Book 8
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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.
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Next, let E, H by multiplying M make N, O respectively.
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Now, since A is a solid number, and C, D, E are its sides, therefore E by multiplying the product of C, D has made A.
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But the product of C, D is K; therefore E by multiplying K has made A.
