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PROP. K. THEOR.

See N.

If there be any number of ratios, and any number of other

ratios such, that the ratio which is compounded of ratios which are the same to the first ratios, each to each, is the same to the ratio which is compounded of ratios which are the same, each to each, to the last ratios ; and if one of the first ratius, or the ratio which is compounded of ratios which are the same to several of the first ratios each to each, be the same to one of the last ratios, or to the ratio which is compounded of ratios which are the same, each to each, to several of the last ratios ; then the remaining ratio of the first, or, if there be more than one, the ratio which is compour ded of ratios which are the same, each to each, to the remaining ratios of the first, shall be the same to the remaining ratio of the last,

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 Sto X is compounded of the

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h, k, 1. A, B; C, D; E, F. S, T, V, X. G, H; K, L; M, N; O, P; Q, R. Y, Z, a, b, c, d, e, f, g. m, n, o, p. 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 Ĝ 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 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 conpounded of the ratios of e to f, and f to g, which by the

hypothesis, are the same to the ratios of G to H, and K to L, two of the other ratios ; and let the ratio of h to l be that which is compounded of the ratios of h to k, and k to l, 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 l shall be the same to the ratio of m to p; or h shall be to l, as m

to p.

+ B.5.

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 t ex æquali, e is to g, as Y to a: and by the + 22. 5. hypothesis, A is to B, that is, S to T, as e to g; whereforet, S is to T, as Y to a; and, by inversiont, T 11.5. is to S, as a to Y: but S is to Xt, as Y to d; there

+ Hyp. fore ex æquali, T is to X, as a to d; alsó t, because h + Hyp. is to k as (C to D, that is, as) T to V; and k is to l as (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 shewn, that T is tọ X, as a to d; therefore* h is to l, " 11. 5. as m to p.

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 shew the true meaning of them when they are so expressed; especially since they are very frequently made use of by geometers.

Q. E. D.

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See N.

Reciprocal figures, viz. triangles and parallelograms,
66 are such as have their sides about 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.

IV.

The altitude of any figure is the straight

line drawn from its vertex perpendicular
to the basc.

ogram CF.

EA

PROP. I. THEOR. Triangles and parallelograms of the same altitude are See N.

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 parallel

Produce BD both ways to the points H, L, and+ + 3. 1. 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* eqnal: there- * 38. 1. fore, whatever miultiple 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 tri

{G B C angle 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 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 shewn, 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 BC is to * 5 Def.5. the base CD, so is the triangle ABC to the triangle ACD.

And because the parallelogram CE is donble of the triangle ABC*, and the parallelogram CF double of • 41. 1. the triangle ACD, and that magnitudes have the same ratio* which their equimultiples have; as the triangle * 15. 5. ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF: and because it has been shewn, that, as the base BC is to the base CD, so is the triangle

38. 1.

* 11. 5.

33. 1.

ABC to the triangle ACD; and as the triangle ABC is to the triangle ACD, so is the parallelogran EC to the parallelogram CF; therefore, as the base BC is to the base CD, so is * the parallelogram EC to the parallelogram 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 perpen† 28. 1. diculars are both equal and parallelt to one another.

Then, if the same construction be made as in the proposition, the demonstration will be the same.

PROP. II. THEOR. See N.

If a straight line be drawn parallel to one of the sides of

a triangle, it shall cut the other sides, or these produced, proportionally: and if the sides, or the sides produced, be cut proportionally, the straight line which joins the points of section shall be parallel to the remaining side of the triangle.

Let DE be drawn parallel to BC, one of the sides of

the triangle ABC: BD shall be to DA, as CE to EA. * 37, 1. Join BE, CD; then the triangle BDE is equal* to

the triangle CDE, because they are on the same base DE, and between the same parallels DE, BC: ADE is another triangle; and equal magnitudes have to the same* the same ratio; therefore, as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE: but as the triangle BDE to the triangle ADE so is* BD to DA, because, 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 ADE, so is CE to EA: therefore, as BD to DA, so is CE to EA*.

Next, let the sides AB, 11. AC, of the triangle ABC,

or these sides produced,
be cut proportionally in

E B!
the points D, E, that is,
so that BD may beto DA

E B
as CE to EA; and join DE: DE shall be parallel to BC.

* 7.5.

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* 11.5.

A

E

D

B

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