D D rec Through the point c draw CE paral A lel (1. 31.) to the straight line AB; and because AB is parallel to CE, and AC meets them, the alternate angles BAC, ACE are equal (1. 29.). Again, because AB is parallel to CE, and BD B C falls upon them, the exterior angle Ecd is equal 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 acp is equal to the two interior and opposite angles CAB, ABC; to these equals add the angle ACB, and the angles ACD, ACB are equal to the three angles CBA, BAC, ACB; but the angles ACD, ACB are equal (1. 13.) to two right angles; therefore also the angles CBA, BAC, ACB are equal to two right angles. Wherefore if a side of a triangle, &c. Q.E.D. CoR. 1. All the interior angles of any tilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides. For any rectilineal figure ABCDE can be divided into as many triangles as the figure B has sides, by drawing straight lines from a point F within the figure to each of its angles. And, by the preceding proposition, all the angles of these triangles are equal to twice as many right angles as there are triangles, that is, as there are sides of the figure; and the same angles are equal to the angles of the figure, together with the angles at the point F, which is the common vertex of the triangles ; that is (Cor. 2. 1. 15.), together with four right angles. 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, are together equal to four right angles. Because every interior angle ABC, with its adjacent exterior ABD, is equal (1. 13.) to two right angles ; therefore all the interior, together with all the exterior angles of the figure, are equal to twice as many с right angles as there are sides of the figure; DB that is, by the foregoing corollary, they are equal to all the interior angles of the figure, together with A A four right angles; therefore all the exterior angles are equal to four right angles. B PROP. XXXIII. THEOR. The straight lines which join the extremities of two equal and parallel straight lines, towards the same parts, are also themselves equal and parallel. Let A B, CD be equal and parallel straight lines, and joined towards the same parts by the straight A lines AC, BD; AC, BD are also equal and parallel. Join BC; and because AB is parallel to CD, and BC meets them, the alternate С angles ABC, BCD are equal (1. 29.); and because AB is equal to CD (Hyp.), and BC common to the two triangles ABC, DC B, the two sides AB, BC are equal to the two DC, CB; and the angle ABC is equal to the angle BCD; therefore the base ac is equal (1. 4.) to the base BD, and the triangle ABC to the triangle BCD, and the other angles to the other angles (1. 4.), each to each, to which the equal sides are opposite; therefore the angle acB is equal to the angle CBD; and because the straight line BC meets the two straight lines AC, BD, and makes the alternate angles ACB, CBD equal to one another, Ac is parallel (I. 27.) to BD; and it was shown to be equal to it. Therefore, straight lines, &c. Q.E.D. ; PROP. XXXIV. THEOR. The opposite sides and angles of parallelograms are equal to one another, and the diameter bisects them, that is, divides them in two equal parts. N. B. - A parallelogram is a four-sided figure, of which the opposite sides are parallel ; and the diameter is the straight line joining two of its opposite angles. Let ACDB be a parallelogram, of which Bc is a diameter the opposite sides and angles of the figure are equal to one another; and the diameter Bc bisects it. Because AB is parallel to cd, and BC meets them, the alternate angles ABC, BCD are equal A B (I. 29.) to one another; and because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal (1. 29.) to one another; wherefore the two С D triangles ABC, CBD have two angles ABC, BCA in one, equal to two angles BCD, CBD in the other, each to each, and one side bc common to the two triangles, which is adjacent to their equal angles; therefore their other sides shall be equal, each to each, and the third angle of the one to the third angle of the other (1. 26.), viz. the side AB to the side CD, and AC to BD, and the angle BAC equal to the angle BDC: And because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, the whole angle ABD is equal to the whole angle ACD: And the angle BAC has been shown to be equal to the angle BDC; therefore the opposite sides and angles of parallelograms are equal to one another ; also, their diameter bisects them; for AB being equal to CD, and BC common, the two AB, BC are equal to the two DC, CB, each to each; and the angle ABC is equal to the angle BCD; therefore the triangle ABC is equal (1. 4.) to the triangle BCD, and the diameter Bc divides the parallelogram ACDB into two equal parts. Q. E. D. PROP. XXXV. THEOR. Parallelograms upon the same base and between the same parallels, are equal to one another. Let the parallelograms ABCD, EBCF (see the 2d and 3d figures) be upon the same base BC, and between the same parallels AF, BC; the parallelogram ABCD shall be equal to the parallelogram EBCF. If the sides AD, DF of the parallelograms ABCD, DBCF opposite to the base bc be ter D minated in the same point D; it is plain that each of the parallelograms is double (L. 34.) of the triangle BDC; and they are therefore equal to one another. But, if the sides AD, EF, opposite to the base Bc of the parallelograms ABCD, EBCF, be not terminated in the same point; then, because ABCD is a parallelogram, Ad is equal (1. 34.) to A E A F B C A D E F D F BC; for the same reason EF is equal to BC; wherefore ad is equal (Ax. 1.) to EF; and DE is common ; therefore the whole, or the remainder, A E is equal (Ax. 2. or 3.) to the whole, or the remainder DF; A B also is equal to DC; and the two EA, AB are therefore equal to the two FD, DC, each to each; and the exterior angle FDC is equal (I. 29.) to the interior E A B, therefore the base EB is equal to the base FC, and the triangle EAB equal (1. 4.) to the triangle FDC; take the triangle Foc from the trapezium ABCF, and from the same trapezium take the triangle EA B; the remainders, therefore, are equal (Ax. 3.), that is, the parallelogram ABCD is equal to the parallelogram EBCF. Therefore, parallelograms upon the same base, &c. Q. E. D. PROP. XXXVI. THEOR. Parallelograms upon equal bases, and between the same parallels, are equal to one another. Let ABCD, EFGH be parallelograms DE H upon equal bases BC, FG, and between the same parallels AH, BG; the parallelogram ABCD is equal to EFGH. Join BE, CH; and because Bc is equal to FG, and Fg to (1. 34.) EH, BC is equal B F to EH; and they are parallels, and joined towards the same parts by the straight lines BE, CH: But straight lines which join equal and parallel straight lines towards the same parts, are themselves equal and parallel (1. 33.); therefore EB, CH are both equal and parallel, and EBC is a parallelogram; and it is equal (1. 35.) to ABCD, because it is upon the same base BC, and between the same parallels BC, AD: For the like reason, the parallelogram EFGH is equal to the same EBCH: Therefore also the parallelogram ABCD is equal to EFGH. Wherefore parallelograms, &c. A PROP. XXXVII. THEOR. Triangles upon the same base, and between the same parallels, are equal to one another. Let the triangles ABC, DBC be upon the same base BC and between the same parallels AD, BC: The E A D triangle ABC is equal to the triangle DBC. Produce ad both ways to the points E, F, and through B draw (1. 31.) BE parallel to CA; and through c draw CF F B parallel to BD: Therefore each of the figures EBCA, DBCF is a parallelogram ; and EBCA is equal (I. 35.) to DBCF, because they are upon the same base BC, and between the same parallels BC, EF; and the triangle ABC is the half of the parallelogram EBCA, because the diameter A B bisects it (I. 34.); and the triangle DBC is the half of the parallelogram DBCF, because the diameter 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. THEOR. Triangles upon equal bases, and between the same parallels, are equal to one another. Let the triangles ABC, DEF be upon equal bases BC, EF, and between the same parallels BF, AD: The triangle ABC is equal to the angle DEF. Produce ad both ways to the point G, H, and through B draw BG parallel (1. 31.) to cs, and through F draw FH parallel to ED: Then each of the figures G A D H GBCA, DEFH is a parallelogram ; and they are equal to (1. 36.) one another, because they are upon equal bases BC, EF, and between the same parallels BF, GH; and the triangle ABC is the half (1. 34.) of CE the parallelogram GBCA, because the diameter AB bisects it; and the triangle DEF is the half (1. 34.) of the parallelogram DEH, because the diameter DF bisects it : But the halves of equal things are equal (Ax. 7.); therefore the triangle ABC is equal to the triangle DEF. Wherefore triangles, &c. Q.E.D. B F PROP. XXXIX. THEOR. Equal triangles upon the same base, and upon the same side of it, are between the same parallels. Let the equal triangles ABC, DBC be upon the same base BC, and upon the same side of it; they are between the same parallels. Join AD: AD is parallel to BC; for, if it is not, through |