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. magnitude : But it is also given in species; because it is S 2 Def

. similara to AG; therefore its sides CA, AD, are given,724.6. and each of the straight lines EA, AF, is given; therefore o 60 Dat. EC, DF are each of them given.

The gnomon and its sides CE, DF, may be found thus in the first case, Let H be the given space to which the gnomon must be made equal, and findd a parallelogram si- • 25. 6. milar to AB and equal to the figures AB and H together, and place its sides AE, AF, from the point A, upon the straight lines AC, AD, and complete the parallelogram AG which is about the same diametere with AB; because e 26. 6. therefore AG is equal to both AB and H, take away

the common part AB, the remaining gnomon ECBDFG is equal to the remaining figure H; therefore a gnomon equal to H, and its sides CE, DF, are found : And in like manner they may be found in the other case, in which the given figure H must be less than the figure FE from which it is to be taken.

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Ir a parallelogram equal to a given space be applied to a given straight line, deficient by a parallelogram given in species; the sides of the defect are given.

Let the parallelogram AC equal to a given space be applied to the given straight line AB, deficient by the parallelogram BDCL given in species, each of the straight lines CD, DB, are given.

Bisect AB in E; therefore EB is given in magnitude; upon

EB describe a the parallelogram EF similar to DL * 18. 6. ** and similarly placed ; therefore EF is given in species, and is about the same

G HF diameter b with DL; let BCG be the

b 26. 6. diameter, and construct the figure;

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CL therefore, because the figure EF given in species is described upon the given straight line EB, EF is given in mag. 4 Е) В nitude, and the gnomon ELH is equald to the given figure AC; therefore e since EF is diminished. 43. 1. by the given gnomon ELH, the sides EK, FH, of the gnomon are given; but EK is equal to DC, and FH to BD; wherefore CD, DB are each of them given.

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c 56 Dat. d 36. and

e 82 Dat

This demonstration is the analysis of the problem in the 28th Prop. of Book 6. the construction and demonstration of which proposition is the composition of the analysis; and because the given space AC, or its equal the gnomon ELH, is to be taken from the figure EF, described upon the half of AB similar to BC, therefore AC must not be greater than EF, as is shown in the 27th Prop. B. 6.

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

Let the parallelogram AC equal to a given space be applied to the given straight line AB, exceeding by the parallelogram BDCL given in species; each of the straight lines CD, DB, are given.

Bisect AB in E; therefore EB is given in magnitude : a 18. 6. Upon BE describe a the parallelogram EF similar to LD,

and similarly placed; therefore EF is given in species, and • 26. 6. is about the same diameter b with LD. Let CBG be the diameter, and con

G FH struct the figure : Therefore, because the figure EF given in species is de

A E

B

D scribed upon the given straight line € 56 Dat. EB, EF is given in magnitude, and

K L C the gnomon ELH is equal to the given 36. and figured AC; wherefore, since EF is increased by the given

gnomon ELH, its sides. EK, FH, are givene ; 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..

Draw the diameter DE of the parallelogram AC, and

43. I. € 82 Dat.

1 24. 6.

construct the figure. Because the parallelogram AK is equal a' to BC which is given, there E G С

1 43. 1. fore AK is given ; and BK is similarb to AC, therefore BK is given in spe. cies. And since the parallelogram H

K AK given in magnitude is applied to the given straight line AB, exceeding A B D by the parallelogram BK given in species, therefore by this proposition, BD, DK, the sides of the excess, are given, and the 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 applye the parallelogram • 29. 6. 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.

The parallelogram BC is equal a to AK, that is, to the given space; and the parallelogram AC is similar b to BK; 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 be given, they shall each of thein be given.

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 А.

T 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 B the angle ABD is given, therefore the parallelogram AD is given in species; and because the given parallelogram AC is applied to the given straight line DC, exceeding by the

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85.

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parallelogram AD given in species, the sides of the excess * 84 Dat. are given a : 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.

PROP. LXXXVI. If two straight lines contain a parallelogram giyen iu 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 DB equal to BA, and complete the parallelogram ABDE. Because DB is equal to BA, and the angle ABD given, because 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

D given: And because the given parallelogram AC is applied to the given straight line DC, deficient

by the parallelogram AD given in species, the sides AB, 43 Dat. BD, of the defect are given a; and DC is given, wherefore

the remainder 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 lesser 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 are each of them given.

Let the given excess of the square of BCabove the

of BA be the rectangle CB, BD; take this from the square 3 2. 2. 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 1 69 Dat. of the sides AB, BC, to the parallelogram AC is given b;

and AC is given, therefore the rectangle AB,BC, is given;

87.

square

and the rectangle CB, BD, is given; therefore the ratio of the rectangle CB, BD, to the rectangle AB, BC, that is ©, *1.6. the ratio of the straight line DB to BA is given; therefore d d 54 Dat. 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 square of BD; and, by compo

B PDC sition, the ratio of four times the rectangle BC, CD to..7 Dat, gether 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 equalf to the square of the straight lines f 8. 2. BC, CD taken together: therefore the ratio of the square of BC, CD, together to the square of BD is given ; wherefores the ratio of the straight line BC, together with CD 558 Dat. 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; there. fore h the ratio of BC to BA is given; and BC is given, h 9 Dat. wberefore 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

MN any point E in FE, draw EG perpendicular to FG; let the

K rectangle EG, GH, be the given space to which the parallelogram AC is to be made equal; ÉG L 0

HN and the rectangle HG, GL, be the given excess of the squares of BC, BA.

Take, in the straight line GE, GK equal to FE, and make GM double of GK; join ML, and in GL produced, take LN equal to LM: Bisect GN in 0, and between GH,

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