is equal to the angle BAC: But DAC is a right angle; c8. 1. therefore also BAČ is a right angle. Therefore, if the square, &c. Q. E. D. ELEMENTS OF GEOMETRY. BOOK II. DEFINITIONS. I. VERY right angled parallelogram, or rectangle, is Book II. 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 46 AB, &c. For the sake of brevity, instead of the rect"angle 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 occasionally write "AD2." 66 "The sign+placed between the names of two magni"tudes, signifies that those magnitudes are to be added "together: and the sign-placed between them, signi"fies 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. E a 11. 1. b 3. 1. In every parallelogram, any of the parallelograms about a diameter, together with the two complements isA of the pa-H "the ❝rallelogram AC. This "gnomon may also, for "the sake of brevity, be B "called the gnomon AGK or EHC." 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 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 into any parts in the points D, E; the rectangle A.BC is equal to the several rectangles A.BD, A.DE, A.EC. b From the point B draw a BF at right angles to BC, and make BG equal to c 31. 1. A; and through & drawe GH parallel to BC; and through D, E, C, drawe DK, EL, CH parallel to BG; then BH, BK, DL, and EH are rectangles, and BH BK+DL+EH. But BHBG.BCA.BC, because BG=A: Also BKBG.BD A.BD, because BG=A; and DL = d 34. 1. DK.DE A.DE, becaused DK BG=A. In like manner, EH = A.EC. |