(35.) To divide a trapezium which has two sides parallel into any number of equal parts, by lines drawn parallel to those sides. K L A N M I/E G B P D R Let ABCD be the given trapezium having the sides AB, DC parallel. On AB the longer side describe a semicircle AFB; from D draw DE parallel to BC; with the centre B, and radius BE, describe the arc EF, and from F let fall the perpendicular FG; and divide AG into the given number of equal parts, e. g. three, in IH and I; and draw HK, IL at right angles to AB. Make BM, BN respectively equal to BL, BK; and draw MO, NP parallel to BC; and PQ, OR parallel to AB; and produce AD, BC to S. Since DC BE = BF, and OR = BM= BL, and PQ=BN=BK, the triangle ORS is to DSC in the duplicate ratio of OR to CD, or of BL to BF, i. e. in the ratio of BI : BG; whence ODCR : In the same manner PDCQ : DSC :: GI : GB. .. ODCR: and ODCR PDCQ :: GI: GH, i. e. in a given ratio of equality. And in a similar manner APQB may be shewn to be equal to PORQ. · of equal parts. COR. In nearly the same manner, the trapezium might be divided into parts having any given ratio. (36.) From one of the angular points of a given square to draw a line meeting one of the opposite sides, and the other produced, in such a manner, that the exterior triangle formed thereby may have a given ratio to the square. Let ABCD be the given square, and M: N the given ratio. From A to DC (produced if necessary) draw a line 40, OC B G such that M: M+N :: DC: AO. With the centre O and radius OA, describe a semicircle meeting DC produced in E and F. Join AF; which will be the line required. Join AE. Then M: M+N:: DC: AO :: ABCD: the rectangle AO, AD. Now the triangle ADE is similar to ABG, and equal to it, since AB = AD ; .. the trapezium AECG is equal to ABCD; and the rectangle AO, AD is equal to the triangle AEF, whence M: M+N :: AECG: AEF, .. M: N :: AECG: GCF :: ABCD: GCF (37.) From a given point in the side produced, of a given rectangular parallelogram, to draw a line which shall cut the perpendicular sides and the other side produced, so that the trapezium cut off, which stands on the aforesaid side, may be to the triangle which stands upon the produced part of the opposite side, in a given ratio. Let AKCD be the given rectangle, and E the given point in the side CD produced. On EC describe a semicircle, and in it place F K A B H M DG EF=ED; join FC; and divide EC in G, so that EG: GC in the given ratio, and draw GH at right angles to EC. In AK produced take BK a fourth proportional to EG, GH and FC. Join BE; it is the line required. For (Eucl. vi. 19.) the triangle ECM EDI :: EC2: ED2, ... div. CDIM: ECM :: EC - ED EC2, but ECM BMK :: EC : BK2, : : .. ex æquo CDIM: BMK :: EC - EF: BK2 (38.) Through a given point, between two straight lines containing a given angle, to draw a line which shall cut off a triangle equal to a given figure. B D I G H Let AB, AC be the lines containing the given angle BAC, and D the given point. Through D draw DE parallel to AC, and describe a parallelogram EG equal to the given figure. Draw GH perpendicular to AC, and equal to DE; and make HC DF; join CD, and produce it to meet AB in B; CB is the line required. For the triangles EBD, DIF, GIC being similar, are to one another in the duplicate ratio of the sides DE, DF, GC; but the square of HC is equal to the squares of HG, GC; and.. the square of DF is equal to the squares of DE, GC; whence the triangle DIF is equal to the triangles DBE, GIC; .. the triangle ABC is equal to AEFG, i. e. to the given figure. (39.) Between two lines given in position, to draw a line equal to a given line, so that the triangle thus formed may be equal to a given rectilineal figure. Let AB, AC be the lines. H given in position, and DE the line whose magnitude is given. Bisect it in F, and on DF describe a rectangular parallelo G K L B gram equal to the given figure. On DE describe a segment of a circle containing an angle equal to the angle at A, and cutting HG in I. Join DI, IE; and make AK = ID, and AL=IE. Join KL; it is the line required. Since AKID, and AL=IE, and the angle at A= DIE, .. KL = DE, and the triangle AKL = IDE= HGFD=the given figure. (40.) From two given lines to cut off two others, so that the remainder of one may have to the part cut off from the other a given ratio; and the difference of the squares of the other remainder and part cut off from the first may be equal to a given square. Perpendicular to AB one of the given lines, draw BC equal to a side of the given square; and take AD to the other given line in the given ratio of the part remaining Ꮓ from the first to the part cut off from the E B G For the difference between the squares of CG, GB is equal to the square of BC, i. e. to the given square; and AG GF :: AD AE, i. e. in the given ratio. (41.) From two given lines to cut off two others which shall have a given ratio, so that the difference of the squares of the remainders may be equal to a given square. Let AC be one of the two given lines. From C draw CD perpendicular to AC, and equal to a side of the given square. Take AE to the other given line in the given ratio of the parts to be cut off. Join ED, and produce it; and with the GE centre A, and radius equal to that other given line, describe a circle cutting ED in B. Join AB; and let it meet DF, which is parallel to AC, in F. Draw FG parallel to CD. CG and BF are the parts required to be cut off. For (DF) CG FB: EA AB, i. e. in the given ratio of the parts to be cut off; and the difference between the squares of FA and AG is equal to the square of GF, i. e. to the square of CD, or the given difference of the squares of the remainders. |