Proof: By Euclid
(related to Proposition: 7.10: Multiples of Alternate Ratios of Equal Fractions)
 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 $F$, also, alternately, which(ever) part, or parts, $AG$ is of $DH$, $C$ is also the same part, or the same parts, of $F$ [Prop. 7.9].
 And so, for the same (reasons), which(ever) part, or parts, $GB$ is of $HE$, $C$ is also the same part, or the same parts, of $F$ [Prop. 7.9].
 And so [which(ever) part, or parts, $AG$ is of $DH$, $GB$ is also the same part, or the same parts, of $HE$.
 And thus, which(ever) part, or parts, $AG$ is of $DH$, $AB$ is also the same part, or the same parts, of $DE$ [Prop. 7.5], [Prop. 7.6].
 But, which(ever) part, or parts, $AG$ is of $DH$, $C$ was also shown (to be) the same part, or the same parts, of $F$.
 And, thus] which(ever) parts, or part, $AB$ is of $DE$, $C$ is also the same parts, or the same part, of $F$.
 (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"