(27.) To divide a given straight line into two such parts, that the square of the one may be to the excess of a given rectangle above the square of the other, in a given ratio. Let AB be the given straight line. From B draw BC at right angles to AB, and make AB : BC2 in the given ratio. Join AC. Find a mean proportional between the sides of the given rectangle; E B and with it as radius, and B as centre describe a circle cutting AC in D. Join BD, and draw DE parallel to BC; E is the point required. For (Eucl. vi. 2.) AE: ED :: AB2: BC. Now the square of ED is equal to the difference of the squares of BD and BE, i. e. to the difference of the given rectangle and the square of BE;.. the square of AE is to the difference between the given rectangle and the square of BE as AB': BC, i. e. in the given ratio. N.B. The given rectangle must not be less than the square of the perpendicular from B upon AC; and when BD is less than BC, there are two points E. (28.) From any angle of a triangle, not isosceles about the angle, to draw a line without the triangle to the opposite side produced, which shall be a mean proportional between the segments of the side. Let ABC be the triangle, and B the angle from which the mean proportional is to be drawn. About the triangle describe a circle, and to the point B Y draw a tangent BD meeting the side CA produced in D. BD is a mean proportional between AD and DC. (Eucl. iii. 36.) the rectangle AD, DC is equal to the square of DB; and .. AD : DB :: DB : DC. (29.) From the obtuse angle of any triangle, to draw a line within the triangle to the opposite side, which shall be a mean proportional between the segments of the side. D E F B Let ABC be a triangle having the obtuse angle ABC. Describe a circle about it, and produce BA to D, making AD=AB. From D draw DE parallel to AC, meeting the circle in E; join BE, cutting AC in F; BF will be a mean proportional between AF and FC. For (Eucl. vi. 2.) BF : FE :: BA : AD, and since BA=AD, :. BF=FE. Now the rectangle AF, FC is equal to the rectangle BF, FE, i. e. to the square of BF; .. AF: FB :: FB : FC. (30.) From the common extremity of the diameters of two semicircles given in magnitude and position; to draw a line meeting the circumferences, so that the rectangle contained by the two chords may be equal to a given square. Let AB, AC be the diameters drawn from A, and given in magnitude and position. With the centre A, and radius equal to a side of the given square, describe a circle, cutting the lesser semicircle in D. Draw DE perpendicular to AC, and meeting the other semicircle in F. Join AF, and produce it to G; AG is the line. required. G D B For joining GC, the triangles AGC, AFE are similar, .. AC : AG :: AF : AE, and.. the rectangle FA, AG is equal to the rectangle CA, AE, i. e. to the square of AD, which is equal to the given square. (31.) To draw a line parallel to a given line, which shall be terminated by two others given in position, so as to form with them a triangle equal to a given rectilineal figure. Let AB, AC be the lines given in position, AD the line to which it is required to draw a parallel. Describe a rectangular parallelogram AEFG equal to the given figure. Produce EF E H to H; and take AK a mean proportional between DH and 2 EF; draw KC parallel to AD; KC is the line required. For the angles DHA, CAK being equal, as also DAH, ACK, the triangles DAH, AKC are equiangular, and similar; whence 2 AKC: AHD :: AK' : DH2 :: 2 EF : DH :: 2 EF × AE : DH × AE. Now the rectangle DH, AE is double of the triangle AHD, .. AKC is equal to the rectangle EF, AE, i. e. to the given rectilineal figure. (32.) To bisect a triangle by a line drawn parallel to one of its sides. Let ABC be a triangle to be bisected by a line parallel to its side AB.. On BC describe a semicircle; bisect BC in O, and draw the perpendicular B A F E OD; join CD; and with C as centre, and radius CD, describe a circle cutting CB in E; draw EF parallel to AB; EF bisects the triangle. (Eucl. vi. 8.) BC : CD :: CD : CÓ, .. BC2 : (CD2=) CE2 :: BC : CO :: 2 : 1; but the triangles ABC, FEC are in the duplicate ratio of BC: CE, and in the ratio of 2: 1, i. e, EFC is half of ABC, and EF bisects the triangle, ג (33.) To divide a given triangle into any number of parts having a given ratio to each other, by lines drawn parallel to one of the sides of the triangle. B M. E K C Let ABC be the given triangle; divide AC into parts AE, EF, FC having the same ratio to one another that the parts of the triangle are to have. On AC describe a semicircle, and draw the perpendiculars EG, FH; and with the centre A, and radii AG, AH, describe circles meeting AC in I and K, from which points draw IL, KM parallel to BC; these will divide the triangle in the ratio required. For the triangles ALI, AKM, ABC are to one another in the duplicate ratio of the sides AI, AK, AC, i. e. in the ratio of the rectangles AC, AE; AC, AF; and the square of AC; or in the ratio of the lines AE, AF, AC; whence ALI, LIKM, MKCB are in the ratio of AE, EF, FC, i. e. in the given ratio. (34.) To divide a given triangle into any number of equal parts by lines drawn parallel to a given line. Let ABC be the given triangle; from the angle CMX draw CD parallel to the N PD given line; and let it be re- LEVHARM OĞ quired to divide the triangle R into five equal parts. On T AD, BD describe semi circles AID, BMD; divide AB into five equal parts in the points E, F, G, H; draw EI, FK, GL, HM perpendicular to AB; and make AN, AO, AP respectively equal to AI, AK, AL, and BQ=BM; and draw NR, OS, PT, QV, parallel to DC; they divide the triangle as required. (Eucl. vi. 1.) the triangle ABC: ADC :: AB: AD, (Eucl. vi. 19.) ACD : ANR :: AD2 : AN2 :: AD : AE. .. ex æquo, ABC ANR: AB: AE :: 5 : 1, i. e. AFR is one fifth of ABC. In the same manner ABC : AOS :: 5 : 2, And by a similar manner, OPTS and BQV, may each be shewn to be one fifth of ABC, .. TPQV will also` be one fifth of ABC. COR. In nearly the same manner the triangle may be divided into any number of parts having a given ratio. |