a. 2. Cor. 15. I. Book I. to each of its angles. And, by the preceeding proposition, all Wthe 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 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. 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 b to two right angles; therefore all the interi- Ac TH "HE straight lines which join the extremities of two equal and parallel straight lines, towards the fame B Join BC; and because AB is pa- С D the alternate angles ABC, BCD are equal”; and becaule AB is equal to CD, and BC common to the two triangles ABC, DCB, the two sides AB, BC are e. qual 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 2 29. I. b 4. I. to the other angles b, each to each, to which the equal sides are Book T. 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. TH 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, A B and BC meets them, the alternate angles ABC, BCD are equal a to one another ; and be a 29. 1, cause AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD, are equal * С D to one another ; wherefore the two triangles ABC, CBD mave 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; there. fore their other fides shall be equal, each to each, and the third angle of the one to the third angle of the other , viz. b 26, 1. 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 Gides and angles of parallelograms are e. qual to one another ; also, their diameter bifects 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 Book I. is equal to the angle BCD; therefore the triangle ABC is emqual to the triangle BCD, and the diameter BC divides the parallelogram ACDB into two equal parts. Q. E. D. C4. 1. See N P4 ARALLELOGRAMS upon the same base and between the same parallels, are equal to one another. & 34. 1. See the ad Let the parallelograms ABCD, EBCF, be upon the same and 3d fic base BC, and between the fame parallels AF, BC; the parallegures. logram ABCD shall be equal to the parallelogram EBCF. If the fides AD, DF of the pa- But, if the sides AD, EF, opposite B С F A E D F b 1. Ax. B B C the two FD, DC, each to each; and the exterior angle FDC d 29. 1. is equal to the interior EAB; therefore the base EB is equal to the base FC, and the triangle EAB equal o to the triangle FDC; take the triangle FDC from the trapezium ABCF, and from the same trapezium take the triangle EAB; the remainf 3. Ax. ' ders therefore are equalf, that is, the parallelogram ABCD is equal to the parallelogram EBCF. Therefore parallelograms upon the fame base, &c. Q. E. D. PROP. Book I. PROP. XXXVI. THEOR. P ARALLELOGRAMS upon equal bases and between the Let ABCDEFGH be DE H Η parallelograms 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 B C F G FG, and FG to * EH, BC is equal to EH; and they are pa- a 34. I. rallels, and joined towards the same parts by the straight lines BE, CH: But straight lines which join equal and parallel ftraight lines towards the fame parts, are themselves equal and parallel b; therefore EB, CH are both equal and parallel, and 33. 8. EBCH is a parallelogram; and it is equal to ABCD, because C 35. 1. it is upon the same base BC, and between the fame 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. TRE TRIANGLES upon the fame base, and between the fame parallels, are equal to one another. Let the triangles ABC, DBC be upon the same base BC and between the fame parallels E A D F Produce AD both ways to the points E, F, and thro' B draw - BE parallel to CA; a 31. d, and thro' C draw CF paral B lel to BD: Therefore each С of the figures EBCA, DBCF is a parallelogram ; and EBCA is equal to DBCF, because they are upon the same base BC, and b 35. I. between the fame parallels BC, EF, and the triangle ABC is the Book I. the half of the parallelogram EBCA, because the diameter AB bifects it; and the triangle DBC is the half of the parallelo gram DBCF, because the diameter DC bifects it : But the d halves of equal things are equal d; therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. Q. E. D. C 34. I. 7. Ax. a 31.1. RIANGLEs upon equal bases, and between the same parallels, are equal to one another. Produce AD both ways to the points G, H, and through B H B CE F 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 halfc 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. b36. I. C 34. 1. d y. Ax. PRO P. XXXIX. THE O R. lame side of it, are between the same 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 tri angle a 31. I |