BC, and 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 ftraight 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 squares of AB, BC are together equal, by construction, to the given rectangle EF, FK. PRO P. LXXXIX. 86. IF F two straight lines contain a given parallelogram in a given angle, and if the excess of the square of one of them above a given space has a given ratio to the square of the other ; each of the straight lines shall be given. Let the two straight lines AB, BC contain the given parallelogram AC in the given angle ABC, and let the excess of the square of BC above a given space have a given ratio to the square of AB; each of the straight lines AB, BC is given. Because the excess of the square of BC above a given space has a given ratio to the square of BA, let the rectangle CB, BD be the given space; take this from the square of BC, the remainder, to wit, the rectangle BC, CD has a given ratio to the square of BA. draw a. 2. 2. AE perpendicular to BC, and let the square of BF be equal to the rectangle BC, CD. then because the angle ABC, as alfo BEA is given, the triangle ABE is A b. 43. Dat. given b in fpecies, and the ratio of AE to AB 1 given, and because the ratio of the rectangle BC, CD, that is of the square of BF to the BED с square of BA is given, the ratio of the straight line BF to BA is given, and the ratio of AE to AB is given, wherefore d the ratio of c. 58. Dat. AE to BF is given, as also the ratio of the rectangle AE, BC, that d. 9. Dat. is of the parallelogram AC to the rectangle FB, BC; and AC is c. 35. 1. given, wherefore the rectangle FB, BC is given. and the excess of the square of BC above the square 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 excess of the square of BC above the N f. 87. Dat. square of ET; FB, BC are each of them given f. and the ratio of TB to BA is given ; therefore AB, BC are given. The Composition is as follows. 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 G made equal, and let LH, HM be the rectangle equal to the given space which is to be taken from the square of one of the sides; and let the HKM ratio of the remainder to the square of the other side he the same with the ratio of the square of the given straight line NHL to the square of the given straight line IIG. By help of the 87. Dat. find two straight lines BC, BF which contain a rectangle equal to the given rectangie NH, HL, and such that the excess of the square of BC above the square of BF be equal to the given rec. A tangle LH, IIM; and join CB, BF in the angle FBC equal to the given angle GHK. and as NH to HG, fo make 1'B to BA, and com-BED C plete the parallelogram AC, and draw AE perpendicular to BC. then AC is equal to the rectangle GK, HL; and if from the square of BC, the given rectangle LH, IIM be taken, the remainder Arall have to the square of BA the same ratio which the square of NH has to the square of HG. Because, by the construction, the square of BC is equal to the square of BF together with the rectangle LH, HM; if from the square of BC there be taken the rectangle LH, HM, there remains 8. 22. 6. the square of EF which has 8 to the square of BA the fame ratio which the square of NH has to the square of HG, because as NH to HG, fo FB was made to BA; but as HG to GK, fo is BA to AE, because the triangle GHK is equiangular to ABE; therefore, ex aequali, as NH to GK, fo is FB to AE. wherefore h 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, k. 14. 5. HL is equal 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 composition of it may be made as that which is in Prop. 87. of this Edition. b. 1. 6. PRO P. XC. . F two straight lines contain a given parallelogram in a given angle, and if the square of one of them be given together with the space which has a given ratio to the square of the other ; each of the straight lines shall 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. Let the square of ED be the space which has the given ratio to the square of AB; therefore, by the hypothesis, the square of BC together with the square of BD is given. froin the point a draw AE perpendicular to BC, and because the angles ABE, BEA are given, the triangle ABE is given ' in species; therefore the ratio a. 43. Dat. of BA to AE is given. and because the ratio of the square of BD to the square of BA is given, the ratio of the straight line BD to to BA is given b; and the ratio of BA to AE is given, therefore b. 58. Dat. c the ratio of AE to BD is given, as also the ratio of the rectangle AE, BC, that is of the parallelogram AC to the rectangle DB, BC. and AC is given, therefore the rectangle DB, BC is given ; and c. 9. Dat. BE IC L the square of BC together with the square of BD is given. therefore d because the rectangle contained by the two firaight lines d. 88. Dat. DB, BC is givell, and the sum of their squares is given ; the straight lines D3, BC are each of them given. and the ratio of DB to BA is given ; therefore AB, BC are given. The Composition 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 TH 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 square of one of the fides of the parallelogram together with the space which has a givea ratio to the square of the other side, is to be made equal; and let this given ratio be the same which the square of the given Itraigar line MG has to the square of GF. By the 83. Dat. find two straight lines DB, BC which contain a rectangle equal to the given rectangle MG, GK, and such that the sum 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 must not be greater than the rectangle KG, GL. let it be fo, and join the straight lines DB, BC in the angle DBC equal to the given angle FGH. and as MG to BE С GH K L GF, fo make DB to BA, and complete the parallelogram AC. AC is equal to the rectangle FH, GK; and the square of BC together with the square of BD which by the construction has to the square of BA the given ratio which the square of MG has to the square 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, GF to FH; ex aequali, as DB to AE, fo is MG to FH. therefore as the rectangle DB, BC to AE, BC, so is the rectangle NG, GK to FII, GK. and the rectangle DB, BC is equal to the rectangle DG, GK; therefore the rectangle AE, BC, that is the parallelogram AC, is equal to the rectangle FH, GK. P R O P. XCI. IF F a straight line drawn within a circle given in mag nitude cuts off a segment which contains a given angle; the straight line is given in magnitude. In the circle ABC given in magnitude, let the straight line AC be drawn cutting off the segment AEC which contains the giren angle AEC ; the straight line AC is given in magnitude. Take D the center of the circle, join AD and produce it to 89. PRO P. XCII. in a circle given in magnitude ; it shall cut off a segment containing a given angle. Let the straight line AC given in magnitude be drawn within the circle ABC given in magnitude; it shall cut off a fegment containing a given angle. Take D the center of the circle, join AD B and produce it to E, and join EC. and because each of the straight lines LA, AC is D given, their ratio is given'; and the angle A C ACE is a right angle, therefore the triargle ACE is given b in species, and confequently the angle AEC is given. a. I. Dat. b. 46. Dat. IT from any point in the circumference of a circle given PRO P. XCIII. go F in position two straight 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 also given. From any point A in the circumference of a circle ABC given in podation, let AB, AC be drawn to the circumference making the given angle BAC; if the point B be gi A ven, the point C is also 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 a in position. and because the angle BAC is given, the angle BDC is given b. therefore because the straight line Ca. 29. Dat. b. 20. 3. |