PROP. XLI. THEOR. GEN. ENUN.-If a parallelogram and a triangle be upon the same base, and between the same parallels, the parallelogram is double of the triangle. DEMONST.-Then the m ABCD = 2 A ABC, because the diam. AC bisects it (Prop. XXXIV.) = 2 EBC, upon the same base BC, and between same || s. (Prop. XXXVII.) Wherefore, if a □m and a &c.—Q. E. D. Hence the areas of ▲ are readily computed; for since the area of a m the base its altitude, that of a base the altitude. Δ = From this Proposition the following are deductions : PROP. VV. THEOR. GEN. ENUN.-The two triangles formed by drawing straight lines from any point within a parallelogram to the extremities of two opposite sides, are together equal to half the parallelogram. PART. ENUN.-Let ABCD be a ", and E any pt. within C B CONST. Through E draw FEG to AC or BD. (Prop. XXXI.) DEMONST.-1. Then the ▲ AEC the ABED = = ACGF, and FGDB (Prop. XLI.); .. ▲ aec + ABCD (AX. 1). BED = AAEB+CED = 2. Subtract these from the wholem ABCD; ABCD. Wherefore the two As formed &c.-Q. E. D. PROP. WW. THEOR. GEN. ENUN.-If two sides of a trapezium be parallel, the triangle contained by either of the other sides, and the two straight lines drawn from its extremities to the bisection of the opposite side, is equal to half the trapezium. PART. ENUN.-Let ABCD be a trapezium, having the side AB to the side CD. Bisect AC in E (Prop. X.), and join EB, ED. F A B Then trapezium Through E C draw FEG to BD, meet G ing CD in G, and BA produced in F. (Prop. XXXI.) DEMONST.-1. Since the alternate FAE = alternate ECG (Prop. XXIX.), the vertical ▲ AEF = vertical CEG (Prop. XV.); and, adjacent to = 4, the side AE = side Ec (Const.); .. the A AEF = A ECG. (Prop. XXVI.) 2. Add to each the figure AEGDB; .. the " FGDB = trapezium ABCD. 3. Hence the ▲ BED = altitude (Prop. XLI.) = Wherefore, if two sides FGDB, with same base and trapezium ABCD. &c.-Q. E. D. COR.-Hence a trapezium, which has two of its sides is = to am formed by drawing, through the bisection of one of the sides which are not , a line to the opposite side, and meeting the sides. PROP. XX. THEOR. GEN. ENUN.-The perimeter of an isosceles triangle is greater than the perimeter of a rectangular parallelogram, which is of the same altitude with, and equal in area to, the given triangle. DEMONST.-1. Because AB = AC, and BD = DC, .. the perimeter of A B ABC = 2 AC + 2 DC = 2 (AC + DC); and the perimeter of the " ADCE = 2 AD + 2 DC = = 2 (AD + DC). 2. But, because ADC is a rt. 4d A. AC is > AD (Prop. XIX.); .'. AC' + DC is > AD + DC; and .. the perimeter of A ABC is > perimeter of ADCE. Wherefore the perimeter &c.-Q. E. D. PROP. YY. PROB. GEN. ENUN.-To describe a parallelogram, of which the area and perimeter shall be respectively equal to the area and perimeter of a given triangle. PART. ENUN.-Let ABC be the gn. A; then it is required K CONST.-Produce BA to D, and make AD AC. (Prop. III.) Bisect BD in E. (Prop. X.) Through a draw AF to BC. (Prop. XXXI.) With cr. B, and rad. BE, describe a ting AF in G (Post. 3). Join BG; bi. sect BC in H, and through a draw нF to BG (Prop. XXXI.); then BGFH is the required. DEMONST. 1. Since HF + BG 2 BE = BD = BA + BG = BE (Def. 15); .'. HF AC. (Const.) 2. And since GF = BH = HC (Const.); .. GF+BH = BH, m BGFH = the perimeter of BGFH = double of a ▲, upon the base and between the same || (Prop. XLI.), (Prop. NN). = Δ ABC Wherefore am has been described, whose area and pethe area and perimeter of the gn. ▲ ABC. rimeter Q. E. F. PROP. XLII. PROB. GEN. ENUN.-To describe a parallelogram which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. ; A PART. ENUN.-Let ABC be the gn. A, D the gn. rectilineal then it is required to describe a m = A ABC, and having one one of its s = ≤ D. CONST.-Bisect BC in E (Prop. X.), and join AE (Post. 1); at the pt. E in the st. line EC make CEF = D (Prop. XXIII.); through a draw AFG || to BC, and through c draw CG || to EF (Prop. XXXI.); then FECG is the m required. DEMONST.-1. Because BE EC, and AG is I to BC, ... the a ABE = ▲ AEC (Prop. XXXVIII.); ... the ▲ ABC = 2 A AEC (AX.9) = m FECG. (Prop. XLI.) 2. And them FECG has the CEF = the <D. (Const.) Wherefore am has been described = the gn. ▲ ABC, and having an ≤ = the gn. ≤ D.-Q. E. F. To this may be added PROP. ZZ. PROB. GEN. ENUN.-To describe a triangle which shall be equal to a given parallelogram, and having an angle equal to a given rectilineal angle. PART. ENUN.-Let ABCD be the gn. m, E the gn. rectilineal ; then it is re = quired to describe a ▲ Om ABCD, and having an CONST.-At the pt. c, in the st. line DC, make the DCF = E (Prop. XXIII.), and let cr meet AB produced in F; pro A B duce CD to G, making DG = CD (Prop. III.), and join FG; then FCG is the ▲ required. Join FD. DEMONST.-Because CD = DG, .'. ▲ FCD = Δ FDG (Prop. XXXVIII.) ; .'. ▲ FCG = 2 ▲ FCD (Ax. 6), = □m ABCD. (Prop. XLI.) |