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]
For the same reason also, as C is to D, so is F to G. [VII. 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]
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]
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]
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.
Complete text
Book I