Proof: By Euclid
(related to Proposition: Prop. 10.015: Commensurability of Sum of Commensurable Magnitudes)
 For let the two commensurable magnitudes $AB$ and $BC$ be laid down together.
 I say that the whole $AC$ is also commensurable with each of $AB$ and $BC$.
 For since $AB$ and $BC$ are commensurable, some magnitude will measure them.
 Let it (so) measure (them), and let it be $D$.
 Therefore, since $D$ measures (both) $AB$ and $BC$, it will also measure the whole $AC$.
 And it also measures $AB$ and $BC$.
 Thus, $D$ measures $AB$, $BC$, and $AC$.
 Thus, $AC$ is commensurable with each of $AB$ and $BC$ [Def. 10.1] .
 And so let $AC$ be commensurable with $AB$.
 I say that $AB$ and $BC$ are also commensurable.
 For since $AC$ and $AB$ are commensurable, some magnitude will measure them.
 Let it (so) measure (them), and let it be $D$.
 Therefore, since $D$ measures (both) $CA$ and $AB$, it will thus also measure the remainder $BC$.
 And it also measures $AB$.
 Thus, $D$ will measure (both) $AB$ and $BC$.
 Thus, $AB$ and $BC$ are commensurable [Def. 10.1] .
 Thus, if two magnitudes, and so on ....
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"