Proof: By Euclid
(related to Proposition: 7.35: Least Common Multiple Divides Common Multiple)
 For if $E$ does not measure $CD$ then let $E$ leave $CF$ less than itself (in) measuring $DF$.
 And since $A$ and $B$ (both) measure $E$, and $E$ measures $DF$, $A$ and $B$ will thus also measure $DF$.
 And ($A$ and $B$) also measure the whole of $CD$.
 Thus, they will also measure the remainder $CF$, which is less than $E$.
 The very thing is impossible.
 Thus, $E$ cannot not measure $CD$.
 Thus, ($E$) measures ($CD$).
 (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"