Proof: By Euclid

For as many times as $B$ measures $A$, so many units let there be in $C$.
Since $B$ measures $A$ according to the units in $C$, and the unit $D$ also measures $C$ according to the units in it, the unit $D$ thus measures the number $C$ as many times as $B$ (measures) $A$.
Thus, alternately, the unit $D$ measures the number $B$ as many times as $C$ (measures) $A$ [Prop. 7.15].
Thus, which(ever) part the unit $D$ is of the number $B$, $C$ is also the same part of $A$.
And the unit $D$ is a part of the number $B$ called the same as it (i.e., a $B$th part).
Thus, $C$ is also a part of $A$ called the same as $B$ (i.e., $C$ is the $B$th part of $A$).
Hence, $A$ has a part $C$ which is called the same as $B$ (i.e., $A$ has a $B$th part).
(Which is) the very thing it was required to show.
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