(i. 31) ae parallel to bc, and join e c. The triangle a bc is equal (i. 37) to d the triangle e b c, because it is upon the same base b c, and between the same parallels b c, a e. But the triangle a bc 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, b it can be demonstrated, that no other line but a d is parallel to bc; ad is therefore parallel to it. Wherefore equal triangles, &c. Q. E. D. 09 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 a b c, def be upon equal bases b c, ef, in the same straight line bf, and towards the a d same parts; they are between the same parallels. 8 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 d e f is equal to the triangle gef, the greater to the less, which is impossible: Therefore a g 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 parallelogramn shall be double of the triangle. LET the parallelogram a bed and the triangle ebc be upon the same base bc, and between the same parallels b c, a e; a d e the parallelogram abcd is double of the triangle e b c 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 b c, ae. But the parallelogram abcd is double (i. 34) of the triangle a b c, because the diameter ac divides it into two equal parts; wherefore a b c d b 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 a b c 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 b c, and have one of its angles equal to d. Bisect (i. 10) b c 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) ag parallel to es, and a f through c draw cg parallel to ef: therefore fecg is a parallelogram. And because be is equal to ec, the triangle a be is likewise equal (i. 38) to the triangle a ec, since they d are upon equal bases be ec, and between the same parallels, bc, a g; therefore the triangle a bc is double of the triangle bo e 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 parallelogram fecg is equal to the triangle a bc, 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 bc, having one of its angles cef equal to the given angle d. Which was to be done. aec. Nu 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 pa- 2 2 d rallelograms about a c, that is, through which a c passes, and bk, kd, the other £. parallelograms which make up the whole figure a bed, which are therefore called the complements. The complement bk is equal to the complement k d. Because a bcd is a parallelogram, and ac its diameter, the triangle a b c is equal b (i. 34) to the triangle adc: and, because ekha is a parallelogram, the diameter of which is a k, the triangle a e k is equal to the triangle a hk: for the same reason, the triangle kgc is equal to the triangle kfc. Then, because the triangle aek is equal to the triangle a hk, and the triangle kgc to kfc; the triangle a ek, together with the triangle kgc is equal to the triangle a hk together with the triangle kfc. But the whole triangle a b c is equal to the whole a dc; therefore the remaining complement bk is equal to the remaining complement kd. 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 a b 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 a b a parallelogram equal to the triangle c, and having an angle equal to d. Make (i. 42) the parallelogram befg equal to the triangle c, and having the angle e bg equal to the angle d, so that be be in the same straight line with a b, and produce f fg to h; and through a draw (i. 31) a h parallel to bg or ef, and join h b. Then, because the straight d line hf falls upon the pa m rallels a h, ef, the angles с a hf, hfe, are together equal (i. 29) to two right angles; wherefore the an b a 1 gles bhf, h fe, are less 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 e a or fh, and produce ha, gb to the points 1, m: then hlkf is a parallelogram, of which the diameter is hk, and ag, me are the parallelograms about h k ; and lb, bf are the complements : therefore lb is equal (i. 43) to bf; but bf is equal to the triangle c; wherefore lb is equal to the triangle c; and because the angle gbe is equal (i. 15) to the angle a bm, and likewise to the angle d; the angle a bm is equal to the angle d. Therefore the parallelogram lb 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 d b, 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; а d f 8 g therefore the angles fkh, khg, are equal to the angles kh g ghm; but fkh, khg are equal (i. 29) to two right angles; therefore also k hg, ghm, are equal to two right an gles; and because at the b k h point h in the straight line gh, the two straight lines k h, h m 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, hgl, are equal (i. 29) to two right angles ; wherefore also the angles hgf, hg1 are equal to two right angles, and fg is therefore in the same straight line with gl; and because kf is parallel to hg, and hg to ml;kfis parallel (i. 30) to ml; and km, fl are parallels ; wherefore kflm 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 kfim 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. с m 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 a b be the given straight line; it is required to describe a square ز upon a b e From the point a draw (i. 11) a c at right angles to a b; and make (i. 3) ad 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, eb, are equal to one another, and the parallelogram a deb is equilateral, likewise all its angles are right angles ; 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 a de is a right angle; but the opposite angles of parallelograms are equal (i. 34); therefore each of the opposite angles a be, bed is a right angle; wherefore the figure a deb is b 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. Which was to be done. COR. Hence every parallelogram that has one right angle has all its angles right angles. 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 a bc be a right-angled triangle having the right angle bac; the square described upon the side b c is equal to the squares described upon ba, a co On bc describe (i. 46) the square bdec, and on ba, ac the squares gb, hc; and through a draw (i. 31) a 1 parallel to b d or ce, and join ad. åd, fc. Then, because each of the angles bac, bag is a right angle (30 def.), the two straight lines a c, 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 fb a, each of them being a right angle, add to each the angle a b c, and the whole angle dba is equal (2 ax.) to the |