For let C by multiplying itself make E, and by multiplying D let it make F; let D by multiplying itself make G, and let the numbers C, D by multiplying F make H, K respectively.

Now, since A is a cube, and C its side, and C by multiplying itself has made E, therefore C by multiplying itself has made E and by multiplying E has made A.

For the same reason also D by multiplying itself has made G and by multiplying G has made B.

And, since C by multiplying the numbers C, D has made E, F respectively, therefore, as C is to D, so is E to F. [[VII. 17](elem.7.17)]

For the same reason also, as C is to D, so is F to G. [[VII. 18](elem.7.18)]

Again, since C by multiplying the numbers E, F has made A, H respectively, therefore, as E is to F, so is A to H. [[VII. 17](elem.7.17)]

But, as E is to F, so is C to D.

Therefore also, as C is to D, so is A to H.

Again, since the numbers C, D by multiplying F have made H, K respectively, therefore, as C is to D, so is H to K. [[VII. 18](elem.7.18)]

Again, since D by multiplying each of the numbers F, G has made K, B respectively, therefore, as F is to G, so is K to B. [[VII. 17](elem.7.17)]

But, as F is to G, so is C to D; therefore also, as C is to D, so is A to H, H to K, and K to B.

Therefore H, K are two mean proportionals between A, B.

I say next that A also has to B the ratio triplicate of that which C has to D.

For, since A, H, K, B are four numbers in proportion, therefore A has to B the ratio triplicate of that which A has to H. [[V. Def. 10](elem.5.def.10)]

But, as A is to H, so is C to D; therefore A also has to B the ratio triplicate of that which C has to D. Q. E. D.