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or, if there be more than one, to the ratio which is compounded of ratios which are the same each to each to these remaining ratios.

Let the ratios of A to B, C to D, E to F, be the first ratios; and the ratios of G to H, K to L, M to N, O to P, Q to R, be the other ratios: and let A be to B, as S to T; and C to D, as T to V ; and E to F, as V to X: therefore, by the definition of compound ratio, the ratio of S to X h, k, l. C, D; E, F.

A, B;

S, T, V, X. G, H; K, L, M, N, O, P, Q, R. Y, Z, a, b, c, d.

e,

m, n, o, p.

f, g. is compounded of the ratios of S to T, T to V, and V to X, which are the same to the ratios of A to B, C to D, E to F: each to each. Also, as G to H, so let Y be to Z; and K to L, as Z to a; M to N, as a to b; O to P, as b to c; and Q to R, as c to d: therefore, by the same definition, the ratio of Y to d is compounded of the ratios of Y to Z, Z to a, a to b, b to c, and c to d, which are the same, each to each, to the ratios of G to H, K to L, M to N, O to P, and Q to R: therefore, by the hypothesis, S is to X, as Y to d. Also, let the ratio of A to B, that is, the ratio of S to T, which is one of the first ratios, be the same to the ratio of e to g, which is compounded of the ratios of e to f, and f to g, which, by the hypothesis, are the same to the ratios of

to H, and K to L, two of the other ratios; and let the ratio of h to 1 be that which is compounded of the ratios of h to k, and k to 1, which are the same to the remaining first ratios, viz. of C to D, and E to F; also, let the ratio of m to p, be that which is compounded of the ratios of m to n, n to o, and o to p, which are the same, each to each, to the remaining other ratios, viz. of M to N, O to P, and Q to R: then the ratio of h to 1 shall be the same to the ratio of m to p; or h shall be to 1, as m to p.

+22.5.

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Because e is to f, as (G to H, that is, as) Y to Z; and f is to g, as (K to L, that is, as) Z to a; therefore ex æquali, e is to g, as Y to a: and by the hypothesis, A is to B, that is, S to T, as e to g; wherefore† S is to T, as Y to a; and + B. 5. by inversion + T is to S, as a to Y: but S is to +Hyp. X, tas Y to d; therefore ex æquali, T is to X, Hyp. as a to d: also, † because h is to k as (C to D,

+11. 5.

that is, as) T to V; and k is to las (E to F, that is, as) V to X; therefore, ex æquali, h is to 1, as T to x: in like manner, it may be demonstrated, that m is to p, as a to d; and it has been *11.5. shown, that T is to X, as a to d; therefore* h is to 1, as m to p. Q. E D.

The propositions Gand K are usually, for the sake of brevity, expressed in the same terms with propositions F and H: and therefore it was proper to show the true meaning of them when they are so expressed; especially since they are very frequently made use of by geometers.

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"Reciprocal figures, viz. triangles and parallelo66 grams, are such as have their sides about

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two of their angles proportionals in such a manner, that a side of the first figure is to a "side of the other, as the remaining side of "this other is to the remaining side of the "" first."

III.

A straight line is said to be cut in extreme and mean ratio, when the whole is to the greater segment, as the greater segment is to the less.

†3. 1.

IV.

The altitude of any figure is the
straight line drawn from its vertex
perpendicular to the base.

PROP. I. THEOR.

Triangles and parallelograms of the same altitude are one to another as their bases.

Let the triangles ABC, ACD, and the parallelograms EC, CF, have the same altitude, viz. the perpendicular drawn from the point A to BD as the base BC, is to the base CD, so shall the triangle ABC be to the triangle ACD, and the parallelogram EC to the parallelogram CF.

Produce BD both ways to the points H, L, and + take any number of straight lines BG, GH, each equal to the base BC; and DK, KL, any number of them, each equal to the base CD; and join AG, AH, AK, AL: then, because CB, BG, GH, are all equal, the triangles AHG, *38. 1. AGB, ABC, are all equal: therefore, whatever multiple the base HC is of the base BC, the same multiple is the triangle AHC of the triangle ABC: for the same reason, whatever multiple the base LC is of the base CD, the same multiple is the triangle ALC of the triangle ADC: and if the base HC be equal to the base CL, the triangle AHC

*38. 1. is also equal to the trian- HG BC gle ALC and if the base

EA

HC be greater than the base CL, likewise the triangle AHC is greater than the triangle ALC; and if less, less: therefore, since there are four magnitudes, viz. the two bases BC, CD, and the two triangles ABC, ACD; and of the base BC, and the triangle ABC the first and third, any

equimultiples whatever have been taken, viz. the base HC and the triangle AHC; and of the base CD and the triangle ACD, the second and fourth, have been taken any equimultiples whatever, viz. the base CL and the triangle ALC; and since it has been shown, that, if the base HC be greater than the base CL, the triangle AHC is greater than the triangle ALC; and if equal, equal; and if less, less: therefore,* as the base *5De£.5. BC is to the base CD, so is the triangle ABC to the triangle ACD.

And because the parallelogram CE is double of the triangle ABC,* and the parallelogram CF *41. 1. double of the triangle ACD, and that magnitudes have the same ratio* which their equimultiples * 15.5. have; as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF: and because it has been shown, that, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD; and as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF; therefore, as the base BC is to the base CD, so is the parallelogram EC to the parallelogram *11.5. CF. Wherefore, triangles, &c. Q. E. D.

COR. From this it is plain, that triangles and parallelograms that have equal altitudes are one to another as their bases.

Let the figures be placed so as to have their bases in the same straight line; and having drawn perpendiculars from the vertices of the triangles to the bases, the straight line which joins the vertices is parallel to that in which their bases are, because the perpendiculars are *35. 1. both equal and parallel† to one another. Then, †28, i. if the same construction be made as in the proposition, the demonstration will be the same.

*

PROP. II. THEOR.

If a straight line be drawn parallel to one of the sides of a triangle, it shall cut the other sides,

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