Proof: By Euclid

For since as $AB$ is to $CD$, so $AE$ (is) to $CF$, (it is) also (the case), alternately, (that) as $BA$ (is) to $AE$, so $DC$ (is) to $CF$ [Prop. 5.16].
And since composed magnitudes are proportional then they will also be proportional (when) separated, (so that) as $BE$ (is) to $EA$, so $DF$ (is) to $CF$ [Prop. 5.17].
Also, alternately, as $BE$ (is) to $DF$, so $EA$ (is) to $FC$ [Prop. 5.16].
And it was assumed that as $AE$ (is) to $CF$, so the whole $AB$ (is) to the whole $CD$.
And, thus, as the remainder $EB$ (is) to the remainder $FD$, so the whole $AB$ will be to the whole $CD$.
Thus, if as the whole is to the whole so the (part) taken away is to the (part) taken away then the remainder to the remainder will also be as the whole (is) to the whole.
(Which is) the very thing it was required to show.
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