Proof: By Euclid

For if (it is) not (the case that) as $AB$ is to $BE$, so $CD$ (is) to $FD$, then it will surely be (the case that) as $AB$ (is) to $BE$, so $CD$ is either to some (magnitude) less than $DF$, or (some magnitude) greater (than $DF$).¹
Let it, first of all, be to (some magnitude) less (than $DF$), (namely) $DG$.
And since composed magnitudes are proportional, (so that) as $AB$ is to $BE$, so $CD$ (is) to $DG$, they will thus also be proportional (when) separated [Prop. 5.17].
Thus, as $AE$ is to $EB$, so $CG$ (is) to $GD$.
But it was also assumed that as $AE$ (is) to $EB$, so $CF$ (is) to $FD$.
Thus, (it is) also (the case that) as $CG$ (is) to $GD$, so $CF$ (is) to $FD$ [Prop. 5.11].
And the first (magnitude) $CG$ (is) greater than the third $CF$.
Thus, the second (magnitude) $GD$ (is) also greater than the fourth $FD$ [Prop. 5.14].
But (it is) also less.
The very thing is impossible.
Thus, (it is) not (the case that) as $AB$ is to $BE$, so $CD$ (is) to less than $FD$.
Similarly, we can show that neither (is it the case) to greater (than $FD$).
Thus, (it is the case) to the same (as $FD$).
Thus, if separated magnitudes are proportional then they will also be proportional (when) composed.
(Which is) the very thing it was required to show.

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Here, Euclid assumes, without proof, that a fourth magnitude proportional to three given magnitudes can always be found (translator's note) ↩