Proof: By Euclid

For since the unit $A$ measures the number $BC$ as many times as $D$ (measures) $EF$, thus as many units as are in $BC$, so many numbers are also in $EF$ equal to $D$.
Let $BC$ have been divided into its constituent units, $BG$, $GH$, and $HC$, and $EF$ into the (divisions) $EK$, $KL$, and $LF$, equal to $D$.
So the multitude of (units) $BG$, $GH$, $HC$ will be equal to the multitude of (divisions) $EK$, $KL$, $LF$.
And since the units $BG$, $GH$, and $HC$ are equal to one another, and the numbers $EK$, $KL$, and $LF$ are also equal to one another, and the multitude of the (units) $BG$, $GH$, $HC$ is equal to the multitude of the numbers $EK$, $KL$, $LF$, thus as the unit $BG$ (is) to the number $EK$, so the unit $GH$ will be to the number $KL$, and the unit $HC$ to the number $LF$.
And thus, as one of the leading (numbers is) to one of the following, so (the sum of) all of the leading will be to (the sum of) all of the following [Prop. 7.12].
Thus, as the unit $BG$ (is) to the number $EK$, so $BC$ (is) to $EF$.
And the unit $BG$ (is) equal to the unit $A$, and the number $EK$ to the number $D$.
Thus, as the unit $A$ is to the number $D$, so $BC$ (is) to $EF$.
Thus, the unit $A$ measures the number $D$ as many times as $BC$ (measures) $EF$ [Def. 7.20] .
(Which is) the very thing it was required to show.
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