R * 8. 6. But if twice the given rectangle be not equal to the given sum of the squares of the sides, it must be less than it, as has been shewn. Let ABCD be the rectangle, join AC, and draw BE perpendicular to it, and complete the rectangle AEBF, and describe the circle ABC about the triangle ABC; AC is its diameter: * Cor. 5. 4. and because the triangle ABC is similar to AEB, as AC to CB, so is AB to BE; therefore the rectangle AC, BE, is equal to AB, BC; and the rectangle AB, BC, is given, wherefore AC, BE, is given: and because the sum of the squares of AB, BC is given, the square of AC which is equal to that sum is given; and AC * 47.1. itself is therefore given in magnitude; let AC be likewise given in position, and the point A; therefore AF is given in position and the rectangle AC, BE is given, as has been shewn, and AC is given, wherefore* BE is given in magnitude, as also AF which is equal to it; and AF is also : * B * 32 Dat. * 61 Dat. HL given in position, and the point A is given, wherefore * * 30 Dat. the point F is given, and the straight line FB in posi * tion and the circumference ABC is given in posi- * 31 Dat. tion, wherefore the point B is given. And the points * 28 Dat. A, C, are given; therefore the straight lines AB, BC, are given in position and magnitude. * * 29 Dat. The sides AB, BC of the rectangle may be found thus: let the rectangle GH, GK be the given space to which the rectangle AB, BC is equal; and let GH, GL, be the given rectangle to which the sum of the squares of AB, BC, is equal: find a square equal to * 14.2. the rectangle GH, GL: and let its side AC be given in position; upon AC as a diameter describe the semicircle ABC, and as AC to GH, so make GK to AF, and from the point A place AF at right angles to AC: therefore the rectangle CA, AF, is equal to GH, GK; 16. 6. and, by the hypothesis, twice the rectangle GH, GK, is less than GH, GL, that is, than the square of AC; wherefore twice the rectangle CA, AF, is less than the square of AC, and the rectangle CA, AF, itself less than half the square of AC, that is, than the rectangle contained by the diameter AC and its half; wherefore AF is less than the semidiameter of the circle, and * * consequently the straight line drawn through the point F parallel to AC must meet the circumference in two points: let B be either of them, and join AB, BC, and complete the rectangle ABCD; ABCD is the rectangle which was to be found: draw BE perpendicular to AC; therefore BE is equal to AF, and because the angle ABC in a semicircle is a right angle, the rectangle AB, BC is equal to AC, BE, that is, to the rectangle CA, AF, which is equal to the given rectangle GH, GK: and the squares of AB, BC, are together equal to the square of AC, that is, to the given rectangle GH, GL * * But if the given angle ABC of the parallelogram AC be not a right angle; in this case, because ABC is a given angle, the ratio of the rectangle contained by the sides AB, BC, to the parallelogram AC is given*; and AC is given, therefore the rectangle AB, BC, is given and the sum of the squares of AB, BC, is given: therefore the sides AB, BC, are given by the preceding case. D 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 ÈG 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, FK, be the given rectangle to which the sum of the squares of the sides is to be equal. And, by the preceding case, find the sides of a rectangle which is equal to the given rectangle EF, FH, and the squares of the sides of which are together equal to the given rectangle EF, FK; therefore as was shewn in that case, twice the rectangle EF, FH must not be greater than the st rectangle EF, FK: let it be so, BL E and let AB, BC, be the sides of the F HG K rectangle joined in the angle ABC equal to the given angle EFG, and complete the parallelogram ÅBCD, 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. -1997 art of erdada H elgastoen novia sd) of larg tadrenot ers -309% moviq edi of PROP. LXXXIX. A to 15sph 440 86. If two straight lines contain a given parallelogram in a given angle, and if the excess of the square of one of el them above a given space have 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. F * 43 Dat. 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 * * 2. 2. 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 triangle ABE is given in species, and the ratio of AE to AB is given: and because the ratio of the rectangle BC, CD, that is, of the square of BF to the 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 the ratio of AE to BF is given; as also the ratio of the rectangle AE, BC, that is *, of * 35. 1. 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 also given, for it is equal to the given rectangle CB, BD: therefore, *2.2. because the rectangle contained by the straight lines FB, * BED C * 58 Dat. * 9 Dat. BC, is given, and also the excess of the square of BC above the square of BF: FB, BC, are each of them i 87 Dat. given and the ratio of FB to BA is given; therefore AB, BC, are given. * 22.6. * The Composition is as follows.ions pito G 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 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 of the given straight line NH to the square of the given straight line HG. HKM F BED 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 above the square 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, 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 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 * to the 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 the rectangle NH, * 1.6. 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. 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. 0. 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 in species: there- * 43 Dat. 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 straight line BD to BA is given *; and * 58 Dat. the ratio of BA to AE is given; therefore* the ratio * 9 Dat. 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 square of BD is given: therefore* because the rectangle contained by the two straight lines DB, BC is given, and the sum of their squares is given; the straight D BE * 88 Dat. |
