First, let it be in that ratio to a less magnitude DG.
Then, since, as AB is to BE, so is CD to DG,
they are magnitudes proportional componendo;
so that they will also be proportional separando. [V. 17]
Therefore, as AE is to EB, so is CG to GD.
But also, by hypothesis, as AE is to EB, so is CF to FD.
Therefore also, as CG is to GD, so is CF to FD. [V. 11]
But the first CG is greater than the third CF;
therefore the second GD is also greater than the fourth FD. [V. 14]
But it is also less: which is impossible.
Therefore, as AB is to BE, so is not CD to a less magnitude than FD.
Similarly we can prove that neither is it in that ratio to a greater;
it is therefore in that ratio to FD itself.
Complete text
Book I