but the triangle ABC is the half of the parallelogram EBCA, because the diameter AB bisects (34. 1.) it; and the triangle DBC is the half of the parallelogram DBCF, because the diameter DC bisects it; and 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. PROP. XXVIII. 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 be tween 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 (31. 1.) to CA, and through F draw FH parallel to ED: Then each of the figures GBCA, DEFH is a paral G (34. 1.) of the parallelogram GBCA, because the diameter AB bisects it; and the triangle DEF is the half (34. 1.) of the parallelogram DEFH, because the diameter DF bisects it; But the halves of equal things are equal (7. Ax); 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 Join AD: AD is parallel to BC; for, if it is not, through the point A draw (31. 1.) AE parallel to BC, and join EC; The triangle ABC, is equal (37. 1.) 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, he greater to the less, which is impossible: Therefore AE is not parallel to BC. In the same manner, it may 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.. B C PROP. XL. THEOR. Equal triangles on the same side of bases, which are equal and in the same straight line, are between the same parallels. Let the equal triangles ABC, DEF be upon equal basis BC, EF, Join AD; AD is parallel to BC: For, if, it is not, through A draw (31.1.) AG parallel to BF, and join GF. The triangle ABC is equal (38. 1.) to the triangle GEF, because they are upon equal bases BC, EF, and between the same parallels BF, AG: But the tri angle 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 may 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 a triangle be upon the same base, and between the same parallel; the parallelogram is double of the triangle. Let the paralellogram ABCD and the triangle EBC be upon the same base BC and between the same parallels BC, AE; the parallelogram ABCD A is double of the triangle EBC. D E Join AC; then the triangle ABC is equal (37. 1.) to the triangle EBC, because they are upon the same base BC, and between the same parallels RC, AE. But the parallelogram ABCD is double (34. 1.) of the triangle ABC, because the diameter AC divides it into two equal parts; wherefore ABCD is also double of the trian gle 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. PROP. XXVIII. 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 be tween 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 (31. 1.) to CA, and through F draw FH parallel to ED: Then each of the figures GBCA, DEFH is a paral G 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 (31. 1.) AE parallel to BC, and join EC; The triangle ABC, is equal (37. 1.) 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, he greater to the less, which is impossible: Therefore AE is not parallel to BC. In the same manner, it may 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.. A B D E PROP. XL. THEOR. Equal triangles on the same side of bases, which are equal and in the same straight line, are between the same parallels. Let the equal triangles ABC, DEF be upon equal basis BC, EF, in the same straight line BF, and towards the same parts ; draw (31. 1.) AG parallel to BF, and join GF. The triangle B A D ABC is equal (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 may 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 a triangle be upon the same base, and between the same parallel; the parallelogram is double of the triangle. Let the paralellogram ABCD and the triangle EBC be upon the same base BC and between the same parallels BC, AE; the parallelogram ABCD A is double of the triangle EBC. DE Join AC; then the triangle ABC is equal (37. 1.) 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 (34. 1.) of the triangle ABC, because the diameter AC divides it into two equal parts; where fore ABCD is also double of the trian gle 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. 48 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 (10. 1.) BC in E, join AE, and at the point E in the straight line EC make (23 1.) the augle CEF equal te D; and through A draw (31. 1.) AG parallel to BC, and through C draw CG (31. 1.) parallel to EF: Therefore FECG is a parallelogram: And because A to FG BE is equal to EC, the triangle between the same parallels BC B D EC 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, 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. Let ABCD be a parallelogram of which the diameter is AC; let EH, FG be the parallelograms about AC, that is, through which AC passes, and let BK, KD be the other parallelograms, which make up the AH D whole figure ABCD and are therefore called the complements: The complement BK is equal to the complement KD. Because ABCD is a parallelogram and AC its diameter, the triangle ABC is equal (34. 1.) to the triangle ADC: And because EKHA is a pa- B rallelogram and AK its diameter, the G C triangle AEK is equal to the triangle AHK: For the same reason, the triangle KGC is equal to the triangle KFC. Then, because the triangle AEK is equal to the triangle AHK, and the triangle KGC to the triangle KFC; the triangle AEK, together with the triangle KGC is equal to the triangle AHK, together with the triangle KFC: But the whole triangle ABC is equal to the whole ADC; therefore the remaining complement BK is equal to the remaining complement KD. Wherefore, the complements, &c. Q. E. D. |