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of BC, the remainder, which is a the rectangle BC, CD is e. a 2. 2. qual to the square of AB; and because the angle ABC of the par Helogram AC is given, the ratio of the rectangle of the sides AB, BC to the parallelogram AC is given b; and AC b 62. da. is given, therefore the rectangle AB, BC is given ; and the rectangle CB, BD is given ; therefore the ratio of the rectangle CB, BD to the rectangle AB, BC, that is, the ratio of the c 1.6. straight line DB to BA is given: therefored the ratio of the d 54. dat: square of DB to the square of BA is given, А and the square of BA is equal to the recto angle BC, CD: wherefore the ratio of the rectangle BC, CD to the square of BD is given, as also the ratio of four times the B PD rectangle BC, CD to the square of BD ; and, by compositione, the ratio of four times the rectangle BC, CD e 7. dat: together with the square of BC to the square of BD is given : but four times the rectangle BC, CD together with the square of BD is equalf to the square of the straight lines BC, CD f8. 2. taken together: therefore the ratio of the square of BC, CD together to the square of BD is given ; wherefore the ratio of 8 58. dat. 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 ; therefore 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 shown to be given; therefore h the ratio of BC to BA is given ; and BC is given, b 9, dat wherefore BA is given.

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 sides 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 FE, draw EC perpendicular to FG ; let the rect M angle EG, GH be the given space to which the parallelogram AC is to be made equal; and the rectan

K gle HG, GL, be the given excess

I of the squares of BC, BA.

Takt, in the straight line GE,
GK equal to FE, and make GM F G L O HN
double of GK: join ML, and in GL produced, take LN equal to
LM : bisect GŃ in 0, and between GH, GO find a mean pro-

portional BC : as OG to GL, so make CB to BD; and make the angle CBA equal to GFE, and as LG to GK so make DB to BA; and complete the parallelogram AC: AC is equal to the recte angle EG, GH, and the excess of the squares of CB, BA is equal to the rectangle HG, GL.

Because as CB to BD, so is OG to GL, the square of CB a 1.6.

is to the rectangle CB, BD as a the rectangle HG, GO to the rectangle HG, GL: and the square of CB is equal to the

rectangle HG, GO, because GO, BC, GH are proportionals; 14. 5. therefore the rectangle CB, BD is equal b to HG, GL. And

because as CB 10 BD, so is 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, so is NL, that is, LM, to LG; c 22. 6. therefore the square of BC together with CD is to the square

of BD, as the square of ML to the square of LG: but the d 8. 2.

square of BC and CD together is equald to four times the 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 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,

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GL to HG, GK; and as the rectangle CB, BA to AP, BC, so is (the straight line BA to AP, and so is FE or GK to

EG, and so is) the rectangle HG, GK to HG, GE; therefore ex æquali, as the rectangle CB, BD to AP, BC, so 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.

PROP. LXXXVIII.

N. :

IF 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 sides shall 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 sum of the squares of AB, BC be given ; AB, BC are each of them given.

· 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 un A D equal, twice the rectangle contained by them is less than the sum of their squares, as is evident from the 7th prop.book 2, Elem ; therefore twice B' C the given space, to which space the rectangle of which the sides are to be found is equal, must not be greater than the given sum 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 rectangle AC is equal to the given space, and the sum of the squares of its sides 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 sum of the squares of the sides, it must be less 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 de

seribe the circle ABC about the triarigle ABC; AC is its diame aCor.5.4.

ter a : and because the triangle ABC is similar bio AEB, as AC to b 8. 6.

CB so is AB to BE ; therefore the rectangle AC, BE is equal to
AB, BC; and the rectangle AB, BC is given, where fore AC,

BE is given: and because the sum of the squares of AB, BC is c 47. 1. given, the square of AC which is equal to that sum is given ; and

AC itself is therefore given in magnitude : let AC be likewise given d 32. dat. in position, and the point A ; therefore AF is given d in position : and the rectangle AC, BE is given, as has A

D e 61. dát. been shown, and AC is given; wherefore e

BE is given in magnitude, as also AF
which is equal to it; and AF is also giv-

F
en in position, and the point A is given;

B f 30. dat. wherefore f the point F is given, and the g 31. dat. straight line FB in position 8: and the cir

cumference ABC is given in position, h 28. dat. wherefore h the point B is given : and the G K

HL points A, C are given ; therefore the straight lines AB, BC are i 29. dat. given i in position and magnitude.

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, Be is equal :

a square equal to the rectangle GH, GL: and let its side AC be given in position ; upon AC as a diameter describe the semicircle 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 rect. 1 16.6. angle CA, AF is equall to GH, GK; and, by the hypothesis,

twice the rectangle GH, GK is less 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 cotitain ed by the diameter AC and its balf; wherefore AF is less than the semidiameter of the circle, and consequently the straight line drawn through the point F parallel to AC must meet the circum. ference in two points: let B be either of them, and joint AB, BC, and complete the rectangle ABCD, ABCD is the rectangle

which was to be found : draw BE perpendicular to AC; there m 34. 1. fore BE is equal m .to AF, and because the angle ABC in a semi

circle is a right angle, the rectangle AB, BC is equal b to AC, BE, that is, to the rectangle CA, AF, which is equal to the given rectangle GH, GK; and the squares of AB, BC are together equal to the square of AC, that is, to the given rectangle GO GL.

k 14. 2. find ik

case.

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 ra. tio of the rectangle contained by the sides AB, BC to the parallelogram AC is given n; and AC is given, therefore the rectan- n 62. dat. gle AB, BC is given ; and the sum of the squares of AB, BC is given ; therefore the sides AB, BC are given by the preceding

The sides 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, A

D FK be the given rectangle to which the sum of the squares of the sides is to be equal. And, by the preceding case, find the sides of a rectangle which is equal to the given BL с rectangle EF, FH, and the squares of the

E sides of which are together equal to the given rectangle EF, FK; therefore, as was shown in that case, twice the rectangle EF, FH must not be greater than the rectangle EF, FK; let it be so, and let AB, BC be the sides of the rectangle joined in the an F HG K gle 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 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 squares of AB, BC are together equal, by construction, to the given rectangle EF, FK.

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