* two straight lines KH, HM upon the opposite sides of it make the adjacent angles equal to two right angles, KH is in the same straight line* with HM: *14. 1. and because the straight line HG meets the parallels KM, FG, the alternate angles MHG, HGF* are * 29. 1. equal, add to each of these the angle HGL: therefore the angles MHG, HGL, are equal to the angles +2 Ax. HGF, HGL: but the angles MHG, HGL, are equal to two right angles; wherefore also the angles 29. 1. HGF. HGL are equal to two right angles, and there- +1 Ax. fore FG is in the same straight linet with GL: and +14. 1. because KF is parallel to HG+, and HG to ML; KF + Constr. is parallel to ML: and KM, FL aret parallels; wherefore KFLM ist a parallelogram: and because + Def. 34. the triangle ABD is equal to the parallelogram HFt, 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 described 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 straight 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* to the given straight line a parallelogram 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 describe a square upon AB. C * 30. 1. + Constr. 1. Constr. +2 Ax. * 44. 1. *11.1. * 31. 1. 1. * 34. 1. From the point A draw* AC at right angles to AB; and make* AD equal to AB: through the point D draw * 3. 1. DE parallel* to AB, and through B draw BE parallel to AD; therefore ADEB is at parallelogram: whence + Def. 34. AB is equal 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 parallel- D ogram ADEB is equilateral: likewise all its angles are right angles; for, since the straight line AD meets the parallels AB, DE, the angles BAD, ADE are A E + Constr. † 1 Ax. * 29. 1. + Constr. + 3 Ax. 34. 1. + 1 Ax. + 30 Def. * 46. 1. * 31. 1. + Hyp. 14. 1. † 11 Ax. † 30 Def. * 2 Ax. + 30 Def. * 4. 1. 41. 1. equal to two right angles: but BAD is at right angle, therefore also ADE is a+ right angle: but the opposite angles of parallelograms* are equal; therefore each of the opposite angles ABE, BED is at right angle; wherefore the figure ADEB is rectangular: and it has been demonstrated that it is equilateral; it is therefore at 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. 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 shall be equal to the squares described upon BA, AC. B D H E L K On BC describe* the square BDEC, and on BA, AC the squares GB, HC; and through A draw* AL parallel to BD, or CE, and join AD, FC. Then, because the angle BAC is at right angle, and that the angle BAG is also a right angle, the two straight lines AC, AG upon the opposite sides of AB, make with it at the point A the adjacent angles F equal to two right angles: therefore CA is in the same straight line* 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 each the angle ABC; therefore the whole angle DBA is 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 triangle FBC: now 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 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 *6 Ax. parallelogram BL is equal to the square GB. In the same manner, by joining AE, BK, it can 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 +2 Ax. BDEC is described upon the straight line BC, and the squares GB, HC upon BA, AC; therefore 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. Let 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 shall be a right angle. 1.1. From the point A draw* AD at right angles to AC, and make+ AD equal to BA, and join DC. Then, + 3. 1. 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 A * 47. 1. D + Constr. + 1 Ax. AC; therefore the squares of DA, AC are equal to +2 Ax. the squares of BA, AC: but the square of DC is equal to the squares of DA, AC, because DAC is at 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 + Constr. DAC, BAC, the two DA, AC are equal to the two BA, AC each to each; and the base DC has been proved equal to the base BC; therefore the angle DAC is equal to the angle BAC: but DAC is at right * 8.1. angle; therefore also BAC is at right angle. There-Constr. fore, if the square, &c. Q. E. D. + 1 Ax. THE ELEMENTS OF EUCLID. BOOK II. DEFINITIONS. I. Every right-angled parallelogram, or rectangle, is said to be contained by any two of the straight lines which contain one of the right angles. II. A E In every parallelogram, any of the parallelograms about a diameter, together with the two complements, is called a Gnomon. Thus the parallelogram HG, together with the complements AF, • FC, is the gnomon, which is more briefly expressed by the letters F H K BG AGK, or EHC, which are at the opposite angles ' of the parallelograms which make the gnomon'. 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 contained by the straight lines A, BC shall be equal to the rectangle contained by A, BD, together with that contained by A, DE, and that contained by A, EC. From the point B draw* BF at right angles to BC, and make BG * 3. 1. 31. 1. equal to A; and through G draw* GH_parallel to BC; and through D, E, C, draw* DK, EL, CH pa- 31. 1. rallel to BG. Then the rectangle BH is equal to the rectangles BK, DL, EH: but 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 † Constr. contained by GB, BD, of which GB is equal to A; and DL is 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 by A, EC. Wherefore, 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 contained by AB, BC, together with the rectangle AB, AC shall be equal to the square of AB. * 34. 1. 46. 1. A св 31. 1. Upon AB describe* the square ADEB, and through C draw* CF, parallel to AD or BE. Then AE is equal to the rectangles AF, CE: but AE is the square of AB; 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 30 Def. by AB, BC, for BE is equal to AB: therefore the rectangle contained by AB, AC, together with the rect D FE N. B.--To avoid repeating the word contained too frequently, the rectangle contained by two straight lines AB, AC is sometimes simply called the rectangle AB, AC. |