Let the parallelogram ABCD and the triangle EBC be upon the same base BC, and between the same parallels BC, AE; the parallelogram ABCD is double of the triangle EBC. D Join AC. Then (I. 37.) 1. B 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 (I. 34.) ABC, 2. The parallelogram ABCD is double of the triangle because the diameter AC divides it into two equal parts; wherefore also 3. ABCD is double of the triangle EBC. Therefore, if a parallelogram, &c. Q.E.D. PROP. XLII.-PROBLEM. 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. Bisect (I. 10.) BC in E, join AE, and at the point E in the straight line EC make (I. 23.) the angle CEF equal to D; and through A draw (I 31.) AFG parallel to EC, and through C draw CG parallel to EF. Therefore (I. 34. Def.) FECG is a parallelogram. A F G B E And because BE is equal to EC, likewise (I. 38.) 1. 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 2. The triangle ABC is double of the triangle AEC. And likewise (I. 41.) 3. The parallelogram FECG is double of the triangle AEC, because they are upon the same base EC, and between the same parallels EC, AG. Therefore (Ax. 6.) 4. The parallelogram FECG is equal to the triangle ABC, and it has one of its angles CEF equal (Constr.) 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.-THEOREM. 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, which are therefore called the complements. The complement BK is equal to the complement KD. A H 1. B G Because ABCD is a parallelogram, and AC its diameter, (I. 34.) The triangle ABC is equal to the triangle ADC; and because EKHA is a parallelogram, the diameter of which is AK, 2. The triangle AEK is equal to the triangle AHK; and for the same reason, 3. 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 KFC; (Ax. 2.) 4. 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 5. The remaining complement BK is equal to the remaining complement KD. Wherefore, the complements, &c. Q.E.D. PROP. XLIV.-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. 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. Make (I. 42.) the parallelogram BEFG equal to the triangle C, and having the angle EBG equal to the angle D, so that BE be in the same straight line with AB; and produce FG to H; and through 4 draw (I. 31.) AH parallel to BG or EF, and join HB. Then because the straight line HF falls upon the parallels AH, EF, (I. 29.) 1. The angles AHF, HFE, are together equal to two right angles; wherefore (Ax. 9.) 2. The angles BHF, HFE, are less than two right angles; but (Ax. 12.) straight lines which with another straight line make the interior angles upon the same side less than two right angles, do meet if produced far enough; therefore 3. HB, FE, shall meet, if produced; let them meet in K, and through K draw KL parallel to EA or FH, and produce HA, GB, to the points L, M. Then HLKF is a parallelogram, of which the diameter is HK, and AG, ME, are the parallelograms about HK, and LB, BF, are the complements; therefore (I. 43.) 4. LB is equal to BF. But BF is equal to the triangle C; wherefore (Ax. 1.) 5. LB is equal to the triangle C. And because the angle GBE is equal (I. 15.) to the angle ABM, and likewise (Constr.) to the angle D, 6. The angle ABM is equal to the angle D. Therefore the parallelogram LB is applied to the straight line AB, is equal to the triangle C, and has the angle ABM equal to the angle D. Which was to be done. PROP. XLV.-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 rectilineal angle. It is required to describe a parallelogram equal to ABCD, and having an angle equal to E. Join DB; and describe (I. 42.) the parallelogram FH equal to the triangle ADB, and having the angle HKF equal to the angle E; and to the straight line GH apply (I. 44.) the parallelogram GM equal to the triangle DBC, having the angle GHM equal to the angle E. The figure FKML shall be the parallelogram required. Because the angle E is equal to each of the angles FKH, GHM, 1. The angle FKH is equal to GHM; add to each of these the angle KHG; therefore 2. The angles FKH, KHG, are equal to the angles KHG, GHM; but FKH, KHG are equal (I. 29.) to two right angles; 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 equal to two right angles, (I. 14.) 4. KH is in the same straight line with HM; and because the straight line HG meets the parallels KM, FG, (I. 29.) The alternate angles MHG, HGF, are equal. 5. Add to each of these the angle HGL: therefore 6. The angles MHG, HGL, are equal to the angles HGF, HGL. But the angles MHG, HGL, are equal (I. 29.) to two right angles; wherefore also 7. The angles HGF, HGL, are equal to two right angles, and therefore (I. 14.) And because KF is parallel to HG, and HG to ML, (I. 30.) 9. KF is parallel to ML: and KM, FL, are parallels; wherefore (I. 34. Def.) 10. KFLM is a parallelogram; and because the triangle ABD is equal to the parallelogram HF, and the triangle DBC to the parallelogram GM; 11. The whole rectilineal figure ABCD is equal to the whole parallelogram KFLM. Therefore the parallelogram KFLM has been described equal to the given rectilineal figure ABCD, having the angle FKM equal to the given angle E. Which was to be done. 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 (I. 44.) to the given straight line a parallelogram equal to the first triangle ABD, and having an angle equal to the given angle. PROP. XLVI.-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. From the point A draw (I. 11.) AC at right angles to AB, and make (I. 3.) AD equal to AB, and through the point D draw DE parallel (I. 31.) to AB, and through B draw BE parallel to AD; therefore (I. 34. Def.) 1. whence (I. 34.) ADEB is a parallelogram; 2. AB is equal to DE, and AD to BE; but BA is equal (Constr.) to AD; therefore the four straight lines BA, AD, DE, EB, are equal to one another, and 3. The parallelogram ADEB is equilateral; likewise all its angles are right angles; for since the straight line AD meets the parallels AB, DE, (I. 29.) 4. The angles BAD, ADE, are equal to two right angles; |