But if the given angle ABC of the parallelogram AC be not à right angle, in this case, because ABC is a given angle, the ratio of the rectangle contained by the fides AB, BC to the parallelogram AC is given"; and AC is given, therefore the rect. n 62. dat. angle AB, BC is given ; and the fum of the squares of AB, BC is given; therefore the sides AB, BC are given by the prece. ding case. The sides AB, BC and the parallelogram AC may be found thus: Let EFG be the given angle of the parallelogram, and from any point E in FE draw EG perpendicular to FG ; and let the rectangle EG, FH be the given space to which the parallelogram is to be made equal, and let EF, A D FK be the given rectangle to which the sum of the squares of the fides is to be equal. And, by the preceding case, find the sides of a rectangle which is equal to the given B L C rectangle EF, FH, and the squares of the fides of which are together equal to the gi E ven rectangle EF, FK; therefore, as was fewn in that case, twice the rectangle EF, FH must not be greater than the rectangle EF, FK; let it be ro, and let AB, BC be the fides of the rectangle joined in the F HG K angle ABC equal to the given angle EFG; and complete the parallelogram ABCD, which will be that which was to be found : Draw AL perpendicular to 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 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 squares of AB, BC are together equal, by construction, to the given rectangle EF, FK. 80. PRO P. LXXXIX. IK two straight lines contain a given parallelogram in a given angle, and if the excess of the square of one of thein above a given space, has a given ratio to the square of the other; each of the traight 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 rečiangle 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 AE perpendicular to BC, and let the square of BF be equal to the rectangle BC, CD, then be. cause the angle ABC, as also BEA, is given, F b 43. dat. the triangle ABE is given 6 in species, and A А given, the ratio of the straight line BF to BA ¢ 58. dat is giveno; and the ratio of AE to AB is given, wherefore d the d 9. dat. ratio of AE to BF is given ; as also the ratio of the rectangle € 35. I. AE, BC, that is, e of the parallelogram AC to the rectangle FB, BC; and AC is given, wherefore the rectangle FB, BC is given. 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 square of BF; FB, BC f 87. dat. are each of them given f; and the ratio of FB 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 red angle angle to which the parallelogram is to be N 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 Gides ; and let the ratio of the remainder to the H KM L . square of the other Gde be the same with the ratio of the square of the given straight line NH to the square of the given straight line HG. By help of the 87th dat. find two straight lines BC, BF which contain a rectangle equal to the given rectangle NH, HL, and such that the excess of the square F of BC above the square of BF be equal to the given rectangle LH; HM; and join CB, BF in che angle FBC equal to the given angle GHK: And as NH to HG, so make B ED C 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 square of BC, the given rectangle LH, HM be taken, the remainder shall have to the square of BA the same ratio which the square of NH has to the square of HG. Because, by the conftruction, 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 the square of BF which has & to the square of 8 22. 6. BA the same ratio which the square 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 aequali, as NH to GK fo is FB to AE; wherefore the rectangle NH, HL is to the rect. h s. 6. angle GK, HL, as the rectangle FB, BC to AE, BC; but by the construction, the rectangle NH, HL is equal to FB, BC; therefore the rectangle GK, HL is equal to the rectangle AE, k 14. S. BC, that is, to the parallelogram AC. The analygis of this problem might have been made as in the 86th prop. in the Greek, and the compofition of it may be made as that which is in prop. 87th of this edition. IF two straight lines contain a given parallelogram in a given angle, and if the square of one of them to. gether with the space which has a given ratio to the square of the other be given, each of the straight lines Tha'l be given. 2 43. dat. C g. dat. Let the two straight lines AB, BC contain the given parallelogram AC in the given angle ABC, and let the square of BC together with the space which has a given ratio to the square of AB be given, AB, BC are each of them given. Let the square of BD 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. From the point A, draw AE perpendicular to BC; and because the angles ABE, BEA are given, the triangle ABE is given a in species; therefore the ratio of BA to AE is given : And because the sa tio of the square of BD to the square of BA is given, the rato 38. dat. tio of the straight line BD to BA is given b; and the ratio of BA to AE is given; therefore the ratio of AE to BD is gi. ven, 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 the square of M, F GH K L C & 88. dat. BC together with the square of BD is given; therefore & be. cause the rectangle contained by the two straigh: lines DB, BC is given, and the sum 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 composition is as follows. Let FGH be the given angle to wbich the angle of the parallelogram is to be made equal, and from any point Fin 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 square of D Α Τ A T of one of the sides of the parallelogram together with the space which has a given ratio to the square of the other lide, is to be made equal; and let this given ratio be the same which the square of the given straight line MG has to the square of GF. By the 88th 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 rect. angle KG, GL; therefore, by the determination of the problem in that propofition, twice the rectangle MG, GK inuft not be greater than the rectangle KG, GL. Let it be so, and join the straight lines DB, BC in the angle DBC equal to the given angle FGH; and, as MG to GF, so make DB to BA, and complete the parallelogram AC: AC is equal to the rect angle FH, GK; and the square of BC together with the square of BD, which, by the construction, has to the square of B A 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, fo 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 MG, GK to FH, GK; and the rectangle DB, BC is equal to the rectangle MG, GK; therefore the rectangle AE, PC, that is, the parallelogram AC, is equal to the rectangle FH, GK. IF a a straight line drawn within a circle given in magni tude 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 straight line AC be drawn, cutting off the segment AEC which contains the given angle ALC; the straight line AC is given in magnitude. Take D the centre of the circle, join AD and produce it a 1. 3. |