PROP. XXIX. (THEOREM.)—If a straight line (EF), fall upon two parallel straight lines (AB, CD) it makes the alternate angles (AGH,GHD) equal to one another; the exterior angle (E GB) equal to the interior and opposite angle (GHD) upon the same side of the straight line (EF); and the two interior angles (BGII, GHD) upon the same side of it together equal to two right angles. A E For, if the alternate angles AGH, GHD be not equal, one of them must be greater than the other. Let the angle AGH be greater than the angle GHD. To each of these unequals add the angle BGH. Then the two angles AGH, BGH, are greater (Ax. 4) than the two angles BGH, GHD. But the two angles AGH, BGH, are equal (I. 13) to two right angles. Therefore the two angles BGH, GHD, are less than two right angles. But those straight lines, which with another straight line falling upon them, make the two interior angles on the same side less than two right angles, will meet together (Ax. 12) if continually produced. Therefore the two straight lines AB, CD, if produced far enough, will meet. But they never meet, since (Hyp.) they are parallel. Therefore the angle AGH is not unequal to the angle GHD; that is, the angle AGH is equal to the angle GHD. с H F Again, the angle AGH is equal (I. 15) to the angle EGB. Therefore the angle EGB is equal (4x. 1) to the angle GHD. Lastly, to each of these equals add the angle BGH. Then the two angles EGB, BGH, are equal (Ax. 2) to the two angles BGH, GHD. But the two angles EG B, BGH, are equal (I. 13) to two right angles. Therefore also the two angles BGH, GHD are equal (Ax. 1) to two right angles. Wherefore, if a straight line, &c. Q. E. D. PROP. XXX. (THEOREM.)-Straight lines (AB, CD) which are parallel to the same straight line (EF) are parallel to each other." Draw the straight line GHK cutting the three straight lines AB, EF, and CD. Because the straight line GHK cuts the parallel straight lines AB, EF, the angle AGH is equal (I. 29) to the alternate angle GHF. Again, because the straight line GHK cuts the parallel straight lines EF, CD, the ex E terior angle GHF is equal (I. 29) to the PROP. XXXI. (PROBLEM.)-To draw through a given point (A) a straight line parallel to a given straight line (BC). In the straight line BC take any point Ď, and join ÁD. At the point A in the straight line AD, make (I. 23) the angle DAE equal to the angle ADC. Produce the straight line EA to F. Then EF is parallel to BC. E B D A F C Because the straight line AD meets the two straight lines EF, BC, and makes the alternate angles EAD, ADC, equal to one another; therefore EF is parallel (I. 27) to BC. Wherefore, through the given point A, a straight line EAF has been drawn parallel to the given straight line BC. Q. E. F. PROP. XXXII. (THEOREM.)-If a side (BC) of any triangle (ABC) be produced (to D), the exterior angle (ACD) is equal to the two interior and opposite angles (CAB, ABC); and the three interior angles (ABC, BCA, CAB) of every triangle are together equal to two right angles. Through the point C draw the straight line (I. 31) CE parallel to the side BA. A E Because CE is parallel to BA, and AC meets them, the angle ACE is equal (I. 29) to the alternate angle BAC. Again, because CE is parallel to AB and BD falls upon them, the exterior angle ECD is equal (I. 29) to the interior and opposite angle ABC. But the angle ACE was shown to be equal to the angle BAC. Therefore the whole exterior angle ACD is equal (Ax. 2) B to the two interior and opposite angles CAB, ABC. To each of these equals, add the angle ACB. Therefore the two angles ACD, ACB are equal (Ax. 2) to the three angles CAB, ABC, ACB. But the two angles ACD, ACB are equal (I. 13) to two right angles. Therefore also the three angles CAB, ABC, ACB are equal (Ax. 1) to two right angles. Wherefore, if a side of any triangle be produced, &c. Q. E. D. D Cor. 1.-All the interior angles (ABC, BCD, &c.) of any rectilineal figure (ABCDE) together with four right angles are equal to twice as many right angles as the figure has sides. Divide the rectilineal figure ABCDE into as many triangles as the figure has sides, by drawing straight lines from a point F within the figure to each of its angles. E A D Because the three interior angles of a triangle are equal (I. 32) to two right angles, and there are as many triangles in the figure as it has sides, all the angles of these triangles are equal to twice as many right angles as the figure has sides. But all the angles of these triangles are equal to the interior angles of the figure, viz., ABC, BCD, &c., together with the angles at the point F, which are equal (I. 15, Cor. 2) to four right angles. Therefore all the angles of these triangles are equal (4x. 1) to the interior angles of the figure together with four right angles. But it has been proved that all the angles of these triangles are equal to twice as many right angles as the figure has sides. Therefore all the angles of the figure together with four right angles are equal to twice as many right angles as the figure has sides. Cor. 2.-All the exterior angles of any rectilineal figure, made by producing the sides successively in the same direction, are together equal to four right angles. Because the interior angle ABC, and its adjacent exterior angle ABD, are (I. 13) together equal to two right angles; therefore all the interior angles, together with all the exterior angles of the figure, are equal to twice as many right angles as the figure has sides. But it has been proved by the foregoing corollary, that all the interior angles together with four right angles are equal to twice as many right angles as the figure has sides. Therefore all the interior angles together with all the exterior angles are equal (Ax. 1) to all the interior angles and four right angles. Take from these equals all the interior angles. Therefore all the exterior angles of the figure are equal (Ax. 3) to four right angles D. B PROP. XXXIII. (THEOREM.)-The straight lines (AC, BD) which join the extremities of two equal and parallel straight lines (AB, CD) towards the same parts, are also themselves equal and parallel. Α Because B Join BC. Because AB is parallel to CD, and BC meets them, the angle ABC is equal (I. 29) to the alternate angle BCD. AB is equal to CD, and BC common to the two triangles ABC, DCB; the two sides AB, BC, are equal to the two DC, CB, each to each. And the angle ABC was proved to be equal to the angle BCD. Therefore the base AC is equal (I. 4) to the base BD, and the triangle ABC to the D triangle BCD. Also, the remaining angles of the one are equal to the remaining angles of the other, each to each, viz., those to which the equal sides are opposite. Therefore the angle ACB is equal to the angle CBD. Because the straight line BC meets the two straight lines AC, BD, and makes the alternate angles ACB, CBD equal to one another. Therefore AC is (I. 27) parallel to BD; and AC was proved to be equal to BD. Therefore the straight lines which, &c. Q. E. D. DEFINITION XXXVI.—A parallelogram is a four-sided figure of which the opposite sides are parallel; and the diagonal is the straight line joining two of its opposite angles. PROP. XXXIV. (THEOREM).-The opposite sides and angles of a parallelogram (AD) are equal to one another, and the diagonal (BC) bisects it, that is, divides it into two equal parts. Because AB is parallel to CD, and BC meets them, the angle ABC is equal (I. 29) to the alternate angle BCD. Because AC is B parallel to BD, and BC meets them, the angle ACB is equal (I. 29) to the alternate angle CBD. Because in the two triangles ABC, CBD, the two angles ABC, BCA, in the one, are equal to the two angles BCD, CBD in the A other, each to each; and one side BC, adjacent to these equal angles, is common to the two triangles; therefore their other sides are cqual, each to each, and the third angle of the one is equal to the third angle of the other (I. 26): viz., the side AB to the side CD, the side AC to the side BD, and the angle BAC to the angle BDC Because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB. Therefore the whole angle ABD is equal (4x. 2) to the whole angle ACD; and the angle BAC has been proved to be equal to the angle BDC. Therefore the opposite sides and angles of a parallelogram are equal to one another. Also the diagonal BC bisects the parallelogram AD. Because in the two triangles ABC, DCB, AB is equal to CD, and BC common, the two sides AB, BC, are equal to the two sides DC, CB, each to each; and the angle ABC has been proved to be equal to the angle BCD; therefore the triangle ABC is equal (I. 4) to the triangle BCD. Wherefore the diagonal BC divides the parallelogram AD into two equal parts. Q. E. D. PROP. XXXV. (THEOREM).-Parallelograms (AC, BF) upon the same base (BC), and between the same parallels (AF, BC), are equal to one another. First, let the sides AD, DF, of the parallelograms AC, BF, oppo site to the base BC, be terminated in the same point D. Because each of the parallelograms AC, BF, is double (I. 34) of the triangle BDC; therefore the parallelogram AC is equal (Ax. 6) to the parallelogram BF. Next, let the sides AD, EF, opposite to the base BC, be not terminated in the same point. Because AC is a parallelogram, AD is equal (I. 34) to BC. For a similar reason, EF is equal to BC. Therefore AD is equal (Ax. 1) to EF; and DE is common to both. Wherefore the whole, or the remainder AE, is equal to the whole, or the remainder DF (Ax. 2 or 3); and AB is equal (I. 34) to DC. Because in the triangles E AB, FDC, the side FD is equal to the side EA, and the side DC to the side AB, and the exterior angle FDC is equal (I. 29) to the interior and opposite angle EAB Therefore the base FC is equal (I. 4) to the base EB, and the triangle FDC to the tri ingle EAB. From the trapezium ABCF, take the triangle FDC, and the remainder is the parallelo gram AC. From the same trapezium take the triangle EAB and the remainder is the parallelogram BF. But when equals are taken from equals, or from the same, the remainders (Ax. 3) are equal. Therefore the parallelogram AC is equal to the parallelogram BF. Therefore, parallelograms upon the same, &c. Q. E. D. PROP. XXXVI. (THEOREM.)-Parallelograms (AC, EG) upon equal bases (BC, FG), and between the same parallels (À H, BG), are equal to one another. A DE H Join BE, CH. Because BC is equal to FG (Hyp.), and FG to EH (1. 34), BC is equal to EH (Ax. 1). But BC and EH are parallel, and joined towards the same parts by the straight lines BE, CH. And straight lines which join the extremities of equal and parallel straight lines towards the same parts, are (I. 33) themselves equal and parallel. Therefore the straight lines BE, CH are both equal and parallel. Wherefore BH is a parallelogram (Def.36). Because the parallelograms AC, BH, are upon the same base BC, and between the same parallels BC, AH, the parallelogram AC is equal (I. 35) to the parallelogram BH. Because the parallelograms GE, HB are upon the same base EH, and between the same parallels GB, HE, the parallelogram EG is equal to the parallelogram BH. Therefore the parallelogram AC is equal (Ax. 1) to the parallelogram EG. Therefore, parallelograms upon equal bases, &c. Q. E. D. B F G PROP. XXXVII. (THEOREM.)-Triangles (ABC, DBC) upon the same base (BC) and between the same parallels (AD, BC), are equal to one another. Produce AD both ways to the points E and F. Through B draw BE parallel to CA (I. 31), and through C draw CF parallel to BD. Then each of the figures EC, BF, is a parallelogram (Def. 36). E B A D The parallelograms EC, BF, are equal (I. 35), because they are upon the same base BC, and between the same parallels BC, EF. The triangle ABC is half of the parallelogram EC (I. 34), because the diagonal AB bisects it. Also, the triangle DBC is half of the parallelogram BF, because the diagonal DC bisects it. But the halves of equal things are equal (Ax. 7). Therefore the triangle ABC is equal to the triangle DBC. Wherefore, triangles, &c. Q. E. D. PROP. XXXVIII. (THEOREM.)—Triangies (ABC, DEF) upon equal bases (BC, EF), and between the same parallels (BF, AD), are equal to one another. Produce AD both ways to the points G and H. Through B draw ! |