is equal to the angle BCD; therefore, "the triangle ABC is Book I. equal to the triangle BCD, and the diameter BC divides the parallelogram ACDB into two equal parts. Q. E. D. c4. 1. Parallelograms upon the same base, and between See N. the same parallels, are equal to one another. Let the parallelograms ABCD, EBCF be upon the same See the base BC, and between the same parallels AF, BC; the paral- 2d and 3d lelogram ABCD is equal to the parallelogram EBCF. If the sides AD, DF of the paral lelograms ABCD, DBCF, opposite But, if the sides AD, EF, opposite figures. ABCD, EBCF, be not terminated in the same point; then because ABCD is a parallelogram, AD is equal to BC; for the same reason EF is equal to BC; wherefore AD is equal to b 1. Ax. EF; and DE is common; therefore the whole, or the remainder AE is equal to the whole, or the remainder DF; AB also is equal to DC; and the two EA, AB are therefore equal to the A DE FAED F c 2. or 3. Ax. B C B C two FD, DC, each to each; and the exterior angle FDC is equal to the interior EAB, therefore the base EB is equal to a 29. 1. the base FC, and the triangle EAB equale to the triangle e 4. 1. 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 f3. Ax. ABCD is equal to the parallelogram EBCF. Therefore, "parallelograms upon the same base," &c. Q. E. D. Book I. b 33. 1. c 35. 1. PROP. XXXVI. THEOR. Parallelograms upon equal bases, and between the same parallels, are equal to one another. Let ABCD, EFGH A Join BE, CH; and B because BC is equal to DE H C F G a 34. 1. FG, and FG to a EH, BC is equal to EH; 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 themselves equal and parallel; therefore EB, CH are both equal and parallel, and EBCH is a parallelogram; and it is equal to ABCD, because it is upon the same base BC, and between the same parallels BC, AH: For the like reason, the parallelogram EFGH is equal to the same EBCH: Therefore also the parallelogram ABCD is equal to EFGH. Wherefore, “pa"rallelograms," &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 paral lels AD, BC: The triangle E ABC is equal to the tri- Produce AD both ways a 31. 1. through B draw a BE parallel to CA; and through C draw CF parallel to BĎ: B AD C F Therefore each of the figures EBCA, DBCF is a paral b 35. 1. lelogram; and EBCA is equal to DBCF, because they are upon the same base BC, and between the same parallels BC, EF; and the triangle ABC is the half of the parallelogram EBCA, because the diameter AB bisects it; and the tri- Book I. angle DBC is the half of the parallelogram DBCF, because d the diameter DC bisects it; but the halves of equal things are c 34. 1. equal; therefore the triangle ABC is equal to the triangle d 7. Ax. 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 draw BG parallel a to CA, and through F draw FH parallel to a 31. 1. ED: Then each of the figures GBCA, DEFH is a parallelogram; and they are equal to b one an other, because they are upon equal bases G BC, EF, and be B tween the same pa rallels BF, GH; and the triangle ABC is the half of the parallelogram GBCA, c 34. 1. 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. 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. d d 7. Ax. 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 a 31. 1. C b 37. 1. b D E Book I. ABC is equal 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; therefore also the triangle BDC is equal to the triangle EBC, the greater to the less, which is impossible: Therefore AE is not parallel to BC. In the same manner, it can be demonstrated, that no B C other line than AD is parallel to BC; AD is therefore parallel to BC. Wherefore," equal triangles upon," &c. Q. E. D. 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, EF, in the same straight B A D G CE F 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, a 31. 1. through A draw a AG parallel to BF, and join GF: the triangle ABC is equal 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 than AD; AD is therefore parallel to BF. Wherefore, " equal triangles," &c. Q. E. D. b 38. 1. PROP. XLI. THEOR. If a parallelogram and a triangle are upon the same base, and between the same parallels; the parallelogram is double the triangle. Let the parallelogram ABCD and the triangle EBC be up- Book I. on the same base BC, and between the same parallels BC, AE; the parallelogram ABCD is double the triangle EBC. a Join AC; then the triangle ABC is equal to the triangle EBC, because they are upon the same base BC, and between the same parallels BC, AE. But the parallelogram ABCD is double the triangle ABC, because the diameter AC divides it into two equal parts; wherefore A B DE a 37. 1. b 34. 1. C ABCD is also double the triangle EBC. Therefore, "if a parallelogram," &c. Q. E. D. PROP. XLII. PROB. To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Let ABC be the given triangle, and D the given rectilineal angle. It is required to describe a parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D. a b Bisect BC in E, join AE, and at the point E in the straight a 10. 1. line EC make the angle CEF equal to D; and through A b 23. 1. draw G parallel to BC, and c C through C draw CG parallel to EF: Therefore FECG is a parallelogram: And because BE is equal to EC, the triangle ABE is likewise equal to the triangle AEC, since they are upon equal bases BE, EC, and between the same parallels BC, AF G c 31. 1. AG: Therefore the triangle B E C ABC is double the triangle AEC. And the parallelogram FECG is likewise double e e 41. 1. the triangle AEC, because it is upon the same base, and between the same parallels: Therefore the parallelogram FECG is equal to the triangle ABC, and it has one of its angles CEF equal to the given angle D: wherefore there has been described a parallelogram FECG equal to a given triangle ABC, |