Proof: By Euclid

For as many times as $AE$ is (divisible) by $CF$, so many times let $EB$ also have been made (divisible) by $CG$.
And since $AE$ and $EB$ are equal multiples of $CF$ and $GC$ (respectively), $AE$ and $AB$ are thus equal multiples of $CF$ and $GF$ (respectively) [Prop. 5.1].
And $AE$ and $AB$ are assumed (to be) equal multiples of $CF$ and $CD$ (respectively).
Thus, $AB$ is an equal multiple of each of $GF$ and $CD$.
Thus, $GF$ (is) equal to $CD$.
Let $CF$ have been subtracted from both.
Thus, the remainder $GC$ is equal to the remainder $FD$.
And since $AE$ and $EB$ are equal multiples of $CF$ and $GC$ (respectively), and $GC$ (is) equal to $DF$, $AE$ and $EB$ are thus equal multiples of $CF$ and $FD$ (respectively).
And $AE$ and $AB$ are assumed (to be) equal multiples of $CF$ and $CD$ (respectively).
Thus, $EB$ and $AB$ are equal multiples of $FD$ and $CD$ (respectively).
Thus, the remainder $EB$ will also be the same multiple of the remainder $FD$ as that which the whole $AB$ (is) of the whole $CD$ (respectively).
Thus, if a magnitude is the same multiple of a magnitude that a (part) taken away (is) of a (part) taken away (respectively) then the remainder will also be the same multiple of the remainder as that which the whole (is) of the whole (respectively).
(Which is) the very thing it was required to show.
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