and Then the figure CEFG is a parallelogram (Def. A). Proof. Because the triangles ABE, AEC are on the equal bases BE, EC, and between the same parallels BC, AG; therefore 1. The triangle ABE is equal to the triangle AEC (I. 38), 2. The triangle ABC is double of the triangle AEC; 3. The parallelogram FECG is double of the triangle AEC (I. 41), but because they are upon the same base EC, and between the same parallels EC, AG; therefore 4. 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, a parallelogram FECG has been described equal to the given triangle ABC, and having one of its angles CEF equal to the given angle D. Q.E.F. PROPOSITION 43.-Theorem. The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another. B Let ABCD be a parallelogram, of which the diameter is AC; and EH, GF the parallelograms about AC, that is, through which AC passes: also BK, KD the other parallelograms which make up the whole figure ABCD, which are therefore called the complements. Then the complement BK shall be equal to the complement KD. H E A D F Demonstration. Because ABCD is a parallelogram, and AC its diameter, therefore 1. The triangle ABC is equal to the triangle ADC (I. 34). Again, because EKHA is a parallelogram, and AK its diameter, therefore Wherefore B E H 2. The triangle AEK is equal to the triangle AHK (L. 34); and for the same reason, 3. The triangle KGC is equal to the triangle KFC. F 4. The two triangles AEK, KGC are equal to the two triangles AHK, KFC (Ax. 2); but the whole triangle ABC is equal to the whole triangle ADC; therefore 5. The remaining complement BK, is equal to the remaining complement KD (Ax. 3). Wherefore, the complements, &c. Q.E.D. PROPOSITION 44.- Problem. 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. JA D A 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 the angle D. Construction (I.) Make the parallelogram BEFG equal to the triangle C, and having the angle EBG equal to the angle D (I. 42), so that BE be in the same straight line with AB; produce FG to H; through A draw AH parallel to BG or EF (I. 31), and join HB. Proof (I.) Then because the straight line HF falls upon the parallels AH, EF, therefore Wherefore 1. The angles AHF, HFE are together equal to two right angles (I. 29); 2. The angles BHF, HFE are less than two right angles (Ax. 9), but straight lines which with another straight line make the two interior angles upon the same side less than two right angles, do meet if produced far enough (Ax. 12); therefore 3. HB and FE shall meet if produced. Construction (II.) Let HB and FE be produced and meet in K; through K draw KL parallel to EA or FH, and produce HA, GB to meet KL in the points L, M. Then LB shall be the parallelogram required. Proof (II.) Because HLKF is a parallelogram, of which the diameter is HK; and AG, ME are the parallelograms about HK ; also LB, BF are the complements; therefore 4. The complement LB is equal to the complement BF (I. 43), but the complement BF is equal to the triangle C (constr.); wherefore 5. LB is equal to the triangle C. And because the angle GBE is equal to the angle ABM (1. 15), and likewise to the angle D (constr.); therefore 6. The angle ABM is equal to the angle D (Ax. 1). Therefore to the given straight line AB the parallelogram LB has been applied, equal to the triangle C, and having the angle ABM equal to the given angle D. Q.E.F. PROPOSITION 45.-Problem. To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle. Let ABCD be the given rectilineal figure, and E the given recti lineal angle. It is required to describe a parallelogram that shall be equal to the figure ABCD, and having an angle equal to the given angle E. K H Construction. Join DB. Describe the parallelogram FH equal to the triangle ADB, and having the angle FKH equal to the angle E (I. 42); to the straight line GH apply the parallelogram GM, equal to the triangle DBC, having the angle GHM equal to the angle E (I. 44); 1. The angle FKH is equal to the angle GHM; add to each of these equals the angle KHG; therefore L M Then the figure FKML shall be the parallelogram required. Proof. Because each of the angles FKH, GHM is equal to the angle E, therefore add to each of these equals the angle HGL; therefore 2. The angles FKH, KHG are equal to the angles KHG, GHM; but FKH, KHG are equal to two right angles (I. 29); therefore also 3. KHG, GHM are equal to two right angles; and because at the point H, in the straight line GH, the two straight lines KH, HM, upon the opposite sides of it, make the adjacent angles KHG, GHM equal to two right angles, therefore 4. HK is in the same straight line with HM (I. 14). And because the line HG meets the parallels KM, FG, therefore 5. The angle MHG is equal to the alternate angle HGF (I. 29); 6. The angles MHG, HGL are equal to the angles HGF. HGL; but the angles MHG, HGL are equal to two right angles (I. 29); therefore also 7. The angles HGF, HGL are equal to two right angles, and therefore 8. FG is in the same straight line with GL (I. 14). And because KF is parallel to HG, and HG to ML, therefore 9. KF is parallel to ML (I. 30); and FL has been proved parallel to KM, wherefore 10. The figure FKML is a parallelogram; and since the parallelogram HF is equal to the triangle ABD, and the parallelogram GM to the triangle BDC, therefore 11. The whole parallelogram KFLM is equal to the whole rectilineal figure ABCD. Therefore the parallelogram KFLM has been described equal to the given rectilineal figure ABCD, having the angle FKM equal to the given angle E. Q.E.F. Cor. From this it is manifest how, to a given straight line, to apply a parallelogram, which shall have an angle equal to a given rectilineal angle, and shall be equal to a given rectilineal figure; viz., by applying to the given straight line a parallelogram equal to the first triangle ABD (I. 44), and having an angle equal to the given angle. D PROPOSITION 46.-Problem. To describe a square upon a given straight line. Let AB be the given straight line. It is required to describe a square upon AB. B Construction. From the point A draw AC at right angles to AB (I. 11); make AD equal to AB (I. 3); through the point D draw |