#42. 1. *44. 1. +Const. Draw DB; and describe the parallelogram FH equal to the triangle ADB, and having the angle FKH equal to the angle E; and to the straight line GH apply the parallelogram GM equal to the triangle DBC, having the angle GHM equal to the angle E: the figure FKML shall be the parallelogram required. Because the angle E is equal to each of the +1 Ax. angles FKH, GHM, the angle FKH is equal† to GHM: add to each of these the angle KHG; therefore the angles FKH, KHG, are equalt FG +2 Ax. to the angles KHG, GHM: but FKH *29. 1. +1 Ax. *14. 1. *29. 1. 12 Ax. KHG are equal* to two right angles; therefore also KHG, GHM, are equalt B 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 of these the angle HGL: therefore the angles MHG, HGL, are equal† to the angles HGF, HGL: but the angles MHG, HGL, 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 linet with GL: and +Const. because KF is parallel to HG,† and HG to ML; *30. 1. KF is parallel to ML: and KM, FL aret parallels; +Const. wherefore KFLM ist a parallelogram: and because + Def. 34. 1. the triangle ABD is equal to the parallelogram +Coust. HF, and the triangle DBC to the parallelogram *2 Ax. GM; the whole rectilineal figure ABCD is equal #29. 1. +1 Ax. +14. 1. to the whole parallelogram KFLM. Again, let the figure be five-sided; then, having drawn DC, there will be, besides the four-sided figure, AČ, a triangle upon DC; a parallelogram equal to this triangle may be applied to LM, just as the parallelogram GM equal to the triangle DBC was applied to GH; and thus, however numerous be the sides, a parallelogram may be described equal to the given rectilineal figure, and having the angle FKM equal to the given angle E. 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 *44. 1. given rectilineal figure.* PROP. XLVI. PROB. To describe a square upon a given finite straight line. Let AB be the given straight line; it is required to describe a square upon AB. #11. 1. *3. 1. *31. 1. +Def. 34. 1. 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 at parallelogram: whence AB is *34. 1. equal to DE, and AD to BE: but C *29. 1. +3 Ax. *34. 1. D 3 B 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 A BAD, ADE are equal to two +Const. right angles: but BAD is at 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 at right angle; wherefore the figure ADEB is rectangular: and it has been demonstrated that it is equilateral; it is +25 Def. 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. +1 Ax. F 2 It is easily shown that every quadrilateral that has one right angle has all its angles right angles, provided the figure be equilateral. For, let AE be equilateral, and have a right angle A; the diagonal DB will divide the figure into equal triangles, by the 8th prop. Hence the angle E is equal to the angle A; that is, E is a right angle; moreover, the two triangles being isosceles as well as right-angled, the angles at their bases are each half a right angle (v. and xxxii.); hence the angles D and B are also right angles. The quadrilateral is therefore a square. 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. #31. 1. +Hyp. On BC describe the square BDEC, and on BA, #46. 1. AC the squares GB, HC; and through A draw AL parallel to BD, or CE: draw also AD, FC. Then, because the angle BAC is at right angle, *26 Def. 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 equal to two right angles; therefore CA is in the same straight F #2 Ax. G H K B E ABC; therefore the whole angle DBA is equal* to the whole FBC: and because the two sides AB, 126 Def. 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 #4. 1. #41. 1. 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 #6 Ax. one another; therefore the parallelogram BL is equal to the square GB. In the same manner, by drawing AE, BK, it can be demonstrated, that the parallelogram CL is equal to the square HC: therefore the whole square BDEC is equalf 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; 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. #2 Ax. 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. #11. 1. 13. 1. +2 Ax. #47. 1. +Const. D From the point A draw* AD at right angles to AC, and maket AD equal to BA, and join D, C. 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: there +1 Ax. +Const. *8. 1. tl Ax. fore 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 at right angle; therefore also BAC is a† right angle. Therefore, if the square, &c. Q. E. D. This, like all converse propositions which hold, may be proved by the reductio ad absurdum method of demonstration; that is, by showing the absurdity of denying the truth affirmed. |