Proof: By Euclid
(related to Proposition: Prop. 10.011: Commensurability of Elements of Proportional Magnitudes)
 Let $A$, $B$, $C$, $D$ be four proportional magnitudes, (such that) as $A$ (is) to $B$, so $C$ (is) to $D$.
 And let $A$ be commensurable with $B$.
 I say that $C$ will also be commensurable with $D$.
 For since $A$ is commensurable with $B$, $A$ thus has to $B$ the ratio which (some) number (has) to (some) number [Prop. 10.5].
 And as $A$ is to $B$, so $C$ (is) to $D$.
 Thus, $C$ also has to $D$ the ratio which (some) number (has) to (some) number.
 Thus, $C$ is commensurable with $D$ [Prop. 10.6].
 And so let $A$ be incommensurable with $B$.
 I say that $C$ will also be incommensurable with $D$.
 For since $A$ is incommensurable with $B$, $A$ thus does not have to $B$ the ratio which (some) number (has) to (some) number [Prop. 10.7].
 And as $A$ is to $B$, so $C$ (is) to $D$.
 Thus, $C$ does not have to $D$ the ratio which (some) number (has) to (some) number either.
 Thus, $C$ is incommensurable with $D$ [Prop. 10.8].
 Thus, if four magnitudes, and so on ....
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"