† 34. 1. † 2 Ax. † 3 Ax. triangle AEK is equal to the triangle AHK; and for the same reason, the triangle KGC is equal to the triangle KFC. Therefore, because the triangle AEK is equal to the triangle AHK, and the triangle KGC to KFC, the triangle AEK together with the triangle KGC is equal † to the triangle AHK together with the triangle KFC: but the whole triangle ABC was proved equal to the whole ADC; therefore the remaining complement BK is equal † to the remaining complement KD. Wherefore, the complements, &c. Q. E. D. * 42. 1. 31. 1. * 29. 1. * 12 Ax. * 43. 1. * 15. 1. † Ax. PROPOSITION XLIV. PROB.-To a given straight line, to apply a parallelogram which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Let AB be the given straight line, and C the given triangle, and D the given rectilineal angle; it is required to apply to the straight line AB, a parallelogram equal to the triangle C, and having an angle equal to D. Make the parallelogram BEFG equal to the triangle C, and having the angle EBG equal to the angle D, so that BE be in the same straight line with AB; and produce FG to H; and through A, F E K G M C B H draw * AH parallel to BG or EF, and join HB. Then, because to to the be done. the parallelogram LB is applied, equal to the triangle C, and PROPOSITION XLV. PROB.-To describe a parallelogram equal to a given recti- See N. lineal figure, and having an angle equal to a given rectilineal angle. Let ABCD be the given rectilineal figure, and E the given rectilineal angle; it is required to describe a parallelogram equal to ABCD, and having an angle equal to E. Join DB; and describe* the parallelogram FH equal to the 42. 1. triangle ADB, and having the angle FKH equal to the angle E; and to the straight line GH, apply the parallelogram 44. 1. GM equal to the triangle DBC, having the angle GHM equal to the angle E: the figure FKML shall be the parallelogram required. * +2 Ax. • 29. 1. F GL E † 1 Ax. KHM * 14. 1. Because the angle E is equal to each of the angles FKH, † Constr. GHM, the angle FKH is equal to GHM: add to each of † 1 Ax. these the angle KHG; therefore the angles FKH, KHG are equal to the angles KHG, GHM: but FKH, KHG are equal to two right angles; therefore also KHG, GHM are equal to 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 the straight line HG meets the parallels KM, FG, the alternate angles MHG, HGF are equal; add to each 29. 1. of these the angle HGL; therefore the angles MHG, HGL are equal to the angles HGF, HGL: but the angles MHG, HGL † 2 Ax. are equal to two right angles; wherefore also the angles HGF, HGL are equal † to two right angles, and therefore FG is in the same straight line with GL: and because KF is parallel to HG †, and HG to ML, KF is parallel to ML: and KM, FL are + parallels; wherefore KFLM is a parallelogram: and because the triangle ABD is equal to the parallelogram HF †, and the triangle DBC to the paral-Constr. lelogram GM; the whole rectilineal figure ABCD is equal † † 2 Ax. * * * 29. I. † 1 Ax. † 14. 1. Constr. 30. 1. * 44. 1. * 11. 1. * 3. 1. * 31. 1. + Def. 34. 1. 34. 1. + Constr. +1 Ax. * 29. 1. 34. 1. † 1 Ax. † 30 Def. 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. PROPOSITION 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. D * 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 ADEB is a † parallelogram: whence AB is equal 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: likewise all its angles are right angles; for, since the straight line AD meets the parallels AB, DE, the angles BAD, ADE are equal * to two right angles: but BAD is a † right angle; therefore also ADE is at right angle: but the opposite angles of parallelograms are equal; 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. PROPOSITION 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. • 46. 1. 31. 1. right angle, the 30 Def. 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 a†† Hyp. right angle, and that the angle BAG is also a two straight 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 same straight line with AG: for the same reason, AB and AH are in the same straight line. angle DBC is equal And because the to the angle FBA, each of them being a right † angle, add F B H K 14. 1. D E † 11 Ax. † 30 Def. 2 Ax. + 30 Def. 4. 1. 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, 41. 1. 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 parallelo- ⚫ 6 Ax. gram 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 +2 Ax. square 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. PROPOSITION XLVIII. THEOR.-If the square described upon one of the sides of a triangle, be equal to the squares described upon the other 1. 1. † 3. 1. † 2 Ax. * 47. 1. + Constr. † 1 Ax. + Constr. * 8. 1. + Constr. † 1 Ax. 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. * A D 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 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 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 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 a† right angle; therefore also BAC is at right angle. Therefore, if the square, &c. Q. E. D. B |