87 C. I. 6. PROP. LXXXVII . IF F two straight lines contain a parallelogram given in magnitude, in a given angle; if the excets of the square of the greater above the square of the lesier 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 ; if the excess of the square of BC above the square of BA be given; AB and BC are each of them given. Let the given excess of the square of BC above the square of BA be the rectangle CB, LD; take this from the square of BC, the remainder, which is a the rectangle BC, CD is equal to the square of AB. and because the angle ABC of the parallelogram AC is given, the ratio of the rectangle of the sides AB, BC to the parallelogram b. 62. Dat. AC is given b; and AC is given, therefore the rectang.e AB, EC is given; and the rectangle CB, BD is given; therefore the ratio of the rectangle CB, ED to the rectangle AB, BC, that is the ratio d. 54. Dat. of the straight line DB to BA is given; therefore the ratio of the square of DB to the square of BA is given. Α. C c. 7. Dat. tangle BC, CD to the square of BD; and, by composition, the ratio of four times the rectangle BC, CD together with the square of BD to the square of BD is given. but four times the rectangle BC, CD together with the square of BD is equal to the square of the straight lines BC, CD taken togetber; therefore the ratio of the square of BC, CD together to the square of BD is given; 5. 58. Dat. wherefore & the ratio of the straight line BC together with CD to BD is given. and, by composition, the ratio of BC together with CD and DB, that is the ratio of twice BC to BD is given; there. fore the ratio of BC to BD is given, as also the ratio of the square of BC to the rectangle CB, BD. but the rectangle CB, BD is given, being the given excess of the squares 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 fewn to be h. 9. Dat. given ; therefore the ratio of BC to BA is given ; and BC is gi ven, wherefore BA is given. 1 8. 2. 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 fides above the square of the other BA being given; to find the sides. and The Composition is as follows, Let EFG be the given angle to which the angle ABC is required to be equal, and from any point E in EE draw EG perpendicular to FG; let the rectangle EG, GH be the given space to which the paralle- M logram AC is to be made equal; and the rectangle HG, GL be the given K excess of the squares of BC, BA. E Take, in the itraight line GE, GK equal to FE, and make GM double of F G L O GK; join DiL, and in GL produced take LN equal to LM. bifect 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 GIE, and as LG to GK, fo 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, fo is OG to GL, the square of CB is to the rectangle CB, BD, as the rectangle HG, GO to the rectangle . 1. 6. 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 b to HG, GL. and because as CB. to BD, so is b. 14. 5. 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, fo is NL, that is LM, to LG. therefore the square of BC together with CD is c. 22. 6. to the square of BD), as the square of ML to the square of LG. but the square of BC and CD together is cqual d to four times the d. 9. 2. 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 b 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 becaule the angle AEP is equal to the angie 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, fo is the rectangle IIG, GL to HG, GK; and as the rectangle CB, BA to M K FG L O HN AP, BC, fo is (the straight line B A to AP, and so is FE or GK to EG, and so is) the rectangle HG, GK to HG, GE; therefore, ex aequali, asthe rectangle CB, BD to AP, BC, fu 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. PRO P. LXXXVIII. IT F 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 lides shall each of them be given. Let the two straight lines AB, BC contain the parallelogram ABCD given in magnitude in the given angle APC, and let the fum of the squares of AB, BC be given; AB, BC are each of them given. First, let ABC be a right angle; and because trive 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 sun of their A squares, as is evident from the 7. Prop. B. 2. Elem. C therefore twice the given space, to which space the fectingle of which the sides are to be found, is equal, must not be greater than the given fum 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 retangle AC is equal to the given space, and the sum of the fquues of its fides 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 fum of the squares of the tides, it must be less than it, as has been thewn. 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 '. a. Cor. 5. 6. and because the triangle ABC is fimilar to AEB, as AC to CB, 6. 8. 6. fo is AB to BE ; therefore the rectangle AC, BE is equal to AB, DC; and the rectangle AB, BC is given, wherefore AC, BE is given. and because the fam of the squares of AB, BC is given, the square of AC which is equal to that sum is given ; and AC c. 47. 1. itself is therefore given in magnitude. let AC be likewise given in position, and the point A ; therefore AF D is given din roîtion. and the rectangle AC, d. 32. Dat. BE is given, as has been shewn, and AC is given, wherefore · BE is given in mag c. 6. Dat. F nitude, as also AF which is equal to it; c and AF is also given in position, and the point A is given ; wherefore f the point F is given, and the straight line FB in posi G K FIL tion 8. and the circumference ABC is given in position, wherefore &. 31. Dat. b the point B is given. and the points A, C are given ; therefore h. 28. Dat. the straight lines AB, BC are given i in position and magni- i. 29. Dat. tude. 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, DC is equal. find k a square equal to k. 14. 2. the rectangle GH, GL, and let its side AC be given in position; upon AC as a diameter defcribe the semicircle ABC, and as AC to GH, fo 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, l. 16.6, GK; and, by the hypothesis, twice the rectangle GH, GK is less f. 30. Dat. m. 34 1. b. 8. 6. 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 A D its half; wherefore AF is less than the fe E F G К. HL to be found. draw BE perpendicular to AC; therefore BE is 6qual to AF, and because the angle ABC in a femicircle is a right angle, the rectangle AB, BC is equal to AC, BE, that is to the retangle CA, AF which is equal to the given rectangle GII, 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 2.62. Dat. AC is given a ; and AC is given, therefore the rectangle AB, BC is given, and the sum of the squares of AB, BC is given; therefore the files AB, BC are given by the preceding cate: Thc fides AB,BC and the parallelogran AC may be fuand thus. Ict ETG be the given angle of the paralelogram, and from in point E in FE draw EG perpendicular to FG. and let the recipe EG, Fil be the given space to which the parallelogian i, to be made equal, and kt ET, FK be the given rec D tangle to which the sum of the squares of the sides is to be equal. and, by the preceding case, And the fides of a rectangle which is equal to C B L the given rectangle EF, FH, and the squares of the lives of which are together equal to the E giren rectangle EF, FK. therefore, as was hewn in that cafe, twice the rectangle EF, FH must pot be greater than the rectangle EF, FK; let it be fo, and let AB, BC be the fides of the rectangle joined in the angle ABC equal to the F HG K given angle EFG ; and complete the parallelogram ABCD, which will be that which was to be found. draw AL perpendicular to |