For let three numbers E, F, G, the least of those which have the same ratio with A, C, D, be taken; [VII. 33 or VIII. 2]
therefore the extremes of them E, G are prime to one another. [VIII. 3]
Now, since one mean proportional number F has fallen between E, G, therefore E, G are similar plane numbers. [VIII. 20]
Let, then, H, K be the sides of E, and L, M of G.
Therefore it is manifest from the theorem before this that E, F, G are continuously proportional in the ratio of H to L and that of K to M.
Now, since E, F, G are the least of the numbers which have the same ratio with A, C, D,
and the multitude of the numbers E, F, G is equal to the multitude of the numbers A, C, D,
therefore, ex aequali, as E is to G, so is A to D. [VII. 14]
But E, G are prime, primes are also least, [VII. 21]
and the least measure those which have the same ratio with them the same number of times, the greater the greater and the less the less,
that is, the antecedent the antecedent and the consequent the consequent; [VII. 20]
therefore E measures A the same number of times that G measures D.
Now, as many times as E measures A, so many units let there be in N.
Therefore N by multiplying E has made A.
But E is the product of H, K;
therefore N by multiplying the product of H, K has made A.
Therefore A is solid, and H, K, N are its sides.
Again, since E, F, G are the least of the numbers which have the same ratio as C, D, B,
therefore E measures C the same number of times that G measures B.
Now, as many times as E measures C, so many units let there be in O.
Therefore G measures B according to the units in O;
therefore O by multiplying G has made B.
But G is the product of L, M;
therefore O by multiplying the product of L, M has made B.
Therefore B is solid, and L, M, O are its sides;
therefore A, B are solid.