therefore the angles * † 1 Ax. two right angles: and because at the point H in the straight line GH, the 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: and because *14. 1. 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† † 2 Ax. to the angles HGF, HGL: but the angles MHG, HGL, are equal to two right angles; * 29. 1. wherefore also the angles HGF, HGL are equal† +1 Ax. to two right angles, and therefore FG is in the same straight line with GL: and because KF +14. 1. is parallel to HG,† and HG to ML; KF is +Const, parallel to ML: and KM, FL are+ parallels; wherefore KFLM is a parallelogram: and because the triangle ABD is equal to the parallelo- 34. 1. gram HF,† and the triangle DBC to the paral- +Const. lelogram GM; the whole rectilineal figure ABCD is equal to the whole parallelogram *2 Ax. 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. * *30. 1. + Const. + Def. 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 *44. L' line a parallelogram equal to the first triangle ABD, and having an angle equal to the given angle. * *11. 1. *3.1. *S1. 1. 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. From the point A draw * AC at right angles to AB; and make* AD equal to AB; through the point D draw DE parallel to AB, and through B draw BE parallel to AD; therefore + Def. ADEB is at parallelogram: whence AB is equal to DE, and AD to BE : 34. 1. *34. 1. +1 Ax. * E B + Const. but BA is equal † to AD; there- C fore the four straight lines BA, AD, DE, EB, are equal to one D another, and the parallelogram ADEB is equilateral: likewise all its angles are right angles; for, since the straight line AD meets the parallels AB, DE, the *29.1. angles BAD, ADE are equal to two right +Const. angles: but BAD is at right angle; therefore +3 Ax. also ADE is at right angle: but the opposite 34. 1. angles of parallelograms are equal; therefore +1 Ax. each of the opposite angles ABE, BED is at right angle; wherefore the figure ADEB is rectangular and it has been demonstrated that +30 Def. 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 shall be equal to the squares described upon BA, AC. * On BC describe the square BDEC, and on *46. 1. ̧ BA, AC the squares GB, HC; and through A G draw* AL parallel to BD, or CE, and join AD, *31. 1. FC. Then, because the angle BAC is at right +Hyp. angle, and that the angle BAG is also a* right 30 Def. angle, the two straight lines AC, AG upon the opposite sides of AB, make with it at the point F A the adjacent angles equal to two right angles; therefore CA is in the same straight line * with H K B C *14. 1. AG: for the same reason, AB and AH are in the same straight line. And because the angle DBC is equal to the angle +11 Ax. FBA, each of them being a right † angle, add to †30Def. each the angle ABC; therefore the whole angle * DBA is equal to the whole FBC: and because *2 Ax. the two sides AB, BD, are equal† to the two +30Def. 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 *4. ?. to the triangle FBC; now the parallelogram BL is double* of the triangle ABD, because they *41. 1. 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, †2 Ax. HC: and the square BDEC is described upon *1. 1. +3.1. 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. 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. From the point A draw* 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 +2 Ax. the squares of DA, AC are equal † to the squares of BA, AC: but *47. 1. the square of DC is equal to the squares of DA, AC, because B † Const. DAC is a† right angle; and the A D square of BC, by hypothesis, is equal to the squares of BA, AC: therefore the square of +1 Ax. DC is equal† to the square of BC; and therefore also the side DC is equal to the side BC. And + Const. 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, 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 +Const. angle; therefore also BAC is at right angle Therefore, if the square, &c. Q.E.D. *8. 1. +1 Ax. 49 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 H ' with the complements AF, FC, is the gnomon, F K BG C which is more briefly expressed by the let 'ters AGK, or EHC, which are at the opposite angles of the parallelograms which make the gnomon.' F |