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rectangle CB, BD is given, being the given excess of the square of BC, BA; therefore the square of BC, and the straight line BC, is given: and the ratio of BC to BD, as also of BD to BA, has been shown to be given; therefore (9. dat.) the ratio of BC to BA is given; and BC is given, wherefore BA is given.

The preceding demonstration is the analysis of this problem, viz. A parallelogram AC which has a given angle ABC being given in magnitude, and the excess of the square of BC one of its sides above the square of the other BA being given; to find the sides: and the composition is as follows.

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Let EFG be the given angle to which the angle ABC is required to be equal, and from any point E in EF, draw EG perpendicular to FG; let the rectangle EG, GH be the given space to which the parallelogram AC is to be made equal; and the rectangle HG, GL be the given excess of the squares of BC, BA.

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Take, in the straight line GE, GK equal to FE, and make GM double of GK: join ML, and in GL produced, take LN equal to LM: bisect GN in O, and between GH, GO find a mean proportional BC: as OG to GL, so make CB to BD; and make the angle CBA equal to GFE, and as LG to GK, so make DB to BA; and complete the parallelogram AC: AC is equal to the rectangle EG, GH, and the excess of the squares of CB, BA is equal to the rectangle HG, GL.

Because as CB to BD, so is OG to GL, the square of CB is to the rectangle CB, BD as (1. 6.) the rectangle HG, GO to the rectangle HG, GL and the square of CB is equal to the rectangle HG, GO, because GO, BC, GH are proportionals; therefore the rectangle CB, BD is equal (14. 5.) to HG, GL. And because as CB to BD, so is OG to GL; twice CB is to BD, as twice OG, that is GN, to GL: and, by division, as BC together with CD is to BD, so is NL, that is, LM, to LG: therefore (22. 6.) the square of BC together with CD is to the square of BD, as the square of ML to the square of LG: but the square of BC and CD together is equal (8. 2.) to four times the rectangle BC, CD together with the square of BD; therefore four times the rectangle BC, CD together with the square of BD is to the square of BD, as the square of ML to the square of LG: and, by division, four times the rectangle BC, CD is to the square of BD, as the square of MG to the square of GL; wherefore the rectangle BC, CD is to the square of BD as (the square of KG the half of MG to the square of GL, that is, as) the square of AB to the square of BD, because as LG to GK, so DB was made to BA: therefore (14. 5.) the rectangle BC, CD is equal to the square of AB. To each of these add the rectangle CB, BD, and the square of BC becomes equal to the square of AB together with the rectangle CB, BD; therefore this rectangle, that is, the given rectangle KG, GL, is the excess of the squares of BC, AB. From the point A, draw AP perpendicular to BC, and because the angle ABP is equal to

the angle EFG, the triangle ABP is equiangular to EFG: and DB was made to BA, as LG to GK; therefore as the rectangle CB, BD to CB, BA, so is the rectangle HG, GL to HG, GK; and as the rectangle CB, BA to AP, BC, so is (the straight

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line BA to AP, and so is FE or GK to EG, and so is) the rectangle HG, GK to HG, GE; therefore ex æquali, as the rectangle CB, BD to AP, BC, so is the rectangle HG, GL to EG, GH: and the rectangle C3, BD is equal to HG, GL: therefore the rectangle AP, BC, that is, the parallelogram AC, is equal to the given rectangle EG, GH.

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Ir two straight lines contain a parallelogram given in magnitude, in a given angle; if the sum of the squares of its sides be given, the sides shall each of them be given.

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Let the two straight lines AB, BC contain the parallelogram ABCD given in magnitude in the given angle ABC, and let the sum of the squares of AB, BC be given; AB, BC are each of them given. First, let ABC be a right angle; and because twice the rectangle contained by two equal straight lines is equal to both their squares; but if two straight lines are unequal, twice the rectangle contained by them is less than the sum of their squares, as is evident from the 7th prop. book 2, Elem.; therefore twice the given space, to which space the rectangle of which the sides are to be found is equal, must not be greater than the given sum of the squares of the sides: and if twice that space be equal to the given sum of the squares, the sides of the rectangle must necessarily be equal to one another; therefore in this case describe a square ABCD equal to the given rectangle, and its sides AB, BC are those which were to be found: for the rectangle AC is equal to the given space, and the sum of the squares of its sides AB, BC is equal to twice the rectangle AC, that is, by the hypothesis, to the given space to which the sum of the squares was required to be equal

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 shown. 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 (8. 6.) to AEB, as AC to CB, so is AB to BE; therefore the rectangle

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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 (47. 1.) to that sum is given; and AC itself is therefore given in magnitude: let AC be likewise given in position, and the point A; therefore AF is given (32. dat.) in position: and the rectangle AC, BE is given, as has been shown, and AC is given, wherefore (61. dat.) BE is given in magnitude, as also AF which is equal to it; and AF is also given in position, and the point A is given; wherefore (30. dat.) the point F is given, and the straight line FB in position (31. dat.): and the circumference ABC is given in position, wherefore (28. dat.) the point B is given and the points A, C are given; therefore the straight lines AB, BC are given, (29. dat.) in position and magnitude.

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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 (14. 2.) a square equal to 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 (16. 6.) to GH, GK; 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 (34. 1.) to AF and because the angle ABC in a semicircle is a right angle, the rectangle AB, BC is equal (8. 6.) 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 (47. 1.) 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 (62. dat.); and AC is given, therefore the rectangle AB, BC is given; and the sum of the square of AB, BC is given; therefore the sides AB, BC are given by the preceding case.

The sides AB, BC and the parallelogram AC may be found thus: let EFG be the given angle of the parallelogram, and from

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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, FK be the given rectangle to which the sum of the square 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 shown in that case, twice the rectangle EF, FH must not be greater than the rectangle EF, FK; let it be so, and ⚫let AB, BC be the sides of the rectangle joined in the 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.

PROP. LXXXIX.

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86

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.

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 (2. 2.) BC, CB 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 (43. dat.) 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 (58. dat.); and the ratio of AE to AB is given, wherefore (9. dat.) the ratio of AE to BF is given; as also the ratio of the rectangle AE to BC, that is, (35. 1.) of the paral

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lelogram 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 (3. 2.) 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 of them given (87. dat.); and the ratio of FB to BA is given; therefore, AB, BC are given.

The composition is as follows:

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HKM

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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. 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, AL; 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.

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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 (22. 6.) 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 (1. 6.) the rectangle NH, HM 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 (14. 5.) 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 86th prop. in the Greek, and the composition of it may be made as that which is in prop. 87th of this edition.

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Ir two straight lines contain a given parallelogram in a given angle, and if the square of one of them together with the space

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