Proof: By Euclid

For since which(ever) part $A$ is of $BC$, $D$ is the same part of $EF$, thus as many numbers as are in $BC$ equal to $A$, so many numbers are also in $EF$ equal to $D$.
Let $BC$ have been divided into $BG$ and $GC$, equal to $A$, and $EF$ into $EH$ and $HF$, equal to $D$.
So the multitude of (divisions) $BG$, $GC$ will be equal to the multitude of (divisions) $EH$, $HF$.
And since $BG$ is equal to $A$, and $EH$ to $D$, thus $BG$, $EH$ (is) also equal to $A$, $D$.
So, for the same (reasons), $GC$, $HF$ (is) also (equal) to $A$, $D$.
Thus, as many numbers as [are] in $BC$ equal to $A$, so many are also in $BC$, $EF$ equal to $A$, $D$.
Thus, as many times as $BC$ is (divisible) by $A$, so many times is the sum $BC$, $EF$ also (divisible) by the sum $A$, $D$.
Thus, which(ever) part $A$ is of $BC$, the sum $A$, $D$ is also the same part of the sum $BC$, $EF$.
(Which is) the very thing it was required to show.
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