86. PROP. LXXXIX. IF 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. a 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 rea 2.2. mainder, to wit, the rectangle a 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, because the angle ABC, as also BEA, is given, the F 7 43. dat. triangle ABE is given b in species, and the ratio of AE to AB given : and because the B E D C ven, the ratio of the straight line BF to BA c 58. dat is given ; 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 e 35.1. AE, BC, that is, 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 a 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 are each f 87. dat. of them given f; and the ratio of FB to BA is given; therefore, AB, BC are given. * с The composition is as follorue; Let CHK be the given angle to which the angle of the paral lelogram 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 che parallelogram is to be made equal, and N 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 ratio of the remainder to the square of the other side be the same with the ratio of the square H KM L 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 of BC F above the square of BF bc 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, so make FB to BA, and complete the parallelogram AC, B E D C 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 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 the square of BF which has 8 to the square of BA the same g 22. 6. ratio which the square of NH has to the square of HG, because, as NH 10 HG, so FB was made to BA; but as HG to GK, so is BA to AE, because the triangle GHK is equiangular lo ABE; therefore, ex æquali, as NH to GK, so is FB to AE ; wherefore hh 1. 6. 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 to FB, BC; therefore i the rectangle GK, ÁL i 14. 5. 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 86th prop. in the Greek, and the composition 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 together with the space which has a given ratio to the square of the other be given, 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 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, a 43. dat. BEA are given, the triangle ABE is given a in species; there fore the ratio 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 b 58. dat. straight line BD to BA is given b; and the ratio of BÀ to AE c 9. dat. is given; therefore 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 the square of BC together with the D M d 88. dat. square of BD is given; therefore d because the rectangle contain. ed by the two straight 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 10 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 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 one of the sides of the parallelogram together with the space which has a given ratio to the square of the other side, is to be made equal ; and let this given ratio be the same which the square l 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 rectangle KG, GL: therefore, by the determination of the problem in that proposition, twice the rectangle MG, GK must 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 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, so is MG to GF; and as BA to AE, so GP to FH ; ex æquali, as DB to AE, so is MG to FH; there. fore, 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, BC, that is, the parallelogram AC, is equal to the rectangle FH, GK. IF a straight line drawn within a circle given in magnitude 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 given angle AEC ; the straight line AC is given in magnitude. Take D the centre of the circle, join AD and produce ita 1. 3. B E to E, and join EC: the angle ACE being 31. 3. a right b angle is given ; and the angle b c 43. dat. AEC is given ; therefore the triangle ACE is given in species, and the ratio of EA to AC is therefore given; and EA is given in d 5. def. magnitude, because the circle is given din 2. dat. magnitude; AC is therefore given e in mag. nitude. D А IF a straight line given in magnitude be drawn within 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 B E a 1. dat. is given, their ratio is given a ; and the angle ACE is a right angle, therefore the tri A b 46. dat. angle ACE is given b in species, and conse quently the angle AEC is given. 90. PROP. XCIII. IF from any point in the circumference of a circle given 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 position, let AB, AC be drawn to the circumference, ma king the given angle BAC ; if the point B А be given, the point C is also given. Take D the centre of the circle, and join BD, DC; and because each of the D a 29. dat. points B, D is given, BD is given a in po sition; and because the angle 'BAC is gi. B b 20. 3. ven, the angle BDC is given b, therefore a |