ABC is half of the parallelogram EBCA, because the diameter AB bisects it (1. 34); and the triangle DBC is 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. 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. A D H B CE F Produce AD both ways to the points G, H; and through B draw BG parallel to CA, and through F draw FH parallel to ED: Then each of the figures GBCA, 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 (1. 36): But the triangle ABC is half of the parallelogram GBCA, because the diameter AB bisects it (1. 34); and the triangle DEF is 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 shall be between the same parallels. Join AD; AD is 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 parallels BC, AE (1. 37) : But the triangle ABC is equal to the triangle DBC (Hyp); therefore also the triangle DBC is equal to the triangle EBC-the greater to the less, which is absurd: Therefore AE is not parallel to BC: And, in like manner, it may be demonstrated that no other line but AD can be parallel to BC; therefore AD is parallel to BC. Wherefore, Equal triangles upon &c. Q. E.D. Equal triangles upon equal bases, in the same straight line, and on the same side of it, are between the same parallels. Let the equal triangles, ABC, DEF, be upon equal bases BC, EF, in the same straight line BF, and on the same side of it: they shall be between the same parallels. Join AD; AD is parallel to BC: For, if it is not, through A draw AG parallel to BF, A D and join GF: Then the triangle ABC is equal to the triangle GEF, because they are upon equal bases B CE F BC, EF, and between the same parallels BF, AG (1. 38): 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 absurd : Therefore AG is not parallel to BF: And, in like manner, it may be demonstrated that no other line but AD can be parallel to BF: Therefore AD is 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 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. A Ꭰ E 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 (1. 37): But the parallelogram ABCD is double of the triangle ABC, because the diameter AC bisects it B (1. 34); Therefore also the parallelogram ABCD is double of the triangle EBC. Wherefore, 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 parallelo gram that shall be equal to the given triangle ABC, and have one of its angles equal to D. Bisect BC in E (1. 10); join AE, and at the point E in the straight line CE make the angle CEF equal to D (1. 23); and through A draw AG parallel to EC (1. 31), and through C draw CG parallel to EF. AT G D Then FECG is a parallelogram: And because BE is equal to EC, the triangle ABE is equal to the triangle AEC (1. 38), 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 (1.41), because they are upon the same base, and between the same parallels: Therefore the parallelogram FECG is equal to the triangle ABC (Ax. 6), 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, and having one of its angles CEF equal to the given angle D. Q. E. F. 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, and EH, FG the parallelograms about AC, that is, through which AC passes, and BK, KD the other parallelograms which make up the whole figure ABCD, and are therefore called the complements: the complement BK is equal to the complement KD. For because ABCD is a parallelogram, and AC its E A H B G D K F diameter, the triangle ABC is equal to the triangle ADC (1. 34): And, because AEKH is a parallelogram, and AK its diameter, the triangle AEK is equal to the triangle AHK: And, for the like reason, the triangle KGC is equal to the triangle KFC : Therefore, 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 triangle ADC; therefore the remainder, the complement BK, is equal to the remainder, the complement KD. Wherefore, The complements &c. Q. E.D. To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Let AB be the given straight line, and C the given triangle, and D the given rectilineal angle: it is required to apply to the straight line AB a parallelogram equal to the triangle C, and having an angle equal to D. F Make the parallelogram BEFG equal to the triangle C, and having the angle EBG equal to the angle D (1. 42), and place it so that BE may be in the same straight line with AB; produce FG to H, and D M through A draw AH parallel to BG or EF, and join HB : Then, because the straight line HF falls upon the parallels AH, EF, the angles AHF, HFE are together equal to two right angles (1.29): Therefore the angles BHF, HFE are |