Book I. 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 2 2. Cor. F, which is the common vertex of the triangles; that is a, 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. 15. 1. A 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 b 13. 1. ABD, is equal to two right angles; therefore all the interior, together with all the exterior angles of the figure, are equal to twice as many right angles as there are sides of the figure : D that is, by the foregoing corollary, they are equal to all the interior angles of the figure, to C B gether with four right angles; therefore all the exterior angles are equal to four right angles. 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 AB, CD be equal and pa- A rallel straight lines, and joined towards the same parts by the straight lines AC, BD; AC, BD are also equal and parallel. B Join BC; and because AB is parallel to CD, and BC meets C D b 4. 1. a 29.1. them, the alternate angles ABC, BCD are equal a; 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; and the angle ABC is equal to the angle BCD; therefore the base AC is equal to the base BD, and the triangle ABC to the triangle BCD, and the other angles to the other angles b, each to each, to which the equal sides are opposite: therefore the angle ACB is equal to the angle CBD; and because the straight Book I. line BC meets the two straight lines AC, BD, and makes the alternate angles ACB, CBD equal to one another, AC is parallel to BD; and it was shown to be equal to it. Therefore, straight c 27. 1. 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 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 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, A and BC meets them, the alternate angles ABC, BCD are equal 4 to one another; and because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal a to one C B D a 29. 1. another; wherefore the two 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 b, viz. the b 26. 1. 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 Book I. equal to the angle BCD; therefore the triangle ABC is equal ✔ to the triangle BCD, and the diameter BC divides the parallelo с 4. 1. See N. See the gram 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 be upon the same base 2d and 3d BC, and between the same parallels AF, BC; the parallelogram figures. ABCD shall be equal to the parallelogram EBCF. If the sides AD, DF of the parallelograms ABCD, DBCF opposite to A the base BC be terminated in the same point D, it is plain that each of the a 34. 1. parallelograms is double a of the triangle BDC; and they are therefore equal to one another. D F b 1. Ax. But, if the sides AD, EF, opposite B 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 a to BC; for the same reason, EF is equal to BC; wherefore AD is equal b to EF, and DE is common; therefore the whole, or the remainc 2. or 3. der AE, is equal to the whole, or the remainder DF; AB also Ax. is equal to DC; and the two EA, AB are therefore equal to the C A D E FA EDF 4. 1. B C B C two FD, DC, each to each; and the exterior angle FDC is d 19. 1. equal d to the interior EAB; therefore the base EB is equal to the base FC, and the triangle EAB equal to the triangle FDC; take the triangle FDC from the trapezium ABCF, and from the same trapezium take the triangle EAB; the remainders theref3. Ax. fore are equal f, 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 A parallelograms upon equal bases BC, FG, and between the same parallels AH, BG; the parallelogram ABCD is equal to EFGH. DE C F H G Book I. Join BE, CH; and be- B cause BC is equal to FG, and FG to a EH, BC is equal to EH; a 34 1. 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 b; therefore EB, CH are both equal and pa- b 33. 1. rallel, and EBCH is a parallelogram; and it is equale to ABCD, c 35. 1. 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. Q. E. D. 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 E AD F W a 31. 1. C Produce AD both ways to the points E, F, and through B draw a BE parallel to CA; and through C draw CF parallel to BD: therefore each of the figures EBCA, DBCF is a parallelogram; and EBCA is equal to DBCF, because b 35. 1. they are upon the same base BC, and between the same parallels BC, EF; and the triangle ABC is the half of the parallelo F B Book I. gram EBCA, because the diameter AB bisects cit; and the triangle DBC is the half of the parallelogram DBCF, because the diameter DC bisects it: but the halves of equal things are d 7. Ax. equal d; therefore the triangle ABC is equal to the triangle DBC. Wherefore, triangles, &c. Q. E. D. c 34. 1. 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 triangle DEF. Produce AD both ways to the points G, H, and through B a 31. 1. draw BG parallel a to CA, and through F draw FH parallel to ED: then each of G the figures GBCA, DEFH is a parallel ogram, and they b 36. 1. are equal b to one another, because they are upon equal bases BC, EF, and be tween the same pa c 34.1. rallels BF, GH; and the triangle ABC is the half of the parallelogram GBCĄ, because the diameter AB bisects it; and the triangle DEF is the half of the parallelogram DEFH, because the diameter DF bisects it: but the halves of equal things are d 7. Ax. equal d; therefore the triangle ABC is equal to the triangle DEF. Wherefore, triangles, &c. Q. E. D. 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 the 31. 1. point A draw a AE parallel to BC, and join EC: the triangle |