# 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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