Proof: By Euclid

For $A$ and $BC$ are either prime to one another, or not.
Let $A$ and $BC$, first of all, be prime to one another.
So separating $BC$ into its constituent units, each of the units in $BC$ will be some part of $A$.
Hence, $BC$ is parts of $A$.
So let $A$ and $BC$ be not prime to one another.
So $BC$ either measures, or does not measure, $A$.
Therefore, if $BC$ measures $A$ then $BC$ is part of $A$.
And if not, let the greatest common measure, $D$, of $A$ and $BC$ have been taken [Prop. 7.2], and let $BC$ have been divided into $BE$, $EF$, and $FC$, equal to $D$.
And since $D$ measures $A$, $D$ is a part of $A$.
And $D$ is equal to each of $BE$, $EF$, and $FC$.
Thus, $BE$, $EF$, and $FC$ are also each part of $A$.
Hence, $BC$ is parts of $A$.
Thus, any number is either part or parts of any (other) number, the lesser of the greater.
(Which is) the very thing it was required to show.
∎

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