Proof: By Euclid
(related to Proposition: 5.15: Ratio Equals its Multiples)
 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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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"