PROP. K. THEOR.

If there be any number of ratios, and any number of other ratios such, that the ratio compounded of ratios which are the same with the first ratios, each to each, is the same with the ratio compounded of ratios which are the same, each to each, with the last ratios; and if one of the first ratios, or the ratio which is compounded of ratios which are the same with several of the first ratios, each to each, be the same with one of the last ratios, or with the ratio compounded of ratios which are the same, each to each, with several of the last ratios: Then the ratio compounded of ratios which are the same with the remaining ratios of the first, each to each, or the remaining ratio of the first, if but one remain; is the same with the ratio compounded of ratios which are the same with those remaining of the last, each to each, or with the remaining ratio of the last.

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 is compounded

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

of the ratios of S to T, T to V, and V to X, which are the same with 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

Book V.

Book V. c to d, which are the same, each to each, with 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 with 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 with the ratios of G to H, and K to L, two of the other ratios; and let the ratios of h to be that which is compounded of the ratios of h to k, and k to l, which are the same with 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, with the remaining other ratios, viz. of M to N, O to P, and Q to R: Then the ratio of h to l is the same with the ratio of m to p, or h is to l, as m to p.

h, k, l,

A, B; C, D; E, F.

G, H; K, L;
e, f, g.

S, T, V, X.

M, N; O, P; Q, R. Y, Z, a, b, c, d. m, n, o, p.

a 11. 5.

Because e is to f, as (G to H, that is, as) Y to Z; and fis 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, by inversion, T is to S, as a to Y; and S is to X, as Y tod; therefore, ex æquali, T is to X, as a to d: Also because h is to k as (C to D, that is, as) T to V; and k is to 1, as (E to F, that is, as) V to X; therefore ex æquali, h is to l, as T to X: In like manner, it may be demonstrated, that m is to p, as a to d: And it has been shown, that T is to X, as a to d; therefore h is to l, as m to p. Q. E. D.

a

The propositions G and 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.

"Reciprocal figures, viz. triangles and parallelograms, are such See N. "as have their sides about two of their angles proportionals "in such 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 the extreme and mean ratio, when the whole is to the greater segment, as the greater segment is to the less.

IV.

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

K

Book VI.

See N.

a 38. 1.

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; Then, as the base BC is to the base CD; so is the triangle ABC 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, AGB, ABC are all equal : a 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 is also equal to the triangle ALC; and if the base

EAF

HC be greater than the HGBCD
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 triangle AHC; and of the base CD and 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; b 5. def. 5. and if equal, equal; and if less, less: Therefore as the base BC is to the base CD, so is the triangle ABC to the triangle ACD.

And because the parallelogram CE is double the triangle

OF EUCLID.

d

c 41. 1.

ABC, and the parallellogram CF double the triangle ACD, Book VI. and that magnitudes have the same ratio which their equimultiples 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 parallelo- e 11. 5. gram 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 f, f 33. 1. because the perpendiculars are both equal and parallel to one another. Then, 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 See N. sides of a triangle, it will cut the other sides, or the other sides produced, proportionally: And if the sides, or the sides produced, be cut proportionally, the straight line which joins the points of section will be parallel to the remaining side of the triangle.

Let DE be drawn parallel to BC, one of the sides of the triangle ABC; then BD is to DA, as CE to EA.

Join BE, CD; then the triangle BDE is equal to the triangle CDE, because they are on the same base DE, and a 37. 1. between the same parallels DE, BC: But ADE is another triangle, and equal magnitudes have to the same, the same ratio; therefore, as the triangle BDE to the triangle ADE, b 7. 5. so is the triangle CDE to the triangle ADE; but as the triangle BDE to the triangle ADE, so is BD to DA, because c 1. 6. having the same altitude, viz. the perpendicular drawn from the point E to AB, they are to one another as their bases; and for the same reason, as the triangle CDE to the triangle