Proof: By Euclid
(related to Proposition: 5.18: Magnitudes Proportional Separated are Proportional Compounded)
 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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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes