1 59. a 18. 6. b 26. 6. c 56. dat. d 36. and 43. I € 82. dat. This demonftration is the analysis of the problem in the 28th prop. of book 6. the construction and demonftration of which propofition is the compofition of the analysis; and be caufe the given fpace AC or its equal the gnomon ELH is to be taken from the figure EF defcribed upon the half of AB fimilar to BC, therefore AC must not be greater than LF, as is fhewn in the 27th prop. b. 6. IF Let the parallelogram AC equal to a given space be applied to the given ftraight line AB, exceeding by the parallelogram BDCL given in species; each of the straight lines CD, DB are given. a GFH Bifect AB in E; therefore EB is given in magnitude: Upon EB defcribe the parallelogram EF fimilar to LD, and fimilarly placed; therefore EF is given in fpecies, and is about the fame diameter with LD. Let CBG be the diameter, and conftruct the figure: Therefore, because the figure EF given in fpecies is described upon the given A ftraight line EB, EF is given in magnitude, and the gnomon ELH is equal to the given figured AC; wherefore, E B D KLC fince EF is increafed by the given gnomon ELH, its fides EK, FH are given; but EK is equal to CD, and FH to BD; therefore CD, DB are each of them given. This demonftration is the analysis of the problem in the 29th prop. book 6. the conftruction and demonstration of which is the compofition of the analysis. COR. If a parallelogram given in fpecies be applied to a gi ven ftraight line, exceeding by a parallelogram equal to a given fpace; the fides of the parallelogram are given. Let the parallelogram, ADCE given in fpecies be applied to the given straight line AB exceeding by the parallelogram BDCG equal to a given fpace; the fides AD, DC of the parallelogram are given. Draw E a Draw the diameter DE of the parallelogram AC, and conftruct the figure. Because the parallelogram AK is equal to a 43. I. BC which is given, therefore AK is given; and BK is fimilar to AC, therefore BK is given in fpecies. And fince the parallelogram AK given in magnitude is applied to the given ftraight line AB, exceeding by the parallelogram BK given in fpecies, therefore, by this pro- A H G C b 24. 6. K BD pofition, BD, DK the fides of the excefs are given, and the ftraight line AB is given; therefore the whole AD, as alfo DC, to which it has a given ratio, is given. PROB. To apply a parallelogram fimilar to a given one to a given ftraight line AB, exceeding by a parallelogram equal to a given fpace. с To the given straight line AB apply the parallelogram AK c 29. 6. equal to the given space, exceeding by the parallelogram BK fi• milar to the one given. Draw DF the diameter of BK, and through the point A draw AE parallel to BF meeting DF produced in E, and complete the parallelogram AC. b The parallelogram BC is equal to AK, that is, to the given fpace; and the parallelogram AC is fimilar to BK; therefore the parallelogram AC is applied to the ftraight line AB fimilar to the one given and exceeding by the parallelogram BC which is equal to the given space. PROP. LXXXV. IF two ftraight lines contain a parallelogram given in magnitude, in a given angle; if the difference of the ftraight lines be given, they fhall each of them be given. Let AB, BC contain the parallelogram AC given in magnitude, in the given angle ABC, and let the excefs of BC above AB be given; each of the straight lines AB, BC is given. Let DC be the given excefs of BC above BA, therefore the remainder BD is equal to BA. Complete the parallelogram AD; A E 84. and because AB is equal to BD, the ratio of AB to BD is given; and the angle ABD is given, therefore the parallelogram AD is B D C given in fpecies; and because the given parallelogram AC is applied to the given ftraight line DC, exceeding by the paral ·lelogram AD given in fpecies, the fides of the are given herefo 85. therefore BD is given; and DC is given, wherefore the whole BC is given And AB is given, therefore AB, BC are each of them given. PROP. LXXXVI. IF two ftraight lines contain a parallelogram given in magnitude, in a given angle; if both of them together be given, they fhall each of them be given. Let the two ftraight lines AB, BC contain the parallelogram AC given in magnitude, in the given angle ABC, and let AB, BC together be given; each of the ftraight lines AB, BC is given. E A Produce CB, and make BD equal to AB, and complete the parallelogram ABDE. Because DB is equal to BA, and the angle ABD given, because the adjacent angle ABC is given, the parallelogram AD is given in fpecies: And because AB, BC together are given, and AB is equal to BD; therefore DC is given: And becaufe the gi-D ven parallelogram AC is applied to the given B C ftraight line DC, deficient by the parallelogram AD given in 83. dat. fpecies, the fides AB, BD of the defect are given; and DC is given, wherefore the remainder BC is given; and each of the ftraight lines AB, BC is therefore given." 2. PROP. LXXXVII. IF two ftraight lines contain a parallelogram given in magnitude, in a given angle; if the excefs of the fquare of the greater above the fquare 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. 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 b 69. dat. of the fides AB, BC to the parallelogram 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 the rect angle the ratio of the d 54. dat. A angle CB, BD to the rectangle AB, BC, that is, the ratio of the ex. 6.ftraight line DB to BA is given; therefore fquare of DB to the fquare of BA is gi ven: And the fquare of BA is equal to the rectangle BC, CD: wherefore the ratio of the rectangle BC, CD to the fquare of BD is given, as alfo the ratio of four BPD C times the rectangle BC, CD to the fquare of BD; and, by compofition, the ratio of four times the rect. 7. dat. angle BC, CD together with the fquare of BC to the fquare of BD is given: But four times the rectangle BC, CD together with the fquare of BD is equal to the fquare of the ftraight f 8. 2. lines BC, CD taken together; therefore the ratio of the fquare of BC, CD together to the fquare of BD is given; wherefore the ratio of the ftraight line BC together with CD to BD is 8 58. dát. 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 excefs of the fquares of BC, BA; therefore the fquare of BC, and the ftraight line BC is given: And the ratio of BC to BD, as alfo of BD to BA has been shown to be given; therefore the ratio of BC to BA is h 9. dat. given; and BC is given, wherefore BA is given. The preceding demonftration is the analyfis of this problem, viz. A parallelogram AC which has a given angle ABC being gìven in magnitude, and the excefs of the fquare of BC one of its fides above the fquare of the other BA being given; to find the fides: 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 FGLO HN Take, in the ftraight line GE, GK equal to FE, and make GM double of 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 a z. 6. b 14.5. C 22. 6. d 8. 2. to CFE, 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 fquares of CB, BA is equal to the rec angle HG, GL. Because as CB to BD, fo is OG to GL, the fquare of CB is to the rectangle CB, BD as the rectangle HG, GO to the rectangle HG, GL: And the fquare of CB is equal to the rectangle HG, GO, because GO, BC, GH are proportionals; therefore the rectangle CB, BD is equal to HG, GL. And because as CB to BD, fo is 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 is, LM, to LG: Therefore the fquare of BC together with CD is to the square 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 rectangle BC, CD together with the fquare of BD; therefore four times the rectangle BC, CD together with the square 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 fquare 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 the rectangle BC, CD is equal to the fquare of AB. To each of these 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, so is the rectangle HG, M A B PD C FG LO HN GL to HG, GK; and as the rectangle CB, BA to AP, BC, fo is (the ftraight line BA to AP, and fo is FE or GK to EG, |
