Kitap 8
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I say next that A also has to B the ratio duplicate of that which the corresponding side has to the corresponding side, that is, of that which C has to E or D to F.
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For, since A, G, B are in continued proportion, A has to B the ratio duplicate of that which it has to G. [V. Def. 9]
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And, as A is to G, so is C to E, and so is D to F.
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Therefore A also has to B the ratio duplicate of that which C has to E or D to F. Q. E. D.
PROPOSITION 19.
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Between two similar solid numbers there fall two mean proportional numbers; and the solid number has to the similar solid number the ratio triplicate of that which the corresponding side has to the corresponding side.
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Let A, B be two similar solid numbers, and let C, D, E be the sides of A, and F, G, H of B.
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Now, since similar solid numbers are those which have their sides proportional, [VII. Def. 21] therefore, as C is to D, so is F to G, and, as D is to E, so is G to H.
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I say that between A, B there fall two mean proportional numbers, and A has to B the ratio triplicate of that which C has to F, D to G, and also E to H.
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For let C by multiplying D make K, and let F by multiplying G make L.
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Now, since C, D are in the same ratio with F, G, and K is the product of C, D, and L the product of F, G, K, L are similar plane numbers; [VII. Def. 21] therefore between K, L there is one mean proportional number. [VIII. 18]
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Let it be M
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Therefore M is the product of D, F, as was proved in the theorem preceding this. [VIII. 18]
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Now, since D by multiplying C has made K, and by multiplying F has made M, therefore, as C is to F, so is K to M. [VII. 17]
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But, as K is to M, so is M to L.
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Therefore K, M, L are continuously proportional in the ratio of C to F.
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And since, as C is to D, so is F to G, alternately therefore, as C is to F, so is D to G. [VII. 13]
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For the same reason also, as D is to G, so is E to H.
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Therefore K, M, L are continuously proportional in the ratio of C to F, in the ratio of D to G, and also in the ratio of E to H.
