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 +2 Ax. aigle BAC has been shewn to be equal to the angle BDC; "therefore the opposite sides and angles 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 are equal to the two DC, CB, each to each; and the angle ABC has been proved equal to the angle BCD: therefore, the triangle ABC is equal to the triangle BCD, and the diameter BC divides the parallelogram ACDB into two equal parts. Q.E.D. PROPOSITION XXXV. 4. 1. THEOR.-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 base BC, and between the same parallels AF, BC: the parallelogram ABCD shall be equal to the parallelogram EBCF. If the sides AD, DF of the parallelograms ABCD, DBCF, opposite to the base BC, be terminated in the same point D, it is plain that each of the parallelograms is double* of the triangle BDC; and they are therefore equal + to one another. * * See the 2d and 3d fi gures. * 34. 1. B † 6 Ax. * 1 Ax. *2 or 3 Ax. † 34. 1. But, if the sides AD, EF, opposite to the base BC of the parallelograms ABCD, EBCF, be not terminated in the same point, then, because ABCD is a parallelogram, AD is equal* 34. 1. to BC; for the same reason, EF is equal to BC; wherefore AD is equal to 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; therefore the two EA, AB are equal to the two FD, DC, each to each; and the exterior angle 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 trape B * 29. 1. DE FAE DF • 4. 1. zium ABCF, and from the same trapezium take the triangle EAB, and the remainders are equal; that is, the parallelo-3 Ax gram ABCD is equal to the parallelogram EBCF. Therefore, parallelograms upon the same base, &c. Q. E. D. + Hyp. 34. 1. † 1 Ax. + Hyp. * 33. 1. PROPOSITION XXXVI. THEOR.-Parallelograms upon equal bases, and between the Let ABCD, EFGH be parallelograms A DE Join BE, CH and because BC is Н 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 the same parts, are themselves* equal and parallel; therefore EB, HC + Def. 34. 1. are both equal and parallel; and therefore EBCH is at parallelogram; and it is equal * to ABCD, because they are upon same base BC, and between the same parallels BC, AH: for the like reason, the parallelogram EFGH is equal to the same EBCH: therefore the parallelogram ABCD is equal to EFGH. Wherefore, parallelograms, &c. Q. E. D. * 35. 1. † 1 Ax. † 2 Post. 31. 1. * 35. 1. PROPOSITION XXXVII. the 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 to the points * B E, F; and through B, draw BE parallel to CA; and through. C, draw CF parallel to BD: therefore each of the figures Def.34. 1. EBCA, DBCF, is a† parallelogram; 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 triangle DBC is the half of the parallelogram DBCF, because the diameter DC bisects it: but the halves of equal things are equal; therefore the triangle ABC is equal to the triangle DBC. Wherefore, triangles, &c. Q. E. D. * 34. 1. * 7 Ax. * PROPOSITION XXXVIII. THEOR.-Triangles upon equal bases, and between the same parallels, aré 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. * 31. 1. Produce + AD both ways to the points G, H; and through + 2 Post. B, draw BG parallel * to CA, and through F, draw FH parallel to ED: then each of the figures GBCA, G B CE H+ Def. 34.1. * 36. I. DEFH, is a † parallelogram; and they are equal to one another, because they are upon equal bases BC, EF, and between the same parallels BF, GH; and the triangle ABC is the half of the parallelogram GBCA, because the diameter AB bisects* it; and 34. 1. 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 7 Ax. triangle DEF. Wherefore, triangles, &c. PROPOSITION XXXIX. Q. E. D. 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. A * 37. i. E + Hyp. Ax. 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. * 31. 1. 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 triangle ABC is equal to the triangle DBC; therefore also the triangle DBC 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 other line but AD is parallel to BC; AD is therefore parallel to it. Wherefore, equal triangles, &c. Q. E. D. B PROPOSITION 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 B parallels. * D CE Join AD: AD shall be parallel to BC. For if it is not, through A, draw* 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 but AD; AD is therefore parallel to BF. Wherefore, equal triangles, &c. Q. E. D. * 37. 1. ⚫ 34. 1. PROPOSITION XLI. THEOR.-If a parallelogram and a 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 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 DE 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, because 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. ** PROPOSITION 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. * AF G 10. 1. 23. 1. * 31. 1. † Def. 34.1. Bisect BC in E, join AE, and at the point E, in the straight line EC, make* the angle CEF equal to D; and through A, draw * AFG parallel to EC, and through C, draw CG parallel to EF; therefore FECG is a ↑ parallelogram. And because BE is equal † to EC, the triangle ABE is equal to the triangle AEC, since they are upon equal bases BE, EC, and between the same parallels BC, AG; therefore the triangle ABC is double of the triangle 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; therefore the parallelogram FECG is equal to the triangle ABC; and it has one of its angles † 6 Ax. CEF equal to the given angle D: wherefore, a parallelogram † Constr. 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. PROPOSITION XLIII. B E THEOR.-The complements of the parallelograms which are about the diameter of any parallelogram, are equal to one another. E A H K F Let ABCD be a parallelogram, of which the diameter is AC; and EH, GF 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 the complements the complement BK shall be equal to the comple ment KD. * 41. 1. Because ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal to the triangle ADC. Again, because 34. 1. EKHA is a parallelogram, the diameter of which is AK, the D |