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 * 1. 6. CB is to the rectangle CB, BD, as 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 prob 14. 5. portionals; therefore the rectangle CB, BD is equal 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, © 22. 6. that is, LM, to LG: Therefore 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 d 8. 2. is equal 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 balf 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 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 HG, 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; MIN É PDC F G L O HN 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 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 CB, BD, is equal to HG, GL; therefore the rectangle AP, BC, that is, the parallelogram AC, is equal to the given rectangle EG, GH. N. PROP. LXXXVIII. If 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. 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 ta both their squares; but if two straight lines are A unequal, twice the rectangle contained by them is less than the sum of their squares, as is evident from the 7th Prop. B. 2. Elem. ; therefore B! С 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: And because the triangle Cor. 5.1 ABC is similar to AEB, as AC to CB, so is AB to BE; 18.6. 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 * 47. 1. given, the square of AC which is equal to that sum is given; and AC itself is therefore given in magnitude : Let AC be likewise given in posi D tion, and the point A ; therefore + 32 Dat. AF is given in position : And the rectangle AC, BE is given, as bas been shown, and AC is given, 61 Dat. whereforee BE is given in magni tude, as also Afwhich is equal to it; G HL * 30 Dat. the point A is given, whereforefthe 8 31 Dat. point F is given, and the straight line FB in positions: And the circumference ABC is given in position, whereb 28 Dat. foreh the point B is given. And the points A, C, are given; i 29 Dat. therefore the straight lines AB, BC are giveni in position and magnitude. 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 wbich the sum of the squares of AB, BC, is k 14. 2. equal: Findk a square equal to the rectangle GH, GL: And let its side AC be given in position ; upon AC as a diameter describe the seinicircle 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 1 16. 6. equal to GH, GK; and, by the hypothesis, twice the rect angle GH, GK, is less than G#, 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 in 34. 1. BE is equal m to AF, and because the angle ABC in a se8.6. micircle is a right angle, the rectangle AB, BC is equalb to AC, BE, that is, to the rectangle CA, AF, which is equal to the given rectangle GH, GK: And the squares of AB, • 47. 1. 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 a ; and AC is given, a 62 Dat. 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. The sides AB, BC, and the parallelogram AC, may be found thus: Let EFG be the givenangle 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, 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 wbich 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 É TIG K 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. 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 Let the two straight lines AB, BC, contain the given parallelogranı 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 rect angle CB, BD, he the given space; take this from the • 2. 2. square of BC, the remainder, to wit, the rectanglea 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 F 43 Dat. givenb in species, and the ratio of AE to AB is given : And because the ratio of А. the rectangle BC, CD, that is, of the square of BF to the square of BA, is BED given, the ratio of the straight line BF to € 58 Dat. BA is giveno; and the ratio of AE to AB is given, where. d9 Dat. fored the ratio of AE to BF is given; as also the ratio of * 35. 1. the rectangle AE, BC, that ise, 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 equala 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 787 Dat. each of them givenf; and the ratio of FB to BA is given; therefore AB, BC, are given. b 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 |