Proof: By Euclid

For since $AB$ and $DE$ are equal multiples of $C$ and $F$ (respectively), thus as many magnitudes as there are in $AB$ equal to $C$, so many (are there) also in $DE$ equal to $F$.
Let $AB$ have been divided into (magnitudes) $AG$, $GH$, $HB$, equal to $C$, and $DE$ into (magnitudes) $DK$, $KL$, $LE$, equal to $F$.
So, the number of (magnitudes) $AG$, $GH$, $HB$ will equal the number of (magnitudes) $DK$, $KL$, $LE$.
And since $AG$, $GH$, $HB$ are equal to one another, and $DK$, $KL$, $LE$ are also equal to one another, thus as $AG$ is to $DK$, so $GH$ (is) to $KL$, and $HB$ to $LE$ [Prop. 5.7].
And, thus (for proportional magnitudes), as one of the leading (magnitudes) will be to one of the following, so all of the leading (magnitudes will be) to all of the following [Prop. 5.12].
Thus, as $AG$ is to $DK$, so $AB$ (is) to $DE$.
And $AG$ is equal to $C$, and $DK$ to $F$.
Thus, as $C$ is to $F$, so $AB$ (is) to $DE$.
Thus, parts have the same ratio as similar multiples, taken in corresponding order.
(Which is) the very thing it was required to show.
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