87. 2.2.2. IF PRO P. LXXXVII. F two ftraight lines contain a parallelogram given in magnitude, in a given angle; if the excefs of the fquare of the greater above the square of the leffer be given, each of the straight lines fhall be given. Let the two ftraight lines AB, BC contain the given parallelogram AC in the given angle ABC; if the excefs of the fquare of BC above the fquare of BA be given; AB and BC are each of them given. 2 Let the given excefs of the fquare of BC above the fquare of BA be the rectangle CB, BD; take this from the fquare of BC, the remainder, which is the rectangle BC, CD is equal to the fquare of AB. and because the angle ABC of the parallelogram AC is given, the ratio of the rectangle of the fides AB, BC to the parallelogram b. 62. Dat. AC is given; and AC is given, therefore the rectangle AB, BC is given; and the rectangle CB, BD is given; therefore the ratio of c. 1. 6. the rectangle CB, BD 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 f. 8. 2. fquare of DB to the fquare of BA is given. A c given, as also the ratio of four times the rect. B PD C e. 7. Dat. angle BC, CD to the fquare of BD; and, by compofition, the ratio of four times the rectangle BC, CD together with the fquare of ED to the fquare of BD is given. but four times the rectangle BC, CD together with the fquare of BD is equal f to the fquare of the ftraight lines BC, CD taken together; therefore the ratio of the fquare of BC, CD together to the fquare of BD is given; whereg. 58. Dat. fore the ratio of the ftraight line BC together with CD to BD is given. and, by compofition, the ratio of BC together with CD and DB, that is the ratio of twice BC to BD is given; therefore the ratio of BC to BD is given, as alfo the ratio of the fquare of BC to the rectangle CB, BD. but the rectangle CB, BD is given, being the given excess of the fquares of BC, BA; therefore the square of BC, and the ftraight line BC is given, and the ratio of BC to BD, b. 9. Dat. as also of BD to BA has been fhewn to be given; therefore the ratio of BC to BA is given; and BC is given, wherefore BA is given. The viz. The preceeding Demonstration is the Analysis of this Problem, A parallelogram AC which has a given angle ABC being given in magnitude, and the excess of the fquare of BC one of its fides above the fquare of the other BA being given; to find the sides. and The Compofition is as follows, M Let EFG be the given angle to which the angle ABC is required to be equal, and from any point E in FE 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 excefs of the fquares of BC, BA. K F Take, in the straight line GE, GK FG LO equal to FE, and make GM double of HN GK; join ML, 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, fo make CB to BD; and make the angle CBA equal to GFE, 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 excefs of the fquares of CB, BA is equal to the rectangle HG, GL. Because as CB to BD, fo is OG to GL, the fquare of CB is to a the rectangle CB, BD, as a the rectangle HG, GO to the rectangle a. 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 b CB, BD is equal to HG, GL. and because as CB to BD, fɔ is b. 14. 5. OG to GL, twice CB is to BD, as twice OG, that is GN, to GL; and, by divifion, as BC together with CD is to BD, fo is NL, that c is LM, to LG. therefore the fquare of BC together with CD is c. 22. 6. to the fquare of BD, as the fquare of ML to the fquare of LG. but the fquare of BC and CD together is equal to four times the d. 8. a. *rectangle BC, CD together with the fquare of BD; therefore four times the rectangle BC, CD together with the fquare of BD is to the fquare of BD, as the fquare of ML to the fquare of LG. and, by divifion, four times the rectangle BC, CD is to the fquare of BD, as the fquare of MG to the square of GL; wherefore the rectangle BC, CD is to the fquare of BD, as (the fquare of KG the half of MG to the fquare of GL, that is as) the fquare of AB to the fquare of BD, because as LG to GK, fo DB was made to BA. therefore b the rectangle BC, CD is equal to the fquare of AB; to each of these N. add the rectangle CB, BD, and the fquare of BC becomes equal to the fquare of AB together with the rectangle CB, BD. therefore this rectangle, that is the given rectangle HG, GL is the excess of the fquares 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, fo is the rectangle HG, GL to HG, GK; and as the rectangle CB, BA to AP, BC, fo is (the straight line BA to AP, and fo is FE or GK to EG, and fo is) the rectangle HG, GK to HG, GE; therefore, ex aequali, as the rectangle CB, BD to AP, BC, fo 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. IF two ftraight lines contain a parallelogram given in magnitude, in a given angle; if the fum of the fquares of its fides be given, the fides fhall each of them be gi ven. Let the two ftraight lines AB, BC contain the parallelogram ABCD given in magnitude in the given angle ABC, and let the fum of the fquares of AB, BC be given; AB, BC are each of them given. A D C 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 ftraight lines are unequal, twice the rectangle contained by them is lefs than the fum of their fquares, as is evident from the 7. Prop. B. 2. Elem. therefore twice the given space, to which fpace the rectangle of which the fides are to be found, is equal, greater than the given fum of the fquares of the fides. B muft not be and if twice that that space be equal to the given fum of the fquares, the fides of the rectangle must neceffarily be equal to one another. therefore in this cafe defcribe a fquare ABCD equal to the given rectangle, and its fides AB, BC are those which were to be found. for the rectangle AC is equal to the given space, and the fum of the fquares of its fides AB, BC is equal to twice the rectangle AC, that is, by the hypothefis, to the given space to which the fum of the fquares was required to be equal. But if twice the given rectangle be not equal to the given fum of the fquares of the fides, it must be lefs than it, as has been fhewn. Let ABCD be the rectangle, join AC and draw BE perpendicular to it, and complete the rectangle A EBF, and describe the circle ABC about the triangle ABC; AC is its dia- a. Cor. 5.4. meter 2. and because the triangle ABC is fimilar b to AEB, as AC b. 8. 6. to CB, fo 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 becaufe the fum of the fquares of AB, d F c d. 32. Dat. c. 61. Dat. BC is given, the fquare of AC which is equal to that fum is gi- c. 47. 1. ven; and AC itself is therefore given in magnitude. let AC be likewife given in pofition, and the point A ; A therefore AF is given in pofition. and the rectangle AC, BE is given, as has been fhewn, and AC is given, wherefore BE is given in magnitude, as alfo AF which is equal to it; and AF is alfo given in pofition, and the point A is given; wherefore f the point F is given, and the straight line FB in pofition . and the circumference ABC is given in pofition, g. 31. Dat. wherefore h the point B is given. and the points A, C are given; h. 28. Dat. therefore the straight lines AB, BC are given i in pofition and mag- i. 29. Dat. nitude. G K HL f. 30. Dat. The fides AB, BC of the rectangle may be found thus; let the rectangle GH, CK be the given space to which the rectangle AB, BC is equal; and let GH, GL be the given rectangle to which the fum of the fquares of AB, BC is equal. find k a fquare equal to k. 14. 2. the rectangle GH, GL, and let its fide AC be given in pofition; upon AC as a diameter defcribe the femicircle 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, 1. 16. 6. GK; and, by the hypothefis, twice the rectangle GH, GK is lefs F B I than GH, CL, that is than the fquare of AC; wherefore twice the rectangle CA, AF is lefs than the fquare of AC, and the rectangle CA, AF itself lefs than half the fquare of AC, that is than the rectangle contained by the diameter AC and its half; wherefore AF is lefs than the femidiameter of the circle, and confequently the ftraight line drawn thro' 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 femicircle 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. c. 47. 1. and the fquares of AB, BC are together equal to the fquare of AC, that is to the given rectangle GH, GL. m. 34. 1. b. 8. 6. G K HL But if the given angle ABC of the parallelogram AC be not a right angle, in this cafe because ABC is a given angle, the ratio of the rectangle contained by the fides AB, BC to the parallelogram n. 62. Dat. AC is given n; and AC is given, therefore the rectangle AB, BC is given. and the fum of the fquares of AB, BC is given; therefore the fides AB, BC are given by the preceeding cafe. A 4 D The fides 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 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 fum of the fquares of the fides is to be equal. and, by the preceeding cafe, find the fides of a rectangle which is equal to the given rectangle EF, FH, and the fquares of the BL fides of which are together equal to the given rectangle EF, FK. therefore, as was fhewn in that cafe, twice the rectangle EF, FH must not 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 given E FHG K angle EFG; and complete the parallelogram ABCD, which will be that which was to be found, draw AL perpendicular to BC, and because |