from these equals take away all the int. s, all the ext. s= 4 rt. S. PROP. XXXII. THEOR. 33. 1 Eu. 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 and || str. lines, joined towards the same parts by the str. lines AC, BD: then shall AC and BD be = and to each PROP. XXXIII. THEOR. 34. 1 Eu. 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. 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 the opp. sides and m ", of which BC is a diam. one another; and BC bisects the figure. Prop. 4. Prop. 26. Prop. 28. Prop. 28 in ▲s ABC, CBD, ../s ABC, BCA = Zs BCD, CBD, ea. to ea. and adjacent side BC common to both As, JAB, AC Prop. 25. = CD, BD, ea. to ea. and :: ABC Ax. 2. Prop. 4. =/ BCD, ▲ CBD = ▲ ACB, ..whole ABD whole ACD; and it has been proved ▲ BAC = ▲ BDC. ..the opp. sides ands of .. diam. BC divides them ACDB into two = parts. Wherefore the opposite sides and angles, &c. PROP. XXXIV. THEOR. 35. 1 Eu. Parallelograms upon the same base, and between the same parallels, are equal to one another. Let the C ms ABCD, EBCF, be upon the same base BC, and between the same ||s, AF, BC; then Om ABCD = "EBCF. 2. If the sides AD, EF, be not terminated in the same point (figs. 2 and 3), Prop. 33. Ax. 6. Take the ext. FDC int. EAB, base EB = base FC, EAB from the trapezium ABCF; and from the same trapezium take the Ax. 3. FDC; then the remainders are =; that is, m ABCD = EBCF. Wherefore parallelograms, &c. COR.-Triangles upon the same base, and between the same parallels, are equal to one another. For if the diameters AC, FB, be drawn (figs. 2 and 3), the As ABC, FBC, Ax. 7. Hyp. Prop. 33. Ax. 1. Prop. 32. Prop. 34. Parallelograms upon equal bases, and between the same parallels, are equal to one another. Let ABCD, EFGH, be ms upon = bases BC, FG, and between the same ||s, AH, BG; then Join BE, CH, BC= EH = FG, BC= EH; str. lines EB, HC, join the extremities of = and || str. lines, they are themselves = and : that is, since they are on the same base BC and be tween the same ||s. For the same reason, EFGH = 7 EBCH, |