(i. 31) ae parallel to bc, and join ec. The triangle a bc is equal (i. 37) to the triangle e bc, because it is upon the same base bc, and between the same parallels bc, ae. But the triangle abc is equal to the triangle bdc, therefore also the triangle bdc is equal to the triangle e bc, the greater to the less, which is impossible. Therefore ae is not parallel to bc. In the same manner, it can be demonstrated, that no b d other line but ad is parallel to bc; ad is therefore parallel to it. Wherefore equal triangles, &c. Q. E. D. PROPOSITION XL.—THEOREM. Equal triangles upon equal bases, in the same straight line and towards the same parts, are between the same parallels. LET the equal triangles abc, def be upon equal bases bc, ef, in the same straight line bf, and towards the a same parts; they are between the same parallels. Join ad; ad is parallel to bc. For, if it is not, through a draw (i. 31) ag parallel to bf, and join gf: the triangle abc b is equal (i. 38) to the triangle gef, because they are upon equal bases bc, ef, and between the same parallels bf, ag: but the triangle a b c is equal to the triangle def; therefore also the triangle def is equal to the triangle gef, the greater to the less, which is impossible: Therefore ag is not parallel to bf. And in the same manner it can be demonstrated that there is no other parallel to it but ad: ad is therefore parallel to bf. Wherefore equal triangles, &c. Q. E. D. PROPOSITION XLI.-THEOREM. If a parallelogram and triangle be upon the same base and between the same parallels, the parallelogram shall be double of the triangle. LET the parallelogram abcd and the triangle ebc be upon the same base bc, and between the same parallels bc, ae; the parallelogram abcd is double of the triangle e bc. Join ac; then the triangle abc is equal (i. 37) to the triangle ebc, because they are upon the same base bc, and between the same parallels bc, ae. But the parallelogram abcd is double (i. 34) of the triangle abc, because the diameter a c divides it into two equal parts; wherefore abcd is also double of the triangle e bc. Therefore, if a parallelogram, &c. Q. E. D. PROPOSITION XLII.-PROBLEM. To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. LET abc be the given triangle, and d the given rectilineal angle. It is required to describe a parallelogram that shall be equal to the given triangle a bc, and have one of its angles equal to d Bisect (i. 10) bc in e, join a e, and at the point e in the straight line ec make (i. 23) the angle cef equal to d; and through a draw (i. 31) a f g ag parallel to e c, and through c draw cg parallel to ef: therefore fecg is a parallelogram. And because be is equal to e c, the triangle a be is likewise equal (i. 38) to the triangle a ec, since they are upon equal bases be, ec, and between the same parallels, bc, ag; therefore the triangle abc is double of the triangle a e c. And the parallelogram fecg is likewise double (i. 41) of the triangle a ec, because it is upon the same base, and between the same parallels therefore the раrallelogram fecg is equal to the triangle abc, and it has one of its angles cef equal to the given angle d; wherefore there has been described a parallelogram fecg equal to a given triangle a b'c, having one of its angles cef equal to the given angle d. Which was to be done. b e PROPOSITION XLIII.-THEOREM. The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another. LET abcd be a parallelogram, of which the diameter is a c, and eh, fg, the parallelograms about a c, that is, through which ac passes, and bk, kd, the other parallelograms which make up the whole figure abcd, which are therefore called the complements. The complement bk is equal to the complement k d. Because abcd is a parallelogram, and ac its diameter, the triangle a bc is equal b (i. 34) to the triangle a dc: and, because : h d £ ekha is a parallelogram, the diameter of which is a k, the triangle a ek is equal to the triangle ahk for the same reason, the triangle kgc is equal to the triangle kfc. Then, because the triangle aek is equal to the triangle ahk, and the triangle kgc to kfc; the triangle aek, together with the triangle kg c is equal to the triangle ahk together with the triangle kfc. But the whole triangle a bc is equal to the whole a dc; therefore the remaining complement bk is equal to the remaining complement k d Wherefore the complements, &c. Q. E. D. PROPOSITION XLIV.-PROBLEM. 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. f d g e k Make (i. 42) 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 a b, and produce fg to h; and through a draw (i. 31) ah parallel to bg or ef, and join h b. Then, because the straight line hf falls upon the parallels ah, ef, the angles ahf, hfe, are together equal (i. 29) to two right angles; wherefore the angles bhf, hfe, are less h a 1 than two right angles. But straight lines which with another straight line make the interior angles upon the same side less than two right angles, do meet (12 ax.) if produced far enough: therefore hb, fe shall meet if produced; let them meet in k, and through k draw kl. parallel to ea or fh, and produce ha, gb to the points 1, m: then hlkf is a parallelogram, of which the diameter is hk, and a g, me are the parallelograms about hk; and 1b, bf are the complements: therefore 1b is equal (i. 43) to bf; but bf is equal to the triangle c; wherefore 1b is equal to the triangle c; and because the angle g be is equal (i. 15) to the angle a bm, and likewise to the angle d; the angle abm is equal to the angle d Therefore the parallelogram 1b is applied to the straight line a b, is equal to the triangle c, and has the angle a bm equal to the angle d. Which was to be done. PROPOSITION XLV.-PROBLEM. To describe a parallelogram equal to a given rectilineal 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 a b c d, and having an angle equal to e. Join db, and describe (i. 42) the parallelogram fh equal to the triangle adb, and having the angle hkf equal to the angle e; and to the straight line gh apply (i. 44) the parallelogram gm equal to the triangle dbc, having the angle ghm equal to the angle e; and because the angle e is equal to each of the angles fkh, ghm, the angle fk h is equal to ghm: add to each of these the angle khg; therefore the angles fkh, khg, are equal to the angles khg, ghm; but fkh, khg are equal (i. 29) 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 (i. 14) with hm; and because the straight line hg meets the parallels km, fg, the alternate angles mhg, hgf are equal (i. 29): add to each of these the angle hgl: therefore the angles mhg, hgl, are equal to the angles hgf, hgl. But the angles mhg, hg1, are equal (i. 29) to two right angles; wherefore also the angles hgf. hgl are equal to two right angles, and fg is therefore in the same straight line with gl; and because kf is parallel to h g, and h g to ml;kfis parallel (i. 30) to m 1; and km, fl are parallels; wherefore kilm is a parallelogram; and because the triangle a bd 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 k flm 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 (i. 44) to the given straight line a parallelogram equal to the first triangle a bd, and having an angle equal to the given angle. PROPOSITION XLVI.-PROBLEM. To describe a square upon a given straight line. LET ab be the given straight line; it is required to describe a square upon a b. From the point a draw (i. 11) a c at right angles to a b; and make (i. 3) a d equal to a b, and through the point d draw de parallel (i. 31) to a b, and through b draw be parallel to a d therefore a deb is a parallelogram: whence a b is equal (i. 34) to de, and a d to be but ba is equal to ad; therefore the C four straight lines ba, a d, de, e b, are equal to one another, and the parallelogram adeb is equilateral, likewise all its angles are right angles; a because the straight line ad meeting the parallels ab, de, the angles bad, ade are equal (i. 29) to two right angles: but bad is a right angle; therefore also ade is a right angle; but the opposite angles of parallelograms are equal (i. 34); 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 a b. the given straight line a b. Which was to be done. COR. Hence every parallelogram that has one right angle has all its angles right angles. a b PROPOSITION XLVII.—THEOREM. 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 is equal to the squares described upon ba, a c. On be describe (i. 46) the square bdec, and on ba, ac the squares gb, hc; and through a draw (i. 31) al parallel to bd, or ce, and join ad, fc. Then, because each of the angles bac, bag is a right angle (30 def.), the two straight lines ac, ag, upon the opposite sides of a b, make with it at the point a the adjacent angles equal to two right angles; therefore ca is in the same straight line (i. 14) with a g; for the same reason, a b and a h 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, and the whole angle dba is equal (2 ax.) to the |