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

PROP. LXXXVIII.

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 given.

Let the two straight 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

Firft, let ABC be a right angle; and because twice the rect angle contained by two equal ftraight lines is equal to both their fquares; 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 7th prop. b. 2. Elem. therefore twice the given space, to which space the rectangle of which the fides are to be found is equal, muft not be greater than the given fum of the fquares of the fides: And if twice that space be equal to the given fum of the fquares, the fides of the rectangle muft 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 thofe which were to be found: For the rectangle AC is equal to the given space, and the sum 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.

N.

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 shown. Let ABCD be the rectangle, join AC and draw BE perpendicular to it, and complete the rectangle. AEBF, and defcribe the circle ABC about the triangle ABC; AC is its diameter: And because the triangle ABC is fimi- a Cor. 5. 4lar to AEB, as AC to CB, fo is AB to BE; therefore the 8. 6. rectangle AC, BE is equal to AB, BC; and the rectangle AB,

BC

€ 47. I.

A

D

BC is given, wherefore AC, BE is given: And because the fun of the fquares of AB, BC is given, the fquare of AC which i equal to that fum is given; and AC itself is therefore give in magnitude: Let AC be likewife given in position, and the 32. dat. point A ; therefore AF is given in pofition: And the rectangle AC, BE is given, as has been fhewn, and AC is 61. dat. given, wherefore BE is given in magnitude, as alfo AF which is equal to it; F and AF is alfo given in pofition, and the point A is given; wherefore the point F is given, and the ftraight line G K FB in pofition: And the circumference ABC is given in pofition, wherefore the point B is given: And the points A, C are given; therefore the ftraight lines AB, BC are given in position and magnitude.

1 30 dat

8 31

Lat.

12. dat.

i 29. dat.

k 14. 2.

1 16. 6.

34. X.

b 8.6.

C

HL

The Gides AB, BC of the rectangle may be found thus: Let the rectangle GH, GK be the given space to which the rect angle AB, BC is equal; and let GH, GL be the given rect. angle to which the fum of the fquares of AB, BC is equal: Find a fquare equal to 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, fo make GK to AF, and from the point A place AF at right angles to AC: There fore the rectangle CA, AF is equal to GH, GK; and, by the hypothefis, twice the rectangle GH, GK is lefs than GH, GL, 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 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 EE 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: And the fquares of AB, BC are together equal to the fquare of AC, that is, to the given rectangle GH, GL.

с

But

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 AC is given"; and AC is given, therefore the rect- n 62. dat. angle AB, BC is given; and the fum of the fquares of AB, BC is given; therefore the fides AB, BC are given by the prece ding cafe.

A

E

C

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 preceding cafe, find the fides of a rectangle which is equal to the given B L rectangle EF, FH, and the fquares of the 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 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 ftraight 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 fquares of AB, BC are together equal, by conftruction, to the given rectangle EF, FK.

F HG K

80.

a 2. 2.

€ 35. I.

PROP. LXXXIX.

I two ftraight lines contain a given parallelogram in a given angle, and if the excefs of the fquare of one of them above a given fpace, has a given ratio to the fquare of the other; each of the traight lines fhall be given.

Let the two ftraight lines AB, BC contain the given paralle logram AC in the given angle ABC, and let the excess of the fquare of BC above a given fpace have a given ratio to the fquare of AB, each of the ftraight lines AB, BC is given.

a

Because the excefs of the fquare of BC above a given space has a given ratio to the fquare of BA, let the rectangle CB, BD be the given space; take this from the fquare of BC, the remainder, to wit, the rectangle BC, CD has a given ratio to the fquare of BA: Draw AE perpendicular to BC, and let the fquare of BF be equal to the rectangle BC, CD, then be cause the angle ABC, as alfo BEA, is given,

A

F

BED

b 43. dat. the triangle ABE is given in fpecies, and the ratio of AE to AB given: And because the ratio of the rectangle BC, CD, that is, of the fquare of BF to the fquare of BA, is given, the ratio of the ftraight line BF to BA c 58. dat. is given; and the ratio of AE to AB is given, wherefore d the d 9. dat. ratio of AE to BF is given; as alfo the ratio of the rectangle AE, BC, that is, of the parallelogram AC to the rectangle FB, BC; and AC is given, wherefore the rectangle FB, BC is given. The excefs of the fquare of BC above the fquare of BF, that is, above the rectangle BC, CD, is given, for it is equal to the given rectangle CB, BD; therefore, because the rectangle contained by the ftraight lines FB, BC is given, and also the excefs of the fquare of BC above the fquare of BF; FB, BC f 87. dat. are each of them given f; and the ratio of FB to BA is given ; therefore, AB, BC are given.

The Compofition is as follows.

Let GHK be the given angle to which the angle of the pa rallelogram is to be made equal, and from any point Gin HG, draw GK perpendicular to HK; let GK, HL be the rect

angle

N

angle to which the parallelogram is to be
made equal, and let LH, HM be the rect-
angle equal to the given fpace which is to
be taken from the fquare of one of the fides;
and let the ratio of the remainder to the HKM
fquare of the other fide be the fame with the

L

ratio of the fquare of the given straight line NH to the fquare of the given ftraight line HG.

F

By help of the 87th dat. find two ftraight lines BC, BF which contain a rectangle equal to the given rectangle NH, HL, and fuch that the excefs of the fquare of BC above the fquare 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, HL; and if from the fquare of BC, the given rectangle LH, HM be taken, the remainder fhall have to the fquare of BA the fame ratio which the fquare of NH has to the fquare of HG.

BED

C

Because, by the conftruction, the fquare of BC is equal to the fquare of BF together with the rectangle LH, HM; if from the square of BC there be taken the rectangle LH, HM, there remains the fquare of BF which has to the fquare of g 22. 6. BA the fame ratio which the fquare of NH has to the fquare of HG, because, as NH to HG, fo FB was made to BA; but as HG to GK, fo is BA to AE, because the triangle GHK is equiangular to ABE; therefore, ex aequali, as NH to GK fo is FB to AE; wherefore the rectangle NH, HL is to the rect- k 1. 6. angle GK, HL, as the rectangle FB, BC to AE, BC; but by the conftruction, the rectangle NH, HL is equal to FB, BC; therefore the rectangle GK, HL is equal to the rectangle AE, k 14. 5. BC, that is, to the parallelogram AC.

h

The analysis of this problem might have been made as in the 86th prop. in the Greek, and the compofition of it may be made as that which is in prop. 87th of this edition.