Book T. to each of its angles. And, by the preceeding proposition, all the angles of these triangles are equal to twice as many right angles as there are triangles, that is, as there are fides of the figure; and the fame 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, 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 fides. a. 2. Cor. 15. 1. Cor. 2. All the exterior angles of any rectilineal figure, are together equal to four right angles. b 13. 1. Because every interior angle ABC, with its adjacent exterior ABD, is equal to two right angles; therefore all the interior, together with all the exterior angles of the figure, are equal is, by the foregoing corollary, D they are equal to all the inte rior angles of the figure, toge ther with four right angles; therefore all the exterior angles are equal to four right angles. PROP. XXXIII. THEOR. T HE straight lines which join the extremities of two the alternate angles ABC, BCD 2 29. I. b 4. 1. are equal; and because AB is equal to CD, and BC common to the two triangles ABC, DCB, the two fides AB, BC are equal to the two DC, CB; and the angle ABC is equal to the angle ECD; therefore the base AC is equal to the base BD, and the triangle ABC to the triangle BCD, and the other angles to to the other angles b, each to each, to which the equal fides are Book J. opposite; therefore the angle ACB is equal to the angle CBD; and because the straight line BC meets the two straight lines b 4. 1. AC, BD, and makes the alternate angles ACB, CBD equal to one another, AC is parallel to BD; and it was shown to be c 27. 1, equal to it. Therefore straight lines, &c. Q. E. D. PROP. XXXIV. THEOR. THE oppofite fides 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 fided figure, of which the opposite fides 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 an other; and the diameter BC bifects it. Because AB is parallel to CD, A and BC meets them, the alter nate angles ABC, BCD are e qual to one another; and because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD, are equal D to one 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 fide BC common to the two triangles, which is adjacent to their equal angles; therefore their other fides shall be equal, each to each, and the third angle of the one to the third angle of the other b, viz. b 26, Z. the fide AB to the fide 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 C4 39 Book I. is equal to the angle BCD; therefore the triangle ABC is e☑ qual to the triangle BCD, and the diameter BC divides the parallelogram ACDB into two equal parts. Q. E. D. C 4. I. Sce N. See the 2d and 3d figures. a 34. I. P PROP. XXXV. THEOR. ARALLELOGRAMS upon the fame base and between the fame parallels, are equal to one another. Let the parallelograms ABCD, EBCF, be upon the fame bafe BC, and between the same parallels AF, BC; the parallelogram ABCD shall be equal to the parallelogram EBCF. If the fides AD, DF of the pa- to the bafe BC be terminated in the D F c 2. or 3. Ax. But, if the fides AD, EF, oppofite B to the bafe BC of the parallelograms ABCD, EBCF, be not terminated in the same point; then, because ABCD is a parallelogram, AD is equal to BC; for the b 1. Ax. fame reason EF is equal to BC; wherefore AD is equal to EF; and DE is common; therefore the whole, or the remainder, AE is equal to the whole, or the remainder DF; AB alfo is equal to DC; and the two EA, AB are therefore equal to C the two FD, DC, each to each; and the exterior angle FDC is 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 fame trapezium take the triangle EAB; the remainders therefore are equal f, that is, the parallelogram ABCD is equal to the parallelogram EBCF. Therefore parallelograms upon the fame bafe, &c. Q. E. D. d 29. 1. € 4 I. f 3. Ax. PROP. 15 P PROP. XXXVI. THEOR. ARALLELOGRAMS upon equal bases and between the Let ABCD, EFGH be A parallelograms upon e-qual bases BC, FG, and between the fame parallels AH, BG; the parallelogram ABCD is equal to EFGH. Join BE, CH; and B DE H CF G Book I. because BC is equal to FG, and FG to EH, BC is equal to EH; and they are pa-a 34. I. rallels, and joined towards the fame parts by the straight lines BE, CH: But straight lines which join equal and parallel straight lines towards the fame parts, are themselves equal and parallel; therefore EB, CH are both equal and parallel, and b 33.. EBCH is a parallelogram; and it is equal to ABCD, because 35. 1. it is upon the fame base BC, and between the fame parallels BC, AD: For the like reason, the parallelogram EFGH is equal to the fame EBCH: Therefore also the parallelogram ABCD is equal to EFGH. Wherefore parallelograms, &c. Q. E. D. PROP. XXXVII. THEOR.. RIANGLES upon TR the fame base, and between the same parallels, are equal to one another. Let the triangles ABC, DBC be upon the fame base BC and between the fame parallels AD, BC: The triangle ABC is equal to the triangle DBC. Produce AD both ways to the points E, F, and thro' B draw BE parallel to CA; and thro' C draw CF paral B C lel to BD: Therefore each of the figures EBCA, DBCF is a parallelogram; and EBCA is a 31. Σα equal to DBCF, because they are upon the fame base BC, and b 35. 1. between the same parallels BC, EF; and the triangle ABC is the C 34. I. Book 1. the half of the parallelogram EBCA, because the diameter AB bisects it; 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; therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. Q. E. D. d 7. Ax. TR PROP. XXXVIII. THEO R. RIANGLES upon equal bases, and between the fame parallels, are equal to one another. Let the triangles ABC, DEF be upon equal bases BC, EF, and between the fame parallels BF, AD: The triangle ABC is equal to the triangle DEF. a 31. 1. b 36. 1. C 34. I. d y. Ax. a 31. 1. Produce AD both ways to the points G, H, and through B draw BG parallel to CA, and through F draw FH parallel to ED: Then each of the figures GBCA, G DEFH is a paralle logram; and they other, because they B ADH BF, GH; and the triangle ABC is the half of the parallelogram GBCA, 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 equald; therefore the triangle ABC is equal to the triangle DEF. Where. fore triangles, &c. Q. E. D. PROP. XXXIX. THEOR. EQUAL triangles upon the fame base, and upon the fame fide of it, are between the same parallels. Let the equal triangles ABC, DBC be upon the fame base BC, and upon the same side of it; they are between the fame parallels. Join AD; AD is parallel to BC; for, if it is not, through the point A draw AE parallel to BC, and join EC: The triangle |