A D E b 37. T. angle ABC is equal to the triangle EBC, because it is upon Book I. the fame base BC, and between the fame parallels BC, AE: But the triangle ABC is equal to the triangle BDC; therefore also the triangle BDC is equal to the triangle EBC, the greater to the lefs, which is impoffible: Therefore AE is not parallel to BC. In the fame manner, it can be demonstrated that no other line but AD is parallel to BC; AD is therefore parallel to it. Wherefore equal triangles upon, &c. Q. E. D. B PROP. XL. THEOR. Estraight line, and towards the fame parts, are be QUAL triangles upon equal bafes, in the fame tween the fame parallels. Let the equal triangles ABC, DEF be upon equal bafes BC, EF, in the fame straight line BF, and towards the fame parts; they are between the fame parallels. Join AD; AD is parallel to BC: For, if it is not, through A draw 2 AG parallel to BF, and join GF: The triangle ABC is equal b to the triangle GEF, because they are upon equal bafes BC, EF, and between the fame parallels BF, AG: But the triangle ABC is equal to the triangle DEF; therefore also the triangle DEF is equal to the triangle GEF, the greater to the lefs, which is impoffible: Therefore AG is not parallel to BF: And in the fame manner it can be demonftrated that there is no other parallel to it but AD; AD is therefore parallel to BF. Wherefore equal triangles, &c. Q. E. D. I a parallelogram and triangle be upon the fame base, and between the fame parallels; the parallelogram shall be double of the triangle. Let Let the parallelogram ABCD and the triangle EBC be upon the fame bafe BC, and between the fame parallels BC, AE = the parallelogram ABCD is double of the triangle EBC. A DE Join AC; then the triangle ABC T B PROP. XLII. PROB. "O defcribe a parallelogram that fhall 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 defcribe a parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D. AF G Bifecta 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 because BE is equal to EC, the triangle ABE is likewife equal to the triangle AEC, fince they are upon equal bafes BE, EC, and between the fame parallels BC, AG; therefore the triangle ABC is double of the B triangle AEC: And the parallelogram FECG is likewife double of the triangle AEC, becaufe it is upon the fame bafe, and between the fame 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 defcribed a paralle e D E logram belogram 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. A H D K F Let ABCD be a parallelogram, of which the diameter is AC, and EH, FG the parallelograms about AC, that is, thro' which AC paffes, and BK, KD the other parallelo- E grams which make up the whole figure ABCD, which are therefore called the complements The complement BK is equal to the complement KD. B G C Because ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal a to a 34. Ka 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 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 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 ftraight line AB a parallelogram equal to the triangle C, and having an angle equal to D. Make b 31. 1. C 29. I. 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 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 far 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 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 angle ABM, and likewife 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 tri angle C, and has the angle ABM equal to the angle D: Which was to be done. C 43. I. f 15. 1. a 42. T. b 44. I. T PROP. XLV. PROB. O describe 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 equal to ABCD, and having an angle equal to E. John DB, and defcribe 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 GMequal to to the triangle DBC, having the angle GHM equal to the angle KHG but FKH, KHG are equal to two right angles; therefore alfo KHG, GHM D F GL E Book I. C 29. I. are equal to two right angles; and becaufe at the B скнм 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; 14. I. 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 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, FL 30. 1. 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. b PROP. |