therefore all the interior angles together with all the exterior angles, are equal to all the interior angles and four right angles (Ax. 1); take from these equals all the interior angles, therefore all the exterior angles of the figure are equal to four right angles (Ax. 3). PROPOSITION 33.-Theorem. 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 AB, CD be equal and parallel straight lines, and joined towards the same parts by the straight lines AC, BD. Then AC, BD shall be equal and parallel. B D C Construction. Join BC. Demonstration. Then, because AB is parallel to CD, and BC meets them, therefore 1. The angle ABC is equal to the alternate angle BCD (I. 29); and because 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 2. The base AC is equal to the base BD (I. 4), and the triangle ABC to the triangle BCD, 4. AC is parallel to BD (I. 27), and AC was shown to be equal to BD. and the other angles to the other angles, each to each, to which the equal sides are opposite; therefore 3. 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, therefore Therefore, straight lines which, &c. Q.E.D. PROPOSITION 34.-Theorem. The opposite sides and angles of parallelograms are equal to one another, and the diameter bisects them, that is, divides them into two equal parts. Let ACDB be a parallelogram, of which BC is a diameter. Then the opposite sides and angles of the figure shall be equal to one another, and the diameter BC shall bisect it. C D Demonstration. Because AB is parallel to CD, and BC meets them, therefore 1. The angle ABC is equal to the alternate angle BCD (I. 29). And because AC is parallel to BD, and BC meets them, therefore 2. The angle ACB is equal to the alternate angle CBD (I. 29). Hence, in the two triangles ABC, CBD, because the two angles ABC, BCA in the one, are equal to the two angles BCD, CBD in the other, each to each; and one side BC, which is adjacent to their equal angles, common to the two triangles; therefore, their other sides are equal, each to each, and the third angle of the one to the third angle of the other (I. 26), namely, 3. The side AB is equal to the side CD, and AC to BD, and the angle BAC to the angle BDC. And because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, therefore 4. The whole angle ABD is equal to the whole angle ACD (Ax. 2), and the angle BAC has been shown to be equal to BDC; therefore, the opposite sides and angles of a parallelogram are equal to one another. Also, the diameter BC bisects it. For since AB is equal to CD and BC common, the two sides AB, BC are equal to the two DC, CB, each to each, and the angle ABC has been proved to be equal to the angle BCD; therefore 5. The triangle ABC is equal to the triangle BCD (I. 4), and the diameter BC divides the parallelogram ACDB into two equal parts. Q.E.D. PROPOSITION 35.-Theorem. Parallelograms upon the same base, and between the same parallels, are equal to one another. Let the parallelograms ABCD, EBCF be upon the same base BC, and between the same parallels AF, BC. B Then the parallelogram ABCD shall be equal to the parallelogram EBCF. and therefore E D W B Demonstration. If the sides AD, DF of the parallelograms ABCD, DBCF, opposite to the base BC, be terminated in the same point D, then it is plain that 1. Each of the parallelograms ABCD, DBCF is double of the triangle BDC (I. 34), 2. The parallelogram ABCD is equal to the parallelogram DBCF (Ax. 6). But if the sides AD, EF, opposite to the base BC, be not terminated in the same point; then, because ABCD is a parallelogram, therefore AD is equal to BC (I. 34); and for a similar reason, EF is equal to BC; wherefore 2. The whole or the remainder, AE, is equal to the whole, or the remainder DF (Ax. 2 or 3), and AB is equal to DC (I. 34); hence in the triangles EAB, FDC, because FD is equal to EA, and DC to AB, and the exterior angle FDC is equal to the interior and opposite angle EAB (I. 29), therefore 3. The base FC is equal to the base EB (I. 4), and the triangle FDC to the triangle EAB. From the trapezium ABCF take the triangle FDC, and from the same trapezium take the triangle EAB, and the remainders are equal (Ax. 3); therefore 4. The parallelogram ABCD is equal to the parallelogram EBCF. Therefore, parallelograms upon the same base, &c. Q.E.D. PROPOSITION 36.-Theorem. Parallelograms upon equal bases, and between the same parallels, are equal to one another. Let ABCD, EFGH be parallelograms upon equal bases BC, FG, and between the same parallels AH, BG. Then the parallelogram ABCD shall be equal to the parallelogram EFGH. B wherefore D Construction. Join BE, CH. Demonstration. Then, because BC is equal to FG (hyp.), and FG to EH (I. 34), therefore therefore F 1. BC is equal to EH (Ax. 1), and these lines are parallels, and joined towards the same parts by the straight lines BE, CH; but straight lines which join the extremities of equal and parallel straight lines towards the same parts, are themselves equal and parallel (I. 33); therefore 2. BE, CH are both equal and parallel ; For the same reason, 3. EBCH is a parallelogram (Def. A). And because the parallelograms ABCD, EBCH are upon the same base BC, and between the same parallels BC, AH; therefore 4. The parallelogram ABCD is equal to the parallelogram EBCH (I. 35). 5. The parallelogram EFGH is equal to the parallelogram EBCH; 6. The parallelogram ABCD is equal to the parallelogram EFGH (Ax. 1). Therefore, parallelograms upon equal, &c. Q.ED. PROPOSITION 37.-Theorem. 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. D |