Proof: By Euclid

For since which(ever) parts $AB$ is of $C$, $DE$ (is) also the same parts of $F$, thus as many parts of $C$ as are in $AB$, so many parts of $F$ are also in $DE$.
Let $AB$ have been divided into the parts of $C$, $AG$ and $GB$, and $DE$ into the parts of $F$, $DH$ and $HE$.
So the multitude of (divisions) $AG$, $GB$ will be equal to the multitude of (divisions) $DH$, $HE$.
And since which(ever) part $AG$ is of $C$, $DH$ is also the same part of $F$, thus which(ever) part $AG$ is of $C$, the sum $AG$, $DH$ is also the same part of the sum $C$, $F$ [Prop. 7.5].
And so, for the same (reasons), which(ever) part $GB$ is of $C$, the sum $GB$, $HE$ is also the same part of the sum $C$, $F$.
Thus, which(ever) parts $AB$ is of $C$, the sum $AB$, $DE$ is also the same parts of the sum $C$, $F$.
(Which is) the very thing it was required to show.
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