to the other angles b, each to each, to which the equal fides are Book I. oppofite; therefore the angle ACB is equal to the angle CBD; and because the ftraight line BC meets the two ftraight lines b 4. I. 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. THE HE oppofite fides and angles of parallelograms are equal to one another, and the diameter bifects them, that is, divides them in two equal parts. N. B. A parallelogram is a four fided figure, of which the oppofite fides are parallel; and the diameter is the Straight line joining two of its oppofite angles. Let ACDB be a parallelogram, of which BC is a diameter; the opposite fides and angles of the figure are equal to one an other; and the diameter BC bifects it. C B D a 29. I. Becaufe AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equal to one another; and becaufe AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equala to one another; wherefore the two triangles ABC, CBD nave 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 fhall be equal, each to each, and the third angle of the one to the third angle of the other b, 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 becaufe 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 oppofite fides and angles of parallelograms are equal to one another; alío, 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 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. See N. See the 2d P ARALLELOGRAMS upon the fame bafe and between the fame parallels, are equal to one another, Let the parallelograms ABCD, EBCF be upon the fame and 3d fi bafe BC, and between the fame parallels AF, BC; the parallelogram ABCD fhall be equal to the parallelogram EBCF. gures. 2. 34. I. If the fides AD, DF of the pa- to the bafe BC be terminated in the But, if the fides AD, EF, oppofite B to the base BC of the parallelograms ABCD, EBCF, be not terminated in the fame point; then, becaufe ABCD is a parallelogram, AD is equal a to BC, for the b I. Ax. fame reafon EF is equal to BC; wherefore AD is equal b to EF and DE is common; therefore the whole, or the remain2. or 3. der, 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 Ax. 29. I. € 4. I. f 3. Ax, с e the two FD, DC, each to each; and the exterior argle FDC is equal to the interior EAB; therefore the bafe EB is equal to the bafe 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, that is, the parallelogram ABCD is equal to the parallelogram EBCF. Therefore parallelograms upon the fame bale, &c. Q. E. D. PROP ARALLELOGRAMS upon equal bafes, and between the PA Let ABCD, EFGH be parallelograms upon e qual bafes BC, FG, and between the fame parallels AH, BG; the paral lelogram ABCD is equal to EFGH. A DE Η Book I. Join BE, CH; and because BC is equal to FG, and FG to a EH, BC is equal to EH; and they are pa- a 34. I. rallels, and joined towards the fame parts by the ftraight lines BE, CH: But ftraight lines which join equal and parallel ftraight lines towards the fame parts, are themselves equal and parallel; therefore EB, CH are both equal and parallel, and b 33. I. EBCH is a parallelogram; and it is equal to ABCD, because c 35. 1. it is upon the fame bafe BC, and between the fame parallels BC, AD: For the like reafon, the parallelogram EFGH is equal to the fame EBCH: Therefore alfo the parallelogram ABCD is equal to EFGH. Wherefore parallelograms, &c. Q. E. D. RIANGLES upon the fame bafe, and between the E A D F 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. kel to BD: Therefore each of the figures EBCA, DBCF is a parallelogram; and EBCA is equal to DBCF, because they are upon the fame base BC, and b 35. 1. between the fame parallels BC, EF; and the triangle ABC is the 42 Book I. € 34. I. the half of the parallelogram EBCA, becaufe the diameter AB bifects e it; and the triangle DBC is the half of the parallelogram DBCF, because the diameter DC bifects it: But the d 7. Ax. halves of equal things are equal; therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. Q. E. D. TR RIANGLES upon equal bases, and between the fame parallels, are equal to one another. Let the triangles ABC, DEF be upon equal bafes BC, EF, and between the fame parallels BF, AD: The triangle ABC is equal to the triangle DEF. Produce AD both ways to the points G, H, and through B draw BG parallel a to CA, and through F draw FH parallel to ED: Then each of G the figures GBCA, B A CE D H F the fame parallels E QUAL triangles upon the fame base, and upon the fame fide of it, are between the fame parallels. Let the equal triangles ABC, DBC be upon the fame bafe BC, and upon the fame fide of it; they 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 b A D E b 37. T angle ABC is equal to the triangle EBC, because it is upon Book I. the fame base BC, and between the fame parallels BC, AE: But the triangle ABC is equal to the triangle BDC; therefore alfo the triangle BDC is equal to the triangle EBC, the greater to the leis, which is impoffible: There. fore AE is not parallel to BC. In the fame manner, it can be demonstrated that no other line but AD is parallel to BC; AD is therefore parallel to it. Wherefore equal triangles upon, &c. Q. E D. PROP. XL.. B THEOR. QUAL triangles upon equal bafes, in the fame tween the fame parallels. Let the equal triangles ABC, DEF be upon equal bases BC, EF, in the fame ftraight line BF, and towards the fame parts; they are be tween the fame parallels. Join AD; AD is parallel to BC: For, if it is not, through A draw AG parallel to BF, and GF: The triangle ABC is equal B Fb 38. CE to the triangle GEF, because they are upon equal bases BC, EF, and between the fame parallels BF, AG But the triangle ABC is equal to the triangle DEF; therefore alfo the triangle DEF is equal to the triangle GEF, the greater to the lefs, which is impoffible: Therefore AG is not parallel to BF: And in the fame manner it can be demonftrated that there is no other parallel to it but AD; AD is therefore parallel to BF. Wherefore equal triangles, &c. Q. E. D. IF a parallelogram and triangle be upon the same base, and between the fame parallels; the parallelogram shall be double of the triangle. Let |