because the angle ABL is equal to EFG, the triangle ABL is equiangular to EFG. and the parallelogram AC, that is the rectangle AL, BC is to the rectangle AB, BC as (the straight line AL to AB, that is as EG to EF, that is as) the rectangle EG, FH to EF, FH. and, by the construction, the rectangle AB, BC is equal to EF, FH, therefore the rectangle AL, BC, or, its equal, the parallelogram AC is equal to the given rectangle EG, FH. and the fquares of AB, BC are together equal, by conftruction, to the given rectangle EF, FK. IF PRO P. LXXXIX. two ftraight lines contain a given parallelogram in a given angle, and if the excefs of the fquare of one of them above a given space has a given ratio to the square of the other; each of the ftraight lines fhall be given. Let the two ftraight lines AB, BC contain the given parallelogram AC in the given angle ABC, and let the excefs of the fquare of BC above a given space have a given ratio to the fquare of AB; each of the straight lines AB, BC is given. 86. Because the excess of the fquare of BC above a given space has a given ratio to the fquare of BA, let the rectangle CB, BD be the given space; take this from the fquare of BC, the remainder, to wit, the rectangle BC, CD has a given ratio to the fquare of BA. draw a. 2. 2. AE perpendicular to BC, and let the fquare of BF be equal to the rectangle BC, CD. then because the angle ABC, as alfo BEA is given, the triangle ABE is F A given in fpecies, and the ratio of AE to AB given. and because the ratio of the rectangle BC, CD, that is of the fquare of BF to the BED b. 43. Dat. C c. 35. I fquare of BA is given, the ratio of the straight line BF to BA is given c. and the ratio of AE to AB is given, wherefore the ratio of c. 58. Dat. d. 9. Dat. AE to BF is given, as also the ratio of the rectangle AE, BC, that is of the parallelogram AC to the rectangle FB, BC; and AC is given, wherefore the rectangle FB, BC is given. and the excefs of the fquare of BC above the fquare of BF, that is above the rectangle BC, CD is given, for it is equal to the given rectangle CB, BD. therefore because the rectangle contained by the straight lines FB, BC is given, and also the excefs of the fquare of BC above the a fquare f. 87. Dat. fquare of BF; FB, BC are each of them given. and the ratio of FB to BA is given; therefore AB, BC are given. The Compofition is as follows. N Let GHK be the given angle to which the angle of the parallelogram is to be made equal, and from any point G in HG draw GK perpendicular to HK; let GK, HL be the rectangle to which the parallelogram is to be made equal, and let LH, HM be the rectangle equal to the given fpace which is to be taken from the fquare of one of the fides; and let the H KM G L ratio of the remainder to the fquare of the other fide be the fame with the ratio of the fquare of the given ftraight line NH to the fquare of the given ftraight line HG. A F By help of the 87. Dat. find two ftraight lines BC, BF which contain a rectangle equal to the given rectangle NH, HL, and fuch that the excess of the fquare of BC above the fquare of BF be equal to the given rectangle LH, HM; and join CB, BF in the angle FBC equal to the given angle GHK. and as NH to HG, fo make FB to BA, and complete the parallelogram AC, and draw AE perpendicular to BC. then AC is equal to the rectangle GK, HL; and if from the fquare of BC, the given rectangle LH, HM be taken, the remainder shall have to the fquare of BA the fame ratio which the fquare of NH has to the fquare of HG. BED C Because, by the conftruction, the fquare of BC is equal to the fquare of BF together with the rectangle LH, HM; if from the fquare of BC there be taken the rectangle LH, HM, there remains g. 11. 6. the fquare of BF which has to the fquare of BA the fame ratio which the fquare of NH has to the square of HG, because as NH to HG, so FB was made to BA; but as HG to GK, so is BA to AE, because the triangle GHK is equiangular to ABE; therefore, ex aeh. 1. 6. quali, as NH to GK, fo is FB to AE. wherefore the rectangle NH, HL is to the rectangle GK, HL, as the rectangle FB, BC to AE, BC. but, by the construction, the rectangle NH, HL is equal k. 14. 5. to FB, BC; therefore k the rectangle GK, HL is equal to the rectangle AE, BC, that is to the parallelogram AC. The Analysis of this Problem might have been made as in the 86. Prop. in the Greek, and the compofition of it may be made as that which is in Prop. 87. of this Edition. PROP. IF F two ftraight lines contain a given parallelogram in a given angle, and if the fquare of one of them be given together with the space which has a given ratio to the fquare of the other; each of the straight lines fhall be given. Let the two straight lines AB, BC contain the given parallelogram AC in the given angle ABC, and let the square of BC be given together with the space which has a given ratio to the square of AB; AB, BC are each of them given. 0. Let the fquare of BD be the space which has the given ratio to the fquare of AB; therefore, by the hypothefis, the square of BC together with the fquare of BD is given. from the point A draw AE perpendicular to BC, and because the angles ABE, BEA are given, the triangle ABE is given in fpecies; therefore the ratio of a. 43. Dat. · BA to AE is given. and because the ratio of the fquare of BD to the square of BA is given, the ratio of the straight line BD to BA is a given b; and the ratio of BA to AE is given, therefore the ratio of b. 58. Dat. AE to BD is given, as alfo the ratio of the rectangle AE, BC, that c. 9. Dat. is of the parallelogram AC to the rectangle DB, BC. and AC is given, therefore the rectangle DB, BC is given; and the fquare of BC together with the fquare of BD is given. therefore because the d. 88. Dat. rectangle contained by the two ftraight lines DB, BC is given, and the fum of their squares is given; the straight lines DB, BC are each of them given. and the ratio of DB to BA is given; therefore AB, BC are given. The Compofition is as follows. Let FGH be the given angle to which the angle of the parallelogram is to be made equal, and from any point F in GF draw FH perpendicular to GH; and let the rectangle FH, GK be that to which the parallelogram is to be made equal; and let the rectangle KG, GL be the space to which the fquare of one of the fides of the parallelo 1 parallelogram together with the space which has a given ratio to the fquare of the other fide, is to be made equal; and let this given ratio be the fame which the fquare of the given straight line MG has to the fquare of GF. By the 88. Dat. find two ftraight lines DB, BC which contain a rectangle equal to the given rectangle MG, GK, and fuch that the fum of their squares is equal to the given rectangle KG, GL. therefore, by the determination of the Problem in that Propofition, twice the rectangle MG, GK muft not be greater than the rectangle KG, GL. let it be fo, and join the ftraight lines DB, BC in the angle DBC equal to the given angle FGH. and as MG to GF, so make 88. DB to BA, and complete the parallelogram AC. AC is equal to the rectangle FH, GK; and the fquare of BC together with the fquare of BD which by the conftruction has to the fquare of BA the given ratio which the fquare of MG has to the fquare of GF, is equal, by the construction, to the given rectangle KG, GL. Draw AE perpendicular to BC. Because as DB to BA, fo is MG to GF; and as BA to AE, so GF to FH; ex aequali, as DB to AE, fo is MG to FH. therefore as the rectangle DB, BC to AE, BC, fo is the rectangle MG, GK to FH, GK. and the rectangle DB, BC is equal to the rectangle MG, GK; therefore the rectangle AE, BC, that is the parallelogram AC, is equal to the rectangle FH, GK. 2. a ftraight line drawn within a circle given in magnitude cuts off a fegment which contains a given angle; the straight line is given in magnitude. In the circle ABC given in magnitude, let the ftraight line AC be drawn cutting off the fegment AEC which contains the given angle AEC; the straight line AC is given in magnitude, Take D the center of the circle, join AD and produce it to E, and 1 and join EC. the angle ACE being a right b B IF PROP. XCII. a ftraight line given in magnitude be drawn within a circle given in magnitude; it shall cut off a fegment containing a given angle. Let the straight line AC given in magnitude be drawn within the circle ABC given in magnitude; it fhall cut off a fegment containing a given angle. Take D the center of the circle, join AD and produce it to E, and join EC. and becaufe each of the ftraight lines EA, AC is given, their ratio is given; and the angle ACE is a right angle, therefore the triangle ACE is given in fpecies, and confequently the angle AEC is given. PROP. XCIII. A B E D a. 1. Dat. C b. 46. Dat. IF from any point in the circumference of a circle given in pofition two ftraight lines be drawn meeting the circumference and containing a given angle; if the point in which one of them meets the circumference again be given, the point in which the other meets it is alfo given. From any point A in the circumference of a circle ABC given in pofition, let AB, AC be drawn to the circumference making the given angle BAC; if the point B be given, the point C is alfo given. Take D the center of the circle, and join BD, DC. and because each of the points B, D is given, BD is given in pofition. and because the angle BAC is given, the angle BDC is given. therefore because the straight line B D 90. Ca. 29. Dat. b. 20.3. DC |