the parallelogram GM equal to the triangle DBC, having GHM BOOK I for one of its angles: and because the straight line GH meets the parallels KM, FG, the alternate angles MHG, HGF are equal: add to each of these the angle HGL; therefore the angles MHG, HGL are equal to the angles HGF, HGL: But MHG, HGL are equal to two right angles; therefore also HGF, HGL are equal to two right angles: and because the two straight lines FG, GL, on opposite fides of GH, make the ad C 29. 1. jacent angles HGF, HGL equal to two right angles, FG is in the same straight line d with GL: And because KF is parallel to HG, e e 30. 1. E is a parallelogram; and because the triangle ABD is equal to the parallelogram HF, and the triangle DBC to the parallelogram GM; the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM; therefore the parallelogram KFLM has been defcribed equal to the given rectilineal figure ABCD, having the angle FKM equal to the given angle E. Which was to be done. COR. From this it is manifest, how to a given ftraight line to apply a parallelogram, which shall have an angle equal to a given rectilineal angle, and shall be equal to a given rectilineal figure, viz. by applying b to the given straight line, a parallelogram b 44. 1. equal to the first triangle ABD, and having an angle equal to the given angle. PROP. XLVI. PROB. To describe a square upon a given straight line. Let AB be the given straight line; it is required to defcribe a square upon AB. b 3. 1. c 31.1. d 34.1. From the point A draw a AC at right angles to AB; and a II. I. make 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 equal d 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 is equilateral: likewise all its angles are right angles; because the straight line AD meeting the parallels AB, DE, the angles BAD, ADE are equal F2 e to two right e 29. 1. BOOK I. right angles; but BAD is a right angle; therefore also ADE is D a right angle; but the opposite angles of C d 34. 1. parallelograms are equald; therefore each of the oppofite 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 defcribed upon the given straight line AB. Which was to be done. a 46. r. b 31. 1. COR. Hence every parallelogram that E B N any right angled triangle, the square which is described upon the fide fubtending the right angle, is equal to the squares described upon the fides which contain the right angle. Let ABC be a right angled triangle having the right angle BAC; the square described upon the fide BC is equal to the squares described upon BA, AC. On BC describe a 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 30. Def. BAG is a right angle, the two ftraight lines AC, AG upon the opposite sides of AB, make with it at the point A the adjacent angles equal to two right angles; therefore CA is in the fame ftraight lined with AG; for the fame reason, AB and AH are in the fame 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 14. 1. f 4. 1. F G A H B C D LE K e 2. Ax. and the whole angle DBA is equal to the whole FBC; and because the two fides 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 triangle FBC: Now the parallelogram BL is double $ 41. 1. 8 of the triangle ABD, because they are upon the fame base BD, and between the fame parallels, BD, AL; and the square GB GB is double of the triangle FBC, because these also are upon Book I. the same base FB, and between the same parallels FB, GC: But the doubles of equals are equal h to one another; therefore the h 6. Ax. parallelogram BL is equal to the square GB: And in the fame manner, by joining AE, BK, it may be 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 described upon the straight line BC, and the squares GB, HC upon BA, AC: Wherefore the square upon the fide BC is equal to the squares upon the fides BA, AC. Therefore, in any right angled triangle, &c. Q. E. D. PROP. XLVIII. THEOR. F the square described upon one of the fides of a triangle, be equal to the squares described upon the other two fides of it; the angle contained by these two fides is a right angle. If the square described upon BC, one of the fides of the triangle ABC, be equal to the squares upon the other fides BA, AC; the angle BAC is a right angle. A D a II. I. b 47. 1. From the point A draw a 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 to the squares of DA, AC, because DAC is a right angle; and the square of BC, by hypothefis, is equal to the squares of BA, AC; therefore the B square of DC is equal to the square of BC; and therefore also the side DC is equal to the fide 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 also equal to the base BC; therefore the angle DAC is equal to the angle BAC: But DAC is a right angle; there- c 8, fore also BAC is a right angle. Therefore, if the square, &c. Q. E. D. C C THE BOOK II. DEFINITIONS. VERY right angled parallelogram is called a rectangle, and it is faid 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 dia meter, together with the two com plements, is called a Gnomon. F G 'FC, is the gnomon, which is more briefly expressed by the let ters AGK, or EHC, which are rallelograms which make the KC F 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 Let A and BC be two straight lines; and let BC be divided Book. II. into any parts in the points D, E; the rectangle contained by A the straight lines A, BC is equal to B the rectangle contained by A, BD, together with that contained by A, DE, and that contained by A, EC. a II. 1. K LH b 3. 1. A C 31. I. From the point B draw a BF at right angles to BC, and make BG G equal to A; and through G draw • GH parallel to BC; and through D, E, C draw DK, 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 contained by A, DE, because DK, that is, dBG, is equal to A; and in like manner, the rectangle EH d 34. 1. 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 ftraight lines, &c. Q. E. D. PROP. II. THEOR. Fa 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 A any two parts in the point C; the rectangle contained by AB, BC, together with the rectangle * AB, AC, shall be equal to the square of AB. CB a 46. 1. b 31. 1. F E Upon AB defcribe a the square ADEB, and through C drawb CF, parallel to AD or BE; then AE is equal to the rectangles AF, CE; and AE is the square of AB; D and AF is the rectangle contained by BA, AC, for 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 * N. B. To avoid repeating the word contained too frequently, the rectangle contained by two straight lines AB, AC is sometimes fimply called the rectangle AB, AC. |