add the rectangle CB, BD, and the fquare 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 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

N.

AP, BC, fo is (the straight line BA to AP, and so 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 rect, angle 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.

I magnitude,

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 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.

D

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 straight lines are unequal, twice the rectangle contained by them is lefs than the fum of their A fquares, as is evident from the 7. Prop. B. 2. Elem. B therefore twice the given space, to which space the rectangle of which the fides are to be found, is equal, must not be

greater than the given sum of the squares of the fides, and if twice

that fpace be equal to the given fum of the fquares, the fides of the rectangle muft neceffarily be equal to one another. therefore in this case describe a square 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 sum of the fquares of its fides AB, BC is equal to twice the rectangle AC, that is, by the hypothesis, 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 AEBF, and defcribe the circle ABC about the triangle ABC; AC is its diameter 2, a. Cor. 5.4 and because the triangle ABC is fimilar to AEB, as AC to CB, b. 8. 6, 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 because the sum of the fquares of AB, BC is given, the fquare of AC which is equal to that fum is given; and AC c. 47. I. itself is therefore given in magnitude. let AC be likewise given

D

d. 32. Dat,

e.

61. Dat.

in position, and the point A; therefore AF is given in pofition. and the rectangle AC, BE is given, as has been fhewn, and AC is given, wherefore e 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 8. and the circumference ABC is given in position, wherefore 8 31. Dat. h 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.

i

G K

TH

f. 30. Dat.

The fides 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 fum of the fquares of AB, BC is equal. find a square equal to k. 14. 2. the rectangle GH, GL, and let its fide AC be given in position; 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. therefore the rectangle CA, AF is equal to GH, L. 16. 6. GK; and, by the hypothefis, twice the rectangle GH, GK is lefs

1

b. 8. 6.

A

D

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
its half; wherefore AF is lefs than the fe-
midiameter of the circle, and confequently
the straight line drawn through the point F
parallel to AC must meet the circumference B
in two points. let B be either of them, and
join AB, BC and complete the rectangle
ABCD; ABCD is the rectangle which was GK

was

F

HL

to be found. draw BE perpendicular to AC; therefore BE is 34. 1. 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 e. 47. 1. GH, GK. and the fquares of AB, BC are together equal to the fquare 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 fides AB, BC to the parallelogram n. 62. Dat. AC is given "; and AC is given, therefore the rectangle AB, BC is given. and the sum of the squares of AB, BC is given; therefore the fides AB, BC are given by the preceding case..

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 sum of the squares of the

A

D

fides is to be equal. and, by the preceding cafe,
find the fides of a rectangle which is equal to

the given rectangle EF, FH, and the fquaresВ L

C

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

F HG K

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 fquares of AB, BC are together equal, by construction, to the given rectangle EF, FK.

PROP. LXXXIX.

F 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 fquare of the other; each of the straight lines fhall be given.

Let the two straight lines AB, BC contain the given parallelo. gram 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 ftraight lines AB, BC is given.

86.

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 square of BC, the remainder, to wit, the rectangle a BC, CD has a given ratio to the fquare of BA. draw a. 2. 2. AE perpendicular to BC, and let the fquare of BF be equal to the rectangle BC, CD. then because the angle

F

ABC, as alfo BEA is given, the triangle ABE is

given bin 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 BED

b. 43. Dat.

C

fquare of BA is given, the ratio of the straight line BF to BA is gi

d

ven C. and the ratio of AE to AB is given, wherefore the ratio of c. 58. Dat. AE to BF is given, as also the ratio of the rectangle AE, BC, that d. 9. Dat. is of the parallelogram AC to the rectangle FB, BC; and AC is e. 35. I. given, wherefore the rectangle FB, BC is given. and the excess of the fquare of BC above the fquare of BF, that is above the rectangle BC, CD is given, for it is equal a to the given rectangle CB, BD. therefore because the rectangle contained by the straight lines FB, BC is given, and alfo the excess of the square of BC above the

f. 87. Dat. fquare of BF; FB, BC are each of them given f. and the ratio of FB to BA is given; therefore AB, BC are given.

g. 22.6.

The Compofition is as follows.

N

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 the rectangle to which the parallelogram is to be made equal, and let LH, HM be the rectangle equal to the given fpace which is to be taken. from the fquare of one of the fides; and let the HKM

G

L

ratio of the remainder to the fquare of the other fide be the same with the ratio of the square of the given straight line NH to the fquare of the given ftraight line HG..

By help of the 87. 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 rec- A

tangle LH, HM; and join CB, BF in the an

gle FBC equal to the given angle GHK. and

F

as NH to HG, fo make FB to BA, and com-BED

C

plete 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 remain der fhall have to the fquare of BA the fame ratio which the fquare of NH has to the fquare of HG.

Because, by the construction, the square of BC is equal to the fquare of BF together with the rectangle LH, HM; if from the fquare of BC there be taken the rectangle LH, HM, there remains the square of BF which has to the fquare of BA the fame ratio which the square 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 rectangle GK, HL, as the rectangle FB, BC to AE, BC. but, by the conftruction, the rectangle NH, k. 14. 5. HL is equal to FB, BC; therefore the rectangle GK, HL is equal to the rectangle AE, BC, that is to the parallelogram AC.

h. 1. 6.

The Analyfis of this Problem might have been made as in the 86. Prop. in the Greek, and the compofition of it may be made as that which is in Prop. 87. of this Edition.