44 Book I. Let the parallelogram ABCD and the triangle EBC be upon w the fame bafe BC, and between the fame parallels BC, AE; a 37. I. b 34. I. a Io. I. 23 I. C 31. I. d 38. I. € 41. I. the parallelogram ABCD is double of Join AC; then the triangle ABC angle EBC. Therefore, if a parallelogram, &c. Q. E. D. PROP. XLII. PROB. Τ To Let ABC be the given triangle, and D the given rectilineal angle It is required to defcr be a parallelogram that fhall be equal to the given triangle ABC, and have one of its angles equal to D. Bife & 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 CAG parallel to EC, and through AF G C draw CG parallel to EF: D C letogim FECG 15 hike wife double of the triangle AEC, becoute it is upon the fame base, and between the fame pafalle's Therefore the parallelogram FECG is equal to the trungle ABC and it has one of its angles CEF equal to the given angie D: Wherefore there has been described a paralle logram logram FECG equal to a given triangle ABC, having one of Book I. its angles CEF equal to the given angle D. Which was to be done. THE 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, and EH, FG the paralle'ograms about AC, that is, A H D through which AC paffes, and BK, KD the other parallelo. E grams which make up the whole figure ABCD, which G C Because ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal to a 34. I. the triangle ADC: And, because EKHA is a parallelogram, the diameter of which is AK, the triangle AEK is equal to the triangle AHK: By the fame reafon, the triangle KGC is equal to the triangle KFC: Then, because the triangle AEK is equal to the triangle AHK, and the triangle KOC to 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. T a given straight line to apply a parallelogram, which thall 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 Book t a 42. I. b 31. I. and produce FG to H; and thro' A drawb AH parallel to B or EF, and join HB. Then, because the straight line HF fall upon the parallels AH, EF, the angles AHF, HFE, are toge ther equal to two right angles; wherefore the angles BHF HFE are leffer than two right angles : But ftraight lines which with another ftraight line make the interior angles upon the d 12. Ax. fame fide lefs than two right angles, do meet if produced fa C 29. I. € 43. I. f 15. I. a 42. I. b 44. I. enough: Therefore HB, FE fhall 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 paral Jelogram, 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 angle ABM, and likewise to the angle D the angle ABM is equal to the angle D: Therefore the paralle logram LB is applied to the ftraight line AB, is equal to the triangle C, and has the angle ABM equal to the angle D Which was to be done. T PROP. XLV. PRO B. O defcribe a parallelogram equal to a given rectili neal 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 defcribe a parallelogram e qual to ABCD, and having an angle equal to E. Join DB, and defcribe a the parallelogram FH equal to the triangle ADB, and having the angle HKF equal to the angle E and to the ftraight line GH apply the parallelogram GM equal to the triangle DBC, having the angle GHM equal to the angle Book I. 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 angles A KHG GHM; but FKH, KHG are equal to two right angles; therefore allo KHG, GHM are equal to two F G L right angles; and because at the point H in the straight line GH, the two ftraight lines KH, HM, upon the oppofite fides of it make the adjacent angles equal to two right angles, KH is in the fame ftraight lined with HM; d 14. I. and because the ftraight 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 alfo the angles HGF, HGL are equal to two right angles, and FG is therefore in the fame ftraight line with GL: And becaufe KF is parallel to HG, and HG to ML; KF is parallel to ML: And KM, FLe 30. İn 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 defcribed 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 manifeft how to a given ftraight line to apply a parallelogram, which fhall have an angle equal to a given rectilineal angle, and shall be equal to a given rectilineal figure, viz. by applying to the given ftraight line, a parallelo- b 44. 1. gram equal to the first triangle ABD, and having an angle equal to the given angle. PROP. Book T. a II. I. b 3. I. € 34. I. d 34. I. € 39. I. a 46. I. To PRO P. XLVI. PRO B. O describe a square upon a given straight line. Let AB be the given ftraight line; it is required to defcrib a fquare upon AB. E 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; there fore ADEB is a parallelogram; whence AB is equal to DE, and AD to BE: But BA is equal to AD; therefore the four ftraight lines BA, AD, DE, EB are equal to one another, and the parallelogram ADEBD is equilateral, likewife all its angles are right angles; because the ftraight line AD meeting the parallels AB, DE, the angles BAD, ADE are equal to two right angles; but BAD is a right angle; therefore alfo ADE is a right angle; but the oppofite A angles of parallelograms are equal d; с B therefore each of the oppofite angles ABE, BED is a right angle; wherefore the figure ADEB is rectangular, and it has been demonftrated that it is equilateral; it is therefore a fquare, and it is defcribed upon the given ftraight line AB: Which was to be done. COR. Hence every parallelogram that has one right angle has all its angles right angles. IN any right angled triangle, the fquare which is defcribed upon the fide fubtending the right angle, is equal to the fquares defcribed upon the fides which con tain the right angle. Let ABC be a right angled triangle having the right angle BAC; the fquare defcribed upon the fide BC is equal to the fquares defcribed upon BA, AC. On BC deicribe the fquare BDEC, and on BA, AC the fquares GB, |