* See the Fgure of Prop. 35 The Demonftration of the other Cafes is the fame, but more plain and fimple. A. B SCHOLIU M. D If the Side AB of a rightlined Parallelogram ABCD, be conceived to be moved perpendicularly along the whole Line BC, or BC along the whole Line AB, the Area of the Rectang. ABCD fhall be produced by C that Motion. Whence a Rectang. is faid to be made by the Multiplication of two contiguous Sides into each other: for Example, if AB be 4 foot, and BC 3, multiply 3 into 4, and there will be produced 12 fquare Feet for the Area of the Rectangle. This being fuppofed, the Dimenfion of any Parallelogram (*EBCF) is found by this Theorem for the Area thereof is produced by multiplying the Altitude BA by the Bafe BC: for the Area of the Rectang. AC Pgr. EBCF, is made by multiplying BA by BC; therefore, &c. = PROP. XXXVI. с the one to the o H ther. Draw BE and CF: because BCGHT= EF; therefore BCFE is a Pgr. Whence the Pgr. BCDA BCFE' GHFE. 3 Q. E. D. PROP. PRO P. XXXVII. Triangles (BCA, BCD) conftituted upon the fame Bafe BC, and between the fame Parallels BC, EF, are equal the one to the other. Draw a BE parallel to CA, and a CF parallel a 31. 1. F to BD; then the Triang. BCA b = 1⁄2 BDFC" = BCD. QE.D. T BCAE b C 34. I. 35.1. & 7 ax. d Draw BG parallel to CA, and FH parallel to ED: then the Triang. BCA the Pgr. BGACEDHF'=EFD. QE.D. SCHOLIU M. If the Bafe BCEF, 'tis evident that the Triang. BAC EFD: and contrari wife. B CE Side, are between the fame Parallels. If you deny F this, let another તે = Line AH be parallel to BF, and draw FH. b a, b 34. I. c 6 ax. Draw the Line AC, then the Triangle. BCA 37. 1. BCE therefore the Pgr. ABCD 2 BCA =2BCE. QE. D. SCHOLIUM. Hence may the Area of any Triang. (BCE) be found for fince the Area of the Pgr. ABCD is produced by multiplying the Altitude into the Bafe, therefore the Area of a Triangle will be produced by multiplying half the Altitude into the Bafe, or half the Bafe into the Altitude: as fuppofe the Bafe BC be 8, and the Altitude 7, there the Area. of the Triang. BCE is 28. d Through A draw AG parallel to BC; make the Angle BCGD; bife&t f the Bafe BC in E, and draw EF parallel to CG, then is the Problem folved. For if AE be drawn, the Ang. ECG = D (by Conftr.) and the Triang. BAC = 2AEC h 38. 1. = Pgr. ECGF. QE. F 11 h 41. I. a 34. I. 3 ax. a For the Triang. ACD ACB, and the Triang. AGHAGE, and the Triang. GCF a =GCI: Therefore the Pgr. DG BG. Q. E. D. b To apply a Parallelogram FL to a given right Line A, in a given right-lined Angle C, equal to a given Triang. B. с = d Make a Pgr. FD=the Triang. B; so that the Ang. GFE be C, and produce GF till FH be A. Through H draw IL, parall. to EF, and produce DE to it in the Point I. Continue out DG to meet a right Line drawn from I in the Point K. Thro' K draw KL parallel to GH, which let EF, IH continued out meet in M, L. Then fhall FL be the Pgr. fought. e For the Pgr. FL- FD=B, and the Ang. MFH GFE C. Q.E. D. www PRO P. XLV. To make a Parallelogram FL in a given rightlined Angle E, on a given right Line FG, that shall be equal to a given right-lined Figure ABDC. Divide the given right-lined Figure into Triangles, as BAD, ACD; and make the Pgr. FH |