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87.

parallelogram ABDE. Because DB is equal to BA, and the
angle ABD is given, since the adjacent
angle ABC is given, the parallelogram AD

A
is given in species; and because AB, BC
together are given, and AB is equal to
BD; therefore DC is given: and because
the given parallelogram AC is applied to B D C
the given straight line DC, deficient by

the parallelogram AD, given in species, the sides AB, BD of a 83. dat. the defect are given a ; and DC is given, wherefore the re

mainder BC is given: and each of the straight lines AB, BC is therefore given.

PROP. LXXXVII. 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 less is given, each of the straight lines is given.

Let the two straight lines AB, BC contain the given paral. lelogram AC in the given angle ABC; if the excess of the square of BC above the square of BA be given : AB and BC are 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 a 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 of b 62. dat. the sides 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 CB, BD to the rectangle AB, BC, that is “, the ratio of d 5t. dat. the straight line DB to BA is given; therefored the ratio of the square of DB to the square of BA is

A
given: and the square of BA 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

с

PD of four times the rectangle BC, CD to the e 7. dat. square of BD; and, by composition, the ratio of four times

the rectangle BC, CD together with the square of BD to the square of BD is given: But four times the rectangle BC, CD. together with the square of BD, is equal to the square of the straight lines BC, CD taken together: therefore the ratio

a 2. 2.

cl. 6.

B

f 8. 2.

of the squares of BC, CD together to the square of BD, is given ; wherefore the ratio of the straight line BC, together g 58. dat. 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 BC to the rectangle CB, BD: cl 6. 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 9. dat. the ratio of BC to BA is given; and BC is given, 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 EG perpendicular to FG; let the rectangle EG, GH be the given space M to which the parallelogram A C is to be made equal; and the rect- K angle HG, GL, be the given ex- E cess of the squares of BC, BA. Take, in the straight line GE,

HN GK equal to FE, and make GM F G L O double of GK; join ML, and in GL produced, take LN equal to LM: Bisect GN in 0, 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 GFE, and as LG to GK, so 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 rectangle HG, GL.

Because, as CB to BD, so is OG to GL, the square of CB is to the rectangle CB, BD as a the rectangle HG, GO to a l. 6. the rectangle HG, GL; and the square of CB is equal to the rectangle HG, GO, because GO, BC, GH are proportionals: therefore the rectangle CB, BD is equal b to HG, b 14. 5. 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 LN, that is LM, to LG: Therefore, the square of BC together with CD is to c 22. 6. the square of BD, as the square of ML to the square of LG:

d 8. 2.

But the 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 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

М.

A

K к
E

B PDC

F G L O HN BD, because as LG to GK, so DB was made to BA: Thereb 14. 5. fore, 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, 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.

N.

PROP. LXXXVIII. If two straight lines contain a parallelogram, given in magnitude, in a given angle ; if the sum of the squares of its sides are given, each of the sides is 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 unequal, twice the rectangle contained by them A D is less than the sum of their squares, as is evident from the 7th Prop. B. 2, Elem.; there. B С fore 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 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 are 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 describe the circle ABC about the triangle ABC; AC is its diameter a : And because the triangle ABC is simi- a Cor. 5.4. lar to AEB, as AC to CB, so is AB to BË; therefore, the b 8. 6. 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 squares of AB, BC is given, the square of AC which is equal to that sum is given ; and AC itself is therefore given c 47. 1. in magnitude : Let AC be likewise given in position, and the point A ; therefore AF is given d in po

d 32. dat.

А sition : And the rectangle AC, BE is

D given, as has been shown, and AC is

E given, wherefore · BE is given in magnitude, as also AF which is equal to it; F

B В and AF is also given in position, and the point A is given, wherefore the point F is given; and the straight line

G FB in position 8 : And the circumfe

K HL

31. dat. rence ABC is given in position, wherefore the point B is h 28. dat. given; and the points A, C are given ; therefore, the straight lines AB, BC are giveni 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 rect

e 61. dat.

f 30. dat.

8

i 29. dat.

b 8. 6.

angle 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. Findk 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; there1 16. 6. fore the rectangle 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 contained by the diameter AC and its half: 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 circumference in two points. Let B be either of them, and join AB, BC, and complete the rect

angle ABCD, ABCD is the rectangle which was to be found : m 34. 1. Draw BE perpendicular to AC; therefore BE is equalm to AF,

and because the angle ABC in a semicircle is a right angle, the rectangle AB, BC is equal to AC, BE, that is, to the rect

angle CA, AF, which is equal to the given rectangle GH, GK: c 47. 1.

And the squares of AB, BC are together equal c to the square 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 sides AB, BC to the n 62. dat. parallelogram 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 sides AB, BC are given, by the preceding case.

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

A D
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 B L
rectangle to which the sum of the squares

E
of the sides is to be equal. And, by the
preceding case, find the sides of a rect-
angle which is equal to the given rect-
angle EF, FH, and the squares of the
sides of which are together equal to the
given rectangle EF, FK; therefore, as

HG K
was shown in that case, twice the rect-
angle EF, FH must not be greater than the rectangle EF,

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