Proof: By Euclid

For since which(ever) part $A$ is of $BC$, $D$ is also the same part of $EF$, thus as many numbers as are in $BC$ equal to $A$, so many 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 the numbers $BG$ and $GC$ are equal to one another, and the numbers $EH$ and $HF$ are also equal to one another, and the multitude of (divisions) $BG$, $GC$ is equal to the multitude of (divisions) $EH$, $HC$, thus which(ever) part, or parts, $BG$ is of $EH$, $GC$ is also the same part, or the same parts, of $HF$.
And hence, which(ever) part, or parts, $BG$ is of $EH$, the sum $BC$ is also the same part, or the same parts, of the sum $EF$ [Prop. 7.5], [Prop. 7.6].
And $BG$ (is) equal to $A$, and $EH$ to $D$.
Thus, which(ever) part, or parts, $A$ is of $D$, $BC$ is also the same part, or the same parts, of $EF$.
(Which is) the very thing it was required to show.
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