ABCF, and from the same trapezium take the triangle EAB, which has been proved equal to the triangle *3 Ax. FDC; and the remainders are equal,* that is, the parallelogram ABCD is equal to the parallelogram EBCF. Therefore parallelograms upon the same base, &c. Q. E. D. * Hyp. 34. 1. PROP. XXXVI. THEOR. Parallelograms upon equal bases, and between the same parallels, are equal to one another. Join BE, CH; and because BC is equal to FG, and FG to* 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 the extremities of equal and parallel straight lines towards * 33. 1. the same parts, are themselves equal and parallel;* therefore EB, CH are both equal and parallel, and *35. 1. therefore EBCH is a parallelogram; and it is equal to ABCD, because they are upon the same base BC, and between the same parallels BC, AH: for the same reason, the parallelogram EFGH is equal to the same EBCH: therefore 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 shall be equal to the triangle DBC. Produce AD both ways, and through B draw* 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* to DBCF, because * 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 parallelogram EBCA, because the diameter AB bisects it; and the triangle DBC is the half of * 34. 1. the parallelogram DBCF, because the diameter DC bisects it: but the halves of equal things are equal;* * 7 Ax. therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. Q. E. D. 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 shall be equal to the triangle DEF. Produce AD both ways, and through B draw BG parallel to CA, and through F draw FH parallel to * 31. 1. ED: then each of the figures GBCA, DEFH is a parallelogram; and they triangle ABC is the half* of the parallelogram GBCA, * 34. 1. because the diameter AB bisects it; and the triangle DEF is the half of the parallelogram DEFH, be- * 34.1 cause the diameter DF bisects it: but the halves of *7 Ax. equal things are equal;* therefore the triangle ABC is equal to the triangle DEF. Wherefore triangles, &c. Q. E. D. * 31. 1. 37. 1. 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 shall be between the same parallels. Join AD; AD shall be parallel to BC; for, if it is not, through the point A draw* AE parallel to BC, and join EC; then the triangle ABC is equal to the triangle EBC, because they are upon the same base BC, and between the same B E to the less, which is impossible: therefore AE is not parallel to BC. In the same manner, it can be demonstrated that no other line but AD is parallel to BC; therefore AD is 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 line BF, and towards the same parts; they shall be between the same parallels. D Join AD: AD shall be parallel to BC; for, if it is not, through A draw* AG parallel to BF, and join 31. 1. GF. The triangle ABC is equal to the triangle GEF, * 38. 1. 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 the triangle DEF is also 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; therefore AD is parallel to BF. Wherefore equal triangles, &c. Q. E. D. If a parallelogram and a triangle be upon the same base, and between the same parallels; the parallelogram shall be double of the triangle. D Let the parallelogram ABCD and the triangle EBC be upon the same base BC, and between the same parallels BC, AE; the parallelogram ABCD shall be double of the triangle EBC. B * 37. 1. 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* of the triangle ABC, be- 34. 1. cause the diameter AC divides it into two equal parts; wherefore ABCD is also double of 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 * 10. 1. 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. Bisect BC in E, join AE, and at the point E in the 23. 1. straight line EC make* the angle CEF equal to D; and through A draw* AG parallel to EC, and through *31. 1. C draw* CG parallel to EF: * 31. 1. * Def. 34. 1. therefore FECG is a parallelo. gram.* And because BE is equal to EC, the triangle ABE is 38. 1. equal* to the triangle AEC, since they are upon equal bases B A F G E BE, EC, and between the same parallels BC, AG; therefore the triangle ABC is double of the triangle 41. 1. AEC. But the parallelogram FECG is likewise double* of the triangle AEC, because they are upon the same base EC, and between the same parallels EC, AG: * 6 Ax. 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 a parallelogram FECG has been described equal to the given triangle ABC, having one of its angles CEF equal to the given angle D. Which was to be done. PROP. XLIII. THEOR. The complements of the parallelograms which are about the diameter of any parallelogram, are equal to one another. D A H K E E Let ABCD be a parallelogram, of which the diameter is AC, and EH, GF the parallelograms about AC, that is, through which AC passes, and BK, KD the other parallelograms which make up the whole figure ABCD, which are therefore called B G |