Proof: By Euclid

For which(ever) part $AE$ is of $CF$, let $EB$ also be the same part of $CG$.
And since which(ever) part $AE$ is of $CF$, $EB$ is also the same part of $CG$, thus which(ever) part $AE$ is of $CF$, $AB$ is also the same part of $GF$ [Prop. 7.5].
And which(ever) part $AE$ is of $CF$, $AB$ is also assumed (to be) the same part of $CD$.
Thus, also, which(ever) part $AB$ is of $GF$, ($AB$) is also the same part of $CD$.
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 which(ever) part $AE$ is of $CF$, $EB$ [is] also the same part of $GC$, and $GC$ (is) equal to $FD$, thus which(ever) part $AE$ is of $CF$, $EB$ is also the same part of $FD$.
But, which(ever) part $AE$ is of $CF$, $AB$ is also the same part of $CD$.
Thus, the remainder $EB$ is also the same part of the remainder $FD$ that the whole $AB$ (is) of the whole $CD$.
(Which is) the very thing it was required to show.
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