to the other angles b, cach to each, to which the equal fides are Book I. 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. T B HE opposite sides and angles of parallelograms are , equal to one another, and the diaineter bisects them, that is, divides them in two equal parts. N. B. A parallelogram is a four sided figure, of which the opposite lades are parallel ; and the diameter is the fraight 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 bisects it. Because AB is parallel to CD, A and BC meets them, the alter. Date angles ABC, BCD are e. qual 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 nave two angles ABC, BCA in one, equal to iwo angles BCD, CBD in the other, each to each, and one Gde BC common to the two triangles, which is adjacent to their equal angles; there. fore their other sides shall be equal, each to each, and the bird angle of the one to the third angle of the other b, viz. b 26. s. the Gde AB to the Gde 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 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 Book I. is equal to the angle BCD; therefore the triangle ABC is emqual c to the triangle BCD, and the diameter BC divides the C 4.1. parallelogram ACDB into two equal parts. Q. E. D. P4 Sec N. ARALLELOGRAMS upon the same base and between the same parallels, are equal to one another, See the ad Let the parallelograms ABCD, EBCF be upon the same and 3d fiu base BC, and between the same parallels AF, BC; the parallegures. logram ABCD shall be equal to the parallelogram EBCF. If the sides AD, DF of the parallelograms ABCD, DBCF oppolite A D F to the bafe BC be terminated in the fame point D; it is plain that each R. 34. I. of the parallelograms is double of the triangle BDC; and they are there- eB caufe ABCD is a parallelogram, AD is equal a to BC ; for the b 1. Ax. fame reason EF is equal to BC ; wherefore AD is equal b to EF; and DE is common; therefore the whole, or the remain< 2. or 3. der, AE is equal to the whole, or the remainder DF; AB al so is equal to DC; and the two EA, AB are therefore equal to А D E F A E D DF Ax. d 29. I. e 4.1. B C B С the two FD, DC, each to each; and the exterior argle FDC is equal 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 therefore are equal', that is, the parallelogram ABCD is equal to the parallelogram EBCF. Therefore parallelograms upon the fame bale, &c. Q. E. D. PROPE i 3. Ax. book I. PRO P. XXXVI. . THE OR. ARALLELOGRAMS upon equal bases, and between the fame parallels, are equal to one another. Let ABCD, EFGH be A DE H parallelograms upon e qual bases BC, FG, and between the same parallels AH, BG ; the paral. lelogram ABCD is equal to EFGH. Join BE, CH; and because BC is equal to B C F G FG, and FG to a EH, BC is equal to EH; and they are pa• a 34. I. sallels, and joined towards the same 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's; 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. I. 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. RIANGLES 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 AD, BC : The triangle ABC E A D F is equal to the triangle DBC Produce AD both ways to the points E, F, and thro' Buraw BE para lel to CA; 31. I. and thro' c draw CF paral. B Id to BD: 'Therefore each of the figures EBCA, DBCF is a parallelogram; and EBCA is equal 6 to DBCF, because they are upon the same base BC, and b 35. I. between the same parallels BC, EF, and the triangle ABC is the Book I. the half of the parallelogram EBCA, because the diameter AB bisects c it; and the triangle DBC is the half of the paralleloC 34. I. gram DBCF, because the diameter DC bisects it : But the d 7. Ax, halves of equal things are equal d ; therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. PRO P. XXXVIII. THEOR. TRI "RIANGLES upon equal bases, and between the same parallels, are equal to one another. 31. I. b 36. 1. 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. 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 A D H the figures GBCA, DEFH is a paralle logram ; and they are equal tobone another, because they are upon equal bases BC, EF,and between the fame parallels B СЕ 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 half of the parallelogram DEFH, because the diameter DF bisects it: But the halves of equal things are equal", therefore the triangle ABC is equal to the triangle DEF; Wherefore triangles, &c. Q. E. D. C 34. I. d 7. Ax. QUAL triangles upon the fame base, and upon the E lame 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 point A draw a AE parallel to BC, and join EC : The tri. angle a 31. I. ngle ABC is equal to the triangle EBC, because it is upon Book 1, he samebale BC, and between the same A D b 37. !, sarallels BC, AE But the triangle ABC is equal to the triangle BDC ; E herefore also the triangle BDC is equal to the triangle EBC, the greater o the leis, which is impossible : There. ore AE is not parallel to BC. In the ame manner, it can be demonstrated B hat no other line but AD is parallel to BC; AD is therefore parallel to it. Wherefore equal triangles upon, &c. Q. E D. PRO P. XL. THE O R. QUAL triangles upon equal bases, in the same straight line, and towards the fame parts, are between the same parallels. Let the equal triangles ABC, DEF be upon equal bases BC, EF, in the same straight line BF, and towards the A D G a 31. To rallel to BF, and join GF: The triangle ABC is equaliB с Е to the triangle GEF, because they are upon equal bases BC, EF, and between the same parallels BF, AG: But the triangle ABC is equal to the triangle DEF; therefore also the triangle DEF is equal to the triangle GEF, the greater to the less, which is impossible: Therefore AG is not parallel to BF : And in the same manner it can be demonstrated 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 bale, and between the fame parallels ; the parallelogram shall be double of the triangle. Lies |