Book I. a. 37. 1. b. 34. I. 2. 10. 1. b. 23. I. C. 31. I. Let the parallelogram ABCD and the triangle EPC be upon the fame bafe BC, and between the fame parallels BC, AE; the parallelogram ABCD is double of the A triangle EBC. DE parts; wherefore ABCD is alfo double of the triangle EBC. therefore if a parallelogram, &c. Q. E. D. PROP. XLII. PRO B. "O defcribe a parallelogram that fhall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. a AF G Let ABC be the given triangle, and D the given rectilineal angle. It is required to defcribe a parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D. Bifect BC in E, join AE, and at the point E in the straight line EC make the angle CEF equal to D; and thro' A draw "AG parallel to EC, and thro'C draw CG parallel to EF. therefore FECG is a parallelogram. and becaufe BE is equal to EC, the trid. 38. 1. angle ABE is likewife equal to the triangle AEC, fince they are upon equal bafes BE, EC and be c. 41. 1. d tween the fame parallels BC, AG;B D E C therefore the triangle ABC is dou- parallelogram 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 HE 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, Because ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal to the triangle ADC, and because EKHA. 34. f. 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 KGC 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 O a given straight line to apply a parallelogram, which fhall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle, Let AB be the given ftraight 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 hav◄ ing an angle equal to D. b. 31. 1. C. 29. I. c ftraight line with AB, and produce FG to H; and thro' A draw b 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 lefs than two right angles. but ftraight lines which with another straight line make the interior angles upon the d. 12. Ax. fame fide lefs than two right angles, do meet if produced far enough. therefore HB, FE fhall meet, if produced; let them meet in K, and thro' 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 equale 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 likewife to the angle D; 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. e. 43. 1. f. 15. 1. 2. 42. I. b. 44. I. T PROP. XLV. PRO B. O defcribe 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 defcribe a parallelogram equal to ABCD and having an angle equal to E. a Join DB, and defcribe 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 b the parallelogram GM equal to the triangle DBC having the angle GHM equal to the angle E. Book I. 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, A KHG are equal to the angies KHG, GHM. but FKH, KHG, GHM are B c CK HM equal to two 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 line with HM. and because the straight lined. 14. 1. 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 straight line with GL. and becaufe KF is parallel to HG, and HG to ML; KF is parallel to ML. and KM, e. 30. 1. 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 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 straight 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 straight line, a parallelogram equal b. 44. 7, to the first triangle ABD, and having an angle equal to the given angle, CA Book I. 2. II. I. b. 3. 1. C. 31. I. d. 34. I. C. 29. I. PROP. XLVI. PROB. To defcribe a square upon a given straight line. Let AB be the given straight line; it is required to describe a fquare upon AB. d c From the point A draw a AC at right angles to AB; and make b AD equal to AB, and thro' the point D draw DE parallel to it, and thro' 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 C ftraight lines BA, AD, DE, EB are equal to one another, and the parallelogram ADEB is equilateral. likewife all D 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 angles A B of parallelograms are equal; therefore each of the opposite 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 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. PROP. XLVII. THEOR. IN any right angled triangle, the square which is de- 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. 2. 46. . On BC defcribe the fquare BDEC, and on BA, AC the squares |