double b of the triangle ABC, because the diameter AC Book I. divides it into two equal parts; wherefore ABCD is also double of the triangle EBC. Therefore, if a parallelo- b 34. 1. gram, &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 parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D. b Bisecta 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 AG parallel to BC, and through C draw CG parallel to A c EF: Therefore FECG is a parallelogram: And because BE is equal to EC, the triangle ABE is likewise equal to F G the triangle AEC, since they are upon equal bases BE, EC, and be B E D tween the same parallels BC, AG; therefore the triangle ABC is double of the triangle AEC. And the paralle logram FECG is likewise double of the triangle AEC, e 41. 1. 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. Book I. 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 H D passes, and let BK, KD be the other parallelograms, which make up the whole E figure ABCD, and are therefore called the comple B G KD. a Because ABCD is a parallelogram, and AC its diameter, the triangle ABC is a 34. 1. equal to the triangle ADC: And, because EKHA is a parallelogram and AK its diameter, the 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. PROP. XLIV. PROB. 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 the parallelogram BEFG equal to the triangle Book I. b 31. 1. c 29. 1. 1. C, having the angle EBG equal to the angle D, and the side BE in the same straight line with AB: produce FG to H, and through A drawb 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; wherefore the angles BHF, HFE are less than two right angles: But straight lines which with another straight line make the interior angles, upon the same side, less than two right angles, do meet if produced: Therefore HB, FE will d cor. 29. 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 LB is equal to BF: But BF is equal to the triangle C; wherefore LB is equal to the triangle C; and because the angle GBE is equal to the f 15, 1. angle ABM, and likewise to the angle D; the angle ABM is equal to the angle D: Therefore the parallelogram LB, which 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. e PROP. XLV. PROB. 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 pa e 43. 1. Book I. a 42. 1. b 44. 1. c 29. 1. rallelogram equal to ABCD, and having an angle equal to E. Join DB, and describe 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 the parallelogram GM equal to the triangle DBC, having the angle GHM equal to the angle E. And because the angle E is equal to each of the angles FKH, GHM, the angle FKH is equal to GHM: add to each of these the angle KHG; therefore the angles FKH, KHG are equal to the angles KHG, GHM; but FKH, KHG are equal to two right angles; therefore also KHG, GHM d 14. 1. e 30. 1. c 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 GH, make the adjacent angles equal to two right angles, KH is in the same straight line with HM. And because the straight line HG meets the parallels KM, FG, the alternate angles MHG, HGF are equal; add to each of these the angle HGL; therefore the angles MHG, HGL, are equal to the angles HGF, HGL: But the angles MHG, HGL, are equal to two right angles; wherefore also the angles HGF, HGL are equal to two right angles, and FG is therefore in the same straight line with GL. And because KF is parallel to HG, and HG to ML, KF is parallel to ML; but KM, FL are parallels; wherefore KFLM is a parallelogram. And because the triangle ABD is equal to the parallelogram HF, and the triangle DBC to the parallelogram GM, 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, e having the angle FKM equal to the given angle E. Book I 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 to the given b 44. 1. straight line a parallelogram equal to the first triangle ABD, and having an angle equal to the given angle. PROP. XLVI. PROB. 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. d C D From the point A draw a AC at right angles to AB; and make AD equal to AB, and through the point D draw DE parallel to AB, and through B draw BE parallel to AD; therefore ADEB is a parallelogram : Whence AB is equal to DE, and AD to BE: but BA is equal to AD; therefore the four straight lines BA, AD, DE, EB are equal to one another, and the parallelogram ADEB is equilateral: it is likewise rectangular; for the straight line AD meeting the parallels AB, DE, makes the angles BAD, ADE equal to two right angles; but BAD is a right angle; A therefore also ADE is a right an E a 11. 1. b 3. 1. c 31. I. d 24. 1. c 29. 1. B gle; now the opposite angles of parallelograms are equala; therefore each of the opposite angles ABE, BED is a right angle; wherefore the figure ADEB is rectangular, and it has been demonstrated that it is equilateral; it is therefore a square, and it is described upon the given straight line AB: Which was to be done. COR. Hence every parallelogram that has one right angle has all its angles right angles. |