Proof: By Euclid

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 $D$.

fig011e

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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