a 18. 6.

b 26. 6.

c 56. dat.

d 36. and

43. 1.

e 82. dat.

a 43. 1.

b 24. 6.

Bisect AB in E; therefore EB is given in magnitude; upon EB describe the parallelogram EF similar to LD, and similarly placed; therefore EF is given in species, and is about the same diameter with LD.

A

G F H

E

B

D

KLC

Let CBG be the diameter, and con-
struct the figure: Therefore, because
the figure EF, given in species, is de-
scribed upon the given straight line
EB, EF is given in magnitude, and
the gnomon ELH is equal to the
given figured AC; wherefore, since
EF is increased by the given gnomon ELH, its sides EK,
FH are given ; but EK is equal to CD, and FH to BD;
therefore CD, DB are each of them given.

This demonstration is the analysis of the problem in the 29th Prop. Book 6; the construction and demonstration of which is the composition of the analysis.

COR. If a parallelogram, given in species, be applied to a given straight line, exceeding by a parallelogram, equal to a given space; the sides of the parallelogram are given.

Let the parallelogram ADCE, given in species, be applied to the given straight line AB, exceeding by the parallelogram BDCG, equal to a given space; the sides AD, DC of the parallelogram are given.

b

E

Draw the diameter DE of the parallelogram AC, and construct the figure: Because the parallelogram AK is equala to BC, which is given, therefore AK is given; and BK is similar to AC, therefore BK is given in species. And since the parallelogram AK, given in magnitude, is applied to the given straight line AB, exceeding by the parallelogram BK, given in species, there

G

C

F

K

fore by this proposition, BD, DK the

A

B D

sides of the excess are given, and the

H

c 29. 6.

straight line AB is given; therefore the whole AD, as also DC, to which it has a given ratio, is given.

PROB.

To apply a parallelogram similar to a given one to a given straight line AB, exceeding by a parallelogram equal to a given

space.

To the given straight line AB apply the parallelogram AK equal to the given space, exceeding by the parallelogram

BK, similar to the one given. Draw DF, the diameter of
BK, and through the point A draw AE parallel to BF,
meeting DF produced in E, and complete the parallelogram
AC.

a

The parallelogram BC is equal to AK, that is, to the a 43. 1. given space; and the parallelogram AC is similar to BK; b 24. 6. therefore the parallelogram AC is applied to the straight line AB, similar to the one given, and exceeding by the parallelogram BC, which is equal to the given space.

PROP. LXXXV.

If two straight lines contain a parallelogram, given in magnitude, in a given angle; if the difference of the straight lines is given, each of them is given.

E A

Let AB, BC contain the parallelogram AC given in magnitude, in the given angle ABC, and let the excess of BC above AB be given; each of the straight lines AB, BC is given. Let DC be the given excess of BC above BA, therefore the remainder BD is equal to BA. Complete the parallelogram AD; and because AB is equal to BD, the ratio of AB to BD is given; and the angle ABD is given, therefore the parallelogram AD is given in species; and because the given paral- D lelogram AC is applied to the given straight

B C

line DC, exceeding by the parallelogram AD given in species,

84.

the sides of the excess are given, therefore BD is given: a 84. dat. and DC is given, wherefore the whole BC is given: And AB is given, therefore AB, BC are each of them given.

PROP. LXXXVI.

If two straight lines contain a parallelogram given in magnitude, in a given angle; if both of them together are given, each of them is 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 BA, and complete the

85.

A E

parallelogram ABDE. Because DB is equal to BA, and the angle ABD is given, since the adjacent angle ABC is given, the parallelogram AD 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 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; and DC is given, wherefore the remainder BC is given: and each of the straight lines AB, BC is therefore given.

a 2. 2.

87.

PROP. LXXXVII.

D C

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 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 are each of them given.

a

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 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 rectangle CB, BD to the rectangle AB, BC, that is, the ratio of d 54. dat. the straight line DB to BA is given; therefore the ratio of the square of DB to the square of BA is 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 of four times the rectangle BC, CD to the

c 1. 6.

B

A

PD

C

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

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: c 1. 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.

M

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 to which the parallelogram AC is to be made equal; and the rectangle HG, GL, be the given excess of the squares of BC, BA.

K

E

HN

Take, in the straight line GE, GK equal to FE, and make GMFG LO 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 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 the rectangle HG, GO to a 1. 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 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:

a

b

d 8. 2.

But the square of BC and CD together, is equal 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: 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.

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