If magnitudes be proportional separando, they will also be proportional componendo.

Let AE, EB, CF, FD be magnitudes proportional separando, so that, as AE is to EB, so is CF to FD; I say that they will also be proportional componendo, that is, as AB is to BE, so is CD to FD.

For, if CD be not to DF as AB to BE, then, as AB is to BE, so will CD be either to some magnitude less than DF or to a greater.

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.

Therefore etc. Q. E. D.