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PROPOSITION XXIII.—THEOREM.

Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides.

LET a c, cf be equiangular parallelograms, having the angle b c d equal to the angle e cg the ratio of the parallelogram ac to the parallelogram cf is the same with the ratio which is compounded of the ratios of their sides.

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Let bc, cg be placed in a straight line; therefore dc and ce are also in a straight line (i. 14); and complete the parallelogram dg; and taking any straight line k, make (vi. 12) as bc to cg, so k to 1; and as dc to ce, so make (vi. 12) 1 to m: therefore the ratios of k to 1, and 1 to m, are the same with the ratios of the sides viz. of But the ratio of k bc to cg, and dc to ce to m is that which is said to be compounded (v. def. A.) of the ratios of k to 1, and 1 to m wherefore also k has to m the ratio compounded of the ratios of the sides. And because as bc to cg, so is the parallelogram ac to the parallelogram ch (vi. 1); but as bc to cg, so is k to 1; therefore k is (v. 11) to 1, as the parallelogram a c to the parallelogram ch. Again, because as dc to ce, so is the parallelogram ch to the parallelogram cf; but as dc to ce, so is 1 to m; wherefore I is (v. 11) to m, as the parallelogram ch to the parallelogram cf: therefore since it has been proved, that as k to 1, so is the parallelogram ac to the parallelogram ch; and as 1 to m, so the parallelogram ch to the parallelogram cf: ex æquali (v. 22), k is to m, as the parallelogram ac to the parallelogram cf: but k has to m the ratio which is compounded of the ratios of the sides; therefore also the parallelogram ac has to the parallelogram cf the ratio which is compounded of the ratios of the sides. Wherefore, equiangular parallelograms, &c. Q. E. D.

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PROPOSITION XXIV.-THEOREM.

The parallelograms about the diameter of any parallelogram are similar to the whole and to one another.

LET abcd be a parallelogram, of which the diameter is a c; and eg, hk the parallelograms about the diameter. The parallelograms eg, hk are similar both to the whole parallelogram a bcd, and to one another.

Because de, gf are parallels, the angle adc is equal (i. 29) to the angle a gf. For the same reason, because b c, e f are parallels, the angle abc is equal to the angle a ef: and each of the angles bcd, efg is equal to the opposite angle da b (i. 34), and therefore are equal to one another wherefore the parallelograms abcd, a efg, are equiangular: and because the angle a bc is equal to the angle a ef, and the angle bac

common to the two triangles bac, eaf, they are equiangular to one

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another; therefore (vi. 4) as ab to bc, so is ae to ef and because the opposite sides of parallelograms are equal to one another (i. 34), ab (v. 7) is to ad as ae to ag; and dc to cb as gf to fe; and also cd to da as fg to ga: therefore the sides of the parallelograms abcd, aefg about the equal angles are proportionals; and they are therefore similar to one another (vi. def. 1); for the same reason the parallelogram abcd is similar to the parallelogram fhck. Wherefore each of the parallelograms ge, kh is similar to db: but rectilineal figures which are similar to the same rectilineal figure are also similar to one another (vi. 21); therefore the parallelogram ge is similar to kh. Wherefore the parallelogram, &c. Q. E. D.

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PROPOSITION XXV.-PROBLEM.

To describe a rectilineal figure which shall be similar to one and equal to another given rectilineal figure.

LET a b c be the given rectilineal figure to which the figure to be described is required to be similar, and d that to which it must be equal. It is required to describe a rectilineal figure similar to a b c, and equal to d.

Upon the straight line bc describe (i. 45, cor.) the parallelogram be equal to the figure a bc; also upon ce describe (i. 45, cor.) the parallelogram cm equal to d, and having the angle fce equal to the angle cbl: therefore bc and c f are in a straight line (i. 29, and i. 14), as also le and em: between bc and cf find (vi. 13) a mean proportional gh, and upon gh describe (vi. 18) the rectilineal figure kgh similar and similarly

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situated to the figure abc and because bc is to g h as gh to cf; and if three straight lines be proportionals, as the first is to the third, so is (vi. 20, cor. 2) the figure upon the first to the similar and similarly described figure upon the second; therefore, as b c to cf, so is the rectilineal figure abc to kgh: but as bc to cf, so is (vi. 1) the parallelogram be to the parallelogram ef: therefore as the rectilineal figure abc is to kgh, so is the parallelogram be to the parallelogram ef (v. 11): and the rectilineal figure abc is equal to the parallelogram be; therefore the rec

tilineal figure kg h is equal (v. 14) to the parallelogram ef: but ef is equal to the figure d; wherefore also k g h is equal to d; and it is similar to a bc. Therefore the rectilineal figure kgh has been described similar to the figure a bc, and equal to d. Which was to be done.

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If two similar parallelograms have a common angle, and be similarly situated, they are about the same diameter.

LET the parallelograms abcd, aefg be similar and similarly situated, and have the angle dab common: abcd and aefg are about the same diameter.

For, if not, let, if possible, the parallelogram b d have its diameter ahc in a different straight line from af, the diameter of the parallelogram eg, and let gf meet ahcin h; and through h draw hk parallel to ad or bc: therefore the parallelograms abcd, akhg being about the same diameter, they are similar to one another (vi. 24): wherefore as da to a b, so is (vi. def. 1) ga to ak but because abcd and aefg are similar parallelograms, as b

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da is to a b, so is ga to ae; therefore (v. 11) as ga to a e, so is ga to ak; wherefore ga has the same ratio to each of the straight lines a e, ak; and consequently ak is equal (v. 9) to a e, the less to the greater, which is impossible: therefore abcd and akhg are not about the same diameter: wherefore abcd and a efg must be about the same diameter. Therefore, if two similar parallelograms, &c. Q. E. D.

To understand the three following propositions more easily, it is to be observed,

1. That a parallelogram is said to be applied to a straight line, when it is described upon it as one of its sides. For example, the parallelogram ac is said to be applied to the straight line ab.

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2. But a parallelogram a e is said to be applied to a straight line ab, deficient by a parallelogram, when ad the base of ae is less than a b, and therefore ae is less than the parallelogram ac described upon ab in the same angle, and between the same parallels, by the parallelogram dc: and dc is therefore called the defect

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of a e. 3. And a parallelogram a g is said to be applied to a straight line a b, exceeding by a parallelogram, when af the base of a g is greater than a b, and therefore ag exceeds a c the parallelogram described upon a b in the same angle, and between the same parallels, by the parallelogram bg.

PROPOSITION XXVII.-THEOREM.

Of all parallelograms applied to the same straight line, and deficient by parallelograms similar and similarly situated to that which is described upon the half of the line, that which is applied to the half, and is similar to its defect, is the greatest.

LET a b be a straight line divided into two equal parts in c, and let the parallelogram ad be applied to the half a c, which is therefore deficient from the parallelogram upon the whole line ab by the parallelogram ce upon the other half cb; of all the parallelograms applied to any other parts of a b, and deficient by parallelograms that are similar and similarly situated to ce, ad is the greatest.

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Let af be any parallelogram applied to a k, any other part of a b than the half, so as to be deficient from the parallelogram upon the whole line ab by the parallelogram kh similar and similarly situated to ce ad is greater than a f

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First, let ak the base of a fbe greater than ac the half of a b; and because ce is similar to the parallelogram kh, they are about the same diameter (vi. 26): draw their diameter db, and complete the scheme because the parallelogram cf is equal (i. 43) to fe, add kh to both, therefore the whole ch is equal to the whole ke but ch is equal (i. 36) to cg,

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because the base a c is equal to the base cb; therefore cg is equal to ke:

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to each of these add cf; then the whole af is equal to the gnomon chl, therefore ce, or the parallelogram a d, is greater than the parallelogram a f

Next, let ak the base of af be less than a c, and the same construction being made, the parallelogram dh is equal to dg (i. 36), for h m is equal to mg (i. 34), because bc is equal to ca; wherefore dh is greater than 1g: but dh is equal (i. 43) to dk; therefore dk is greater than 1g; to each of these add al; then the whole ad is greater than the whole af. Therefore, of all parallelograms applied, &c. Q. E. D.

PROPOSITION XXVIII.-PROBLEM.

To a given straight line to apply a parallelogram equal to a given rectilineal figure, and deficient by a parallelogram similar to a given parallelogram ; but the given rectilineal figure to which the parallelogram to be applied is to be equal must not be greater than the parallelogram applied to half of the given line, having its defects similar to the defect of that which is to be applied, that is, to the given parallelogram.

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LET ab be the given straight line, and c the given rectilineal figure, to which the parallelogram to be applied is required to be equal, which figure must not be greater than the parallelogram applied to the half of the line, having its defect from that upon the whole line similar to the defect of that which is to be applied; and let d be the parallelogram to which this defect is required to be similar. It is required to apply a parallelogram to the straight line a b, which shall be equal to the figure C, and be deficient from the parallelogram upon the whole line by a parallelogram similar to d.

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ag, which must either be equal to C, or greater than it, by the determination: and if a g be equal to c, then what was required is already done: for, upon the straight line a b, the parallelogram ag is applied equal to the figure c, and deficient by the parallelogram ef similar to d: but, if a g be not equal to c, it is greater than it; and ef is equal to a g; therefore ef also is greater than

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Make (vi. 25) the parallelogram klmn equal to the excess, of ef above c, and similar and similarly situated to d; but d is similar to ef, therefore (vi. 21) also km is similar to ef. Let kl be the homologous side to eg, and 1m to gf; and because e f is equal to c and k m together, ef is greater than km; therefore the straight line eg is greater than kl, and gf than 1m: make gx equal to 1k, and go equal to 1m, and complete the parallelogram xgop: therefore xo is equal and similar to km; but km is similar to ef; wherefore also xo is similar to ef, and thereforexo and e f are about the same diameter (vi. 26): let gpb be their diameter, and complete the scheme. Then, because ef is equal to c and km together, and xo a part of the one is equal to km a part of the other, the remainder, viz. the gnomon ero is equal to the remainder c: and because or is equal (i. 34) to xs, by adding sr to each, the whole ob is equal to the whole xb: but xb is equal (i. 36) to te, because the base

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