Proof: By Euclid
(related to Proposition: 7.16: Natural Number Multiplication is Commutative)
 For since $A$ has made $C$ (by) multiplying $B$, $B$ thus measures $C$ according to the units in $A$ [Def. 7.15] .
 And the unit $E$ also measures the number $A$ according to the units in it.
 Thus, the unit $E$ measures the number $A$ as many times as $B$ (measures) $C$.
 Thus, alternately, the unit $E$ measures the number $B$ as many times as $A$ (measures) $C$ [Prop. 7.15].
 Again, since $B$ has made $D$ (by) multiplying $A$, $A$ thus measures $D$ according to the units in $B$ [Def. 7.15] .
 And the unit $E$ also measures $B$ according to the units in it.
 Thus, the unit $E$ measures the number $B$ as many times as $A$ (measures) $D$.
 And the unit $E$ was measuring the number $B$ as many times as $A$ (measures) $C$.
 Thus, $A$ measures each of $C$ and $D$ an equal number of times.
 Thus, $C$ is equal to $D$.
 (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"