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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.

PRO P. LXXXVI.

85.

IF two straight lines contain a parallelogram given in

magnitude, in a given angle ; if both of them together be given, they shall each of them be given.

Let the two straight 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 straight lines AB, BC is given.

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 an- E A gle ABC is given, the parallelogram AD is given in species : And because AB, BC to. gether are given, and AB is equal to BD; therefore DC is given: And because the gia

B C ven parallelogram AC is applied to the given

straight line DC, deficient by the parallelogram AD given in 83, dat. species, the sides AB, BD of the defect are given; and DC

is given, wherefore the remainder BC is given ; and each of the straight lines AB, BC is therefore given.

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IF two straight lines contain a parallelogram given in

magnitude, in a given angle; if the excess of the square of the greater above the square of the lefler be given, each of the straight lines shall be given.

Let the two straight lines AB, BC contain the given parallelogram AC in the given angle ABC; if the excess of the square of BC above the square of BA be given ; AB and BC art each of them given.

Let the given excess of the square of BC above the square of BA be the rectangle CB, BD; take this from the square of BC, the remainder, which is the rectangle BC, CD is equal to the square of AB; and because the angle ABC of

the parallelogram AC is given, the ratio of the rectangle b 62. dat. of the hides 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

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zagie CB, BD to the rectangle AB, BC, that is, the ratio of the e I. 6. Atraight line DB to BA is given; therefore d the ratio of the d 54. dat. square of DB to the square of BA is gi- A А ven : And the square of BĄ is equal to the rectangle 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 rectangle BC, CD to the square

B PDC of BD; and, by compofition, the ratio of four times the rect. 7. dat. angle BC, CD together with the square of BC to the square of BD is given : But four times the rectangle BC, CD toge. ther with the square of BD is equal to the square of the straight ? 8. 2. Jines BC, CD takon together; therefore the ratio of the square of BC, CD together to the square of BD is given ; wherefore z the ratio of the straight line BC together with CỤ to BD is 8 58. dat. 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 bas 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 analysis of this problem, viz.

A parallelogram AC which has a given angle ABC being gi. ven in magnitude, and the excess of the square of BC one of its fides 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 LG perpendicular to FG; let the rect.

MN angle EG, GH be the given space to which the parallelogram AC is

KI to be made equal; and the rectangle

F
HG, GL, be the given excess of the
squares of BC, BA
Take, in the straight line GE,

FG LO HN Gķ equal to FE, and make GM double of GK ; join ML, and in GL produced, take LN equal to LM: Bisect GN in O, and between GH, GO find a mean proportional BC: As OG to GL, so make CB to BD; and make the angle CBA equal

to

to CTE, 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 squares of CB, BA is equal to the reaangle HG, GL.

Because as CB to BD, fo is OG to GL, the square of CB 1 1. 6. is to the rectangle CB, BD as 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 to HG, GL. And because as CB to 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: e 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 8.2.

square of BC and CD together is equal d to four times the
rectangle BC, CD together with the square of BD ; therefore
four times the rectangle BC, CD together with the square
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 rectan-
gle 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,

M
A

KU
F

.

B PD C F G L O HN GL to HG, GK ; and as the rectangle CB, BA to AP, BC, fo is (the ftraight line BA to AP, and so is FE or GK 10

EG

EG, and so is) the rectangle HG, GK to HG, GE; therefore, ex aequali, as the rectangle CB, BD to AP, BC, so 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, che parallelogram AC, is equal to the given rectangle EG, GH.

PRO P. LXXXVIII.

N.

IF
F 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,

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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 recte angle contained by two equal straight lines is equal to both their squares ; but if two straight lines are un. AD equal, twice the rectangle contained by them is jess than the sum of their squares, as is evident from the 7th prop. b. 2. Elem. therefore twice 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 muft necessarily 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 squares of its fides 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 describe the circle ABC about the triangle ABC; AC is its diameter * : And because the triangle ABC is fimi- a Cor. 5. 4. laro to AEB, as AC to CB, fo is AB to BE; therefore the b 8. 6. rectangle AC, BE is equal to AB, BC, and the rectangle AB,

BC

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BC is given, wherefore AC, BE is given : And because the sun

of the squares of AB, BC is given, the square of AC which is c'47. 1. equal to that sum is given ; and AC itself is therefore given

in magnitude : Let AC be likewise given in position, and the i 32. dar. point A ; therefore AF is given a in po- A

D lition : And the rectangle AC, BE is

E given, as has been shewa, and AC is 61. dat. given, wherefore · BE is given in mag

nitude, as also AF which is equal to it; F

and AF is also given in position, and 1 30, dat the point A is given ; wherefore the

point F is given, and the straight line at. FB in position :: And the circumference

Кн. 23 dat. ABC is given in position, wherefore the point B is given :

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

The Goes 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 rect.

angle to which the sum of the squares of AB, BC is equal: k 14. 2.

Find 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, fo make GK to AF,

and from the point A place AF at right angles to AC: There | 16. 6. fore the rectangle CA, AF is equal 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 AÇ, and the rectangle CA, AF itself less than half the square of AC, tha is, than the rectangle contained by the diameter AC and its half; wherefore AF is less than the femidiameter of the circle, and consequently the straight 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 34. I,

EE perpendicular to AC; therefore BE is equal to AF, and

because the angle ABC in a semicircle is a right angle, the rect. b 8. 6.

angle AB, BC is equal to AC, BE, that is, to the rectangle CĂ, AF, which is equal to the given rectangle GH, GK: And the squares of AB, BC are together equal to the iquare of AC, that is, to the given rectangle GH, GL.

But

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