third angle of the one to the third angle of the other (1. 26.), viz. the side as 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; therefore the opposite sides and angles of the parallelograms are equal to one another; also their diameter bisects them ; for AB being equal to CD, and BC common, the two A B, BC are equal to the two DC, CB, each to each ; and the angle ABC is equal to the angle BCD; therefore the triangle ABC is equal (1. 4.) to the triangle BCD, and the diameter Bc divides the parallelogram ACDB into two equal parts. Q. E. D. EBCF. PROP. XXXV. THEOR. Parallelograms upon the same base and between the same parallels, are equal to one another. Let the parallelograms ABCD, EBCF (see the 2d and 3d figures) be upon the same base BC, and between the same parallels A F, BC; the parallelogram ABCD shall be equal to the parallelogram If the sides AD, DF of the parallelograms ABCD, DBCF opposite to the base BC be terminated in the A same point D; it is plain that each of the parallelograms is double (1. 34.) of the triangle BDC; and they are therefore equal to one another. But, if the sides AD, EF, opposite to the base bo of the parallelograms ABCD, EBCF, be not termi D F 1 B E nated in the same point; then, because ABCD is a parallelogram, Ad is equal (1. 34.) to BC; for the А D E F А E D. F 2 3 13 B с same reason EF is equal to BC; wherefore AD is equal (Ax. 1.) to EF; and DE is common; therefore the whole, or the remainder, AE is equal (Ax. 2. or 3.) to the whole, or the remainder DF; A B also is equal to DC; and the two E A, A B are therefore equal to the two, FD, DC, each to each ; and the exterior angle Foc is equal (1. 29.) to the interior E A B, therefore the base EB is equal to the base Fc, and the triangle E AB equal (1. 4.) to the triangle FDC; take the triangle FDC from the trapezium ABCF, and from the same trapezium take the triangle E AB; the remainders, therefore, are equal (Ax. 3.), that is, the parallelogram ABCD is equal to the parallelogram E BC F. Therefore, parallelograms upon the same base, &c. Q.E.D. A DE Н PROP. XXXVI. THEOR. Parallelograms upon equal bases, and between the same parallels, are equal to one another. Let ABCD, EFGH be 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 FG, and FG to (1. 34.) EH, BC is equal to EH; and they are parallels, and joined towards the same parts by the straight lines BE, ch: But B с F G straight lines which join equal and parallel straight lines towards the same parts, are themselves equal and parallel (1. 33.); therefore EB, ch are both equal and parallel, and EBCH is a parallelogram; and it is equal (1. 35.) to ABCD, because 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. 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 is equal to the triangle E А D F DBC. B с Produce AD both ways to the points E, F, and through B draw (1. 31.) BE parallel to CA; and through c draw CF parallel to BD: Therefore each of the figures EBCA, DBCF is a parallelogram ; and EBCA is equal (1. 35.) to DBCF, because they are upon the same base Bc, and between the saine parallels BC, EF; and the triangle ABC is the half of the parallelogram EBCA, because the diameter A B bisects it (1. 34.); 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 (Ax. 7.); 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. G a A D H AN B с E F 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 angle DEF. Produce AD both ways to the point g, h, and through B draw BG parallel (1. 31.) to ca, and through F draw FH parallel to ED: Then each of the figures GBCA, DEFH is parallelogram; and they are equal to (1. 36.) one another, because they are upon equal bases BC, EF, and between the same parallels BF, GH; and the triangle ABC is the half (1. 34.) of the parallelogram GBCA, because the diameter A B bisects it; and the triangle DEF is the half (1. 34.) of the parallelogram DEH, because the diameter DF bisects it: But the halves of equal things are equal (Ax. 7.); therefore the triangle ABC is equal to the triangle DEF. Wherefore triangles, &c. Q.E.D. 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 (1. 31.) A E parallel to BC, and join Ec: The triangle ABC is equal (1. 37.) to the triangle EBC, because it is upon the same base BC, and between the same parallels BC, AE : But the triangle ABC is equal to the triangle BDC (Hyp.); therefore also the triangle BDC is equal to the triangle EBC, the greater to the less, which is A D E B impossible : Therefore A E is not parallel to BC, In the same manner, 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. А D B с F E 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 B C, EF, in the same straight line BF, and towards the same parts; they are between the same parallels. Join AD; Ad is parallel to BC: For if it is not, through a draw (1. 31.) A G parallel to BF, and join GF; the triangle ABC is equal (1. 38.) 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 (Hyp.); 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 upon the same base, and between the same parallels ; the parallelogram shall be double of the triangle. Let the parallelogram ABCD and the triangle E BC |