angle ACB is equal to the angle CBD; and because the Book I. 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 Let ACDB be a parallelogram, of which BC is a diameter; B a 29. с D 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 otherb, 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; therefcre the opposite sides and apgles 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 arv equal to the two DC; CB,etch to cach; and the angle ABC * . - Book I. equal to the angle BCD; therefore the triangle ABC is equal c to the triangle BCD, and the diameter BC divides the paralC 4. 1. lelogram ACDB into two equal parts. Q. E. D., PROP. XXXV. THEOR. See Note. PARALLELOGRAMS upon the same base, and between the same parallels, are equal to one another. See the 2d Let the parallelograms ABCD, ABCF be upon the same and 3d fi başe BC, and between the same parallels AF, BC; the paralgures. lelogram ABCD shall be equal to the parallelogram BCF. А F point D; it is plain that each of the a 4.1.- parallelograms is double of the trian gle BDC; and they are therefore B C to the base BC of the parallelograms ABCD, ERCF, 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. b'1 Ax. to EF ; and DE is common; therefore the whole, or the rec2.or 3. Ax. niainder AE, is equalc to the whole, or the remainder DF; AB also is equal to DC; and the two EA, AB are therefore a equat to the two FD, DC, each to each; and the exterior angle FDC is equald to the interior EAB ; therefore the base EB is equal to the base FC, and the triangle EAB equal e 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 equals, that is, the parallelogram ABCD is equal to the parallelogram EBCF. Therefore, parallelograms upon the same base, &c. Q: E. D. d 19.1. € 4. 1. f 3. Ax. Book I. PROP. XXXVI. THEOR. PARALLELOGRAMS upon equal bases, and between the same parallels, are equal to one another. Let ABCD, EFGH be A D E H parallelograms upon equal bases BC, FG, and be. tween the same parallels AH, BG ; the parallelogram ABCD is equal to EFGH. Join BE, CH; and be- B С G cause BC is equal to FG, and FG to a EH, BC is equal to EH; á 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 them. selves equal and parallel b; therefore EB, CH are both equal b 33. 1. and parallel, and EBCH is a parallelogram ; and it is equal to 35. 1. ABCD, because it is upon the same başe 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. PROP. XXXVII. THEOR. 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 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 through B draw a BE parallel to CA; . 31. 1. and through C draw CF parallel to BD: therefore each of the figures EBCA,DBCF B. C is a parallelogram ; and EBCA is equal b 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 Book I. gram EBCA, because the diameter AB bisects e it; and the triangle DBC is the half of the parallelogram DBCF, because C 34. 1. the diameter DC bisects it: but the halves of equal things are 07. Ax. equal d; therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. Q. E. D. 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 I draw FH parallel to ED : then each of G A D the figures GBCA, H, DEFH is a parallelogram ; and they b 36. 1. are equal b to one another, because they B F rallels BF, GH; and the triangle ABC is the half of the pa rallelogram GBCA, because the diameter AB bisects it; and c 34. 1. the triangle DEF is the half of the parallelogram DEFH, be cause the diameter DF bisects it: but the halves of equal 07. As. things are equal d; therefore the triangle ABC is equal to the triangle DEF. Wherefore triangles, &c. Q. E. D. с 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 point A draw a AE parallel to BC, and join EC : the triangle 2.31. 1. ABC is equal b to the triangle EBC, because it is upon the same Book I. base BC, and between the same paral. A D lels BC, AE: but the triangle ABC is b 37. 1. equal to the triangle BDC; therefore E С PROP. XL. THEOR. EQUAL triangles upon equal bases, in the same straight line, and towards the same parts, are between the same parallels. Let the equal triangles ABC, DEF be upon equal bases BC, D G a 31. 1. rallel to BF, and join GF: B E F the triangle ABC is equalb b 38. 1 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. PROP. XLI. THEOR. IF a parallelogram and triangle be uport the same base, and between the same parallels; the parallel gram shall be double of the triangle. |