Proof: By Euclid

For $CD$ is not parts of $A$.
For, if possible, let it be (parts of $A$).
Thus, $EF$ is also the same parts of $B$ that $CD$ (is) of $A$ [Def. 7.20] , [Prop. 7.13].
Thus, as many parts of $A$ as are in $CD$, so many parts of $B$ are also in $EF$.
Let $CD$ have been divided into the parts of $A$, $CG$ and $GD$, and $EF$ into the parts of $B$, $EH$ and $HF$.
So the multitude of (divisions) $CG$, $GD$ will be equal to the multitude of (divisions) $EH$, $HF$.
And since the numbers $CG$ and $GD$ are equal to one another, and the numbers $EH$ and $HF$ are also equal to one another, and the multitude of (divisions) $CG$, $GD$ is equal to the multitude of (divisions) $EH$, $HF$, thus as $CG$ is to $EH$, so $GD$ (is) to $HF$.
Thus, as one of the leading (numbers is) to one of the following, so will (the sum of) all of the leading (numbers) be to (the sum of) all of the following [Prop. 7.12].
Thus, as $CG$ is to $EH$, so $CD$ (is) to $EF$.
Thus, $CG$ and $EH$ are in the same ratio as $CD$ and $EF$, being less than them.
The very thing is impossible.
For $CD$ and $EF$ were assumed (to be) the least of those (numbers) having the same ratio as them.
Thus, $CD$ is not parts of $A$.
Thus, (it is) a part (of $A$) [Prop. 7.4].
And $EF$ is the same part of $B$ that $CD$ (is) of $A$ [Def. 7.20] , [Prop. 7.13].
Thus, $CD$ measures $A$ the same number of times that $EF$ (measures) $B$.
(Which is) the very thing it was required to show.
∎

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