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.
在两个相似立体数之间，存在两个比例中项数；且立体数与相似立体数之比等于对应边与对应边之比的三重比。
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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. 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. 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. For let C by multiplying D make K, and let F by multiplying G make L. 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] Let it be M Therefore M is the product of D, F, as was proved in the theorem preceding this. [VIII. 18] 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.
设A、B为两个相似立体数，其边分别为C、D、E和F、G、H。由相似立体数定义，边成比例：C:D = F:G，D:E = G:H。
[VII. 17] But, as K is to M, so is M to L. Therefore K, M, L are continuously proportional in the ratio of C to F. 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] For the same reason also, as D is to G, so is E to H. 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. Next, let E, H by multiplying M make N, O respectively. 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.
作K = C×D，L = F×G，则K、L为相似面数，其间有一比例中项M = D×F，使得K:M = M:L，且该比等于C:F。
But the product of C, D is K; therefore E by multiplying K has made A. For the same reason also H by multiplying L has made B. Now, since E by multiplying K has made A, and further also by multiplying M has made N, therefore, as K is to M, so is A to N. [VII. 17] But, as K is to M, so is C to F, D to G, and also E to H; therefore also, as C is to F, D to G, and E to H, so is A to N. Again, since E, H by multiplying M have made N, O respectively, therefore, as E is to H, so is N to O. [VII. 18] But, as E is to H, so is C to F and D to G; therefore also, as C is to F, D to G, and E to H, so is A to N and N to O. Again, since H by multiplying M has made O, and further also by multiplying L has made B, therefore, as M is to L, so is O to B.
由C:D = F:G得C:F = D:G，同理D:G = E:H，故K、M、L连续成比例于C:F、D:G、E:H。再作N = E×M，O = H×M，则A = E×K，B = H×L。
[VII. 17] But, as M is to L, so is C to F, D to G, and E to H. Therefore also, as C is to F, D to G, and E to H, so not only is O to B, but also A to N and N to O. Therefore A, N, O, B are continuously proportional in the aforesaid ratios of the sides. I say that A also has to B the ratio triplicate of that which the corresponding side has to the corresponding side, that is, of the ratio which the number C has to F, or D to G, and also E to H. For, since A, N, O, B are four numbers in continued proportion, therefore A has to B the ratio triplicate of that which A has to N. [V. Def. 10] But, as A is to N, so it was proved that C is to F, D to G, and also E to H.
由比例传递得A:N = C:F = D:G = E:H，N:O = E:H = C:F，O:B = M:L = C:F，故A、N、O、B连续成比例，且A:B为三重比，即(C:F)³。