G) the rectangle to which the parallelogram N is to be 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 ratio of the H KM L remainder to the square of the other side 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 E square of BC above the square of BF A A be equal to the given rectangle LH, HM; and join CB, BF, in the angle FBC equal to the given angle GHK: BED c And as NH to HG, so 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 square of BC, the given rectangle LH, HM, be taken, the remainder shall haye 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 & to the < 22. 6. square of 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 æquali, as NH to GK, so is FB to AE; wherefore h 'the 1. 6. 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 the ' 14. 5. 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 86th Prop. in the Greek, and the composition of it may be made as that which is in Prop. 87th of this edition. 0. PROP, XC. 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 giver 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 be cause the angles ABE, BEA, are given, the triangle ABE * 13 Dat. is given a in species ; therefore 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 straight line BD to 658 Dat. BA is given b; and the ratio of BA to AE is given; there9 Dat. forec 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 D F BE GH ΚΙ ‘gle DB, BC is given ; and the square of BC together with * 38 Dat. the square of BD is given : therefored because the rectan gle contained 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 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 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 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 of the given straight line MG has to the square of GF. M FA 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 GF to FH; ex æquali, as DB to AE, so 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, BG, that is, the parallelogram AC, is equal to the rectangle FH, GK. 88. PROP. XCI. 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 * 1. . it to E, and join EC: The angle ACE В. triangle ACE is given in species, and A 89. : PROP. XCII. : 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 the circle ABC given in magnitude ; it shall cut off a segment containing a given angle. Take D the centre of the circle, join B. D angle, therefore, the triangle ACE is v 46 Dat. givenb in species, and consequently the angle AEC is given. E 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 making the given angle BAC: if the point B be given, D join BD, DC; and because each of the 29 Dat, points B, D, is given, BD is givena in position, and because the angle BAC is given, the angle 1. 20. 3. BDC is given, therefore because the straight line DC is drawn to the given point D, in the straight line BD given in position in the given angle BDC, DC is given in posi-'c 32 Dat. tion: And the circumference ABC is given in position, therefored the point C is given. d 28 Dat. 91. PROP. XCIV. If from a given point a straight line be drawn touching a circle given in position ; the straight line is given in position and magnitude. Let the straight line AB be drawn from the given point A, touching the circle BC given in position; AB is given in position and magnitude. Take D the centre of the circle, and join DA, DB : Because each of the points D, A is given, the straight line AD is givena in position and magni B 29 Dat. tude: And DBA is a rightb angle, wherefore DA is a diameter of the Cor. 5.4. circle DBA, described about the triangle DBA; and that circle is D A therefore given in position : And the circle BC is givend in position, there 16 Def. fore the point B is givene. The point A is also given: € 28 Dat. ... Therefore the straight line AB is givena in position and magnitude. b 18. 3. 99. PROP. XCV. If a straight line be drawn from a given point 'without a circle given in position; the rectangle contained by the segments betwixt the point and the circumference of the circle is given. Let the straight line ABC be drawn from the given point A without the circle BCD given in position, cutting it in B, C; the rectangle BA, AC, D is given. From the point A drawa AD touching the circle ; therefore ADC BA is given in position and magnitude: And because AD is given, the square of AD is givenc, which is equald to the rectangle BA, AC: Therefore the rectangle à 36.3. BA, AC, is given. * 17.3. b 94 Dat. ( 56 Dat. |