Proof: By Euclid
(related to Proposition: Prop. 10.003: Greatest Common Measure of Commensurable Magnitudes)
 Let $AB$ and $CD$ be the two given magnitudes, of which (let) $AB$ (be) the lesser.
 So, it is required to find the greatest common measure of $AB$ and $CD$.
 For the magnitude $AB$ either measures, or (does) not (measure), $CD$.
 Therefore, if it measures ($CD$), and (since) it also measures itself, $AB$ is thus a common measure of $AB$ and $CD$.
 And (it is) clear that (it is) also (the) greatest.
 For a (magnitude) greater than magnitude $AB$ cannot measure $AB$.
 So let $AB$ not measure $CD$.
 And continually subtracting in turn the lesser (magnitude) from the greater, the remaining (magnitude) will (at) some time measure the (magnitude) before it, on account of $AB$ and $CD$ not being incommensurable [Prop. 10.2].
 And let $AB$ leave $EC$ less than itself (in) measuring $ED$, and let $EC$ leave $AF$ less than itself (in) measuring $FB$, and let $AF$ measure $CE$.
 Therefore, since $AF$ measures $CE$, but $CE$ measures $FB$, $AF$ will thus also measure $FB$.
 And it also measures itself.
 Thus, $AF$ will also measure the whole (of) $AB$.
 But, $AB$ measures $DE$.
 Thus, $AF$ will also measure $ED$.
 And it also measures $CE$.
 Thus, it also measures the whole of $CD$.
 Thus, $AF$ is a common measure of $AB$ and $CD$.
 So I say that (it is) also (the) greatest (common measure).
 For, if not, there will be some magnitude, greater than $AF$, which will measure (both) $AB$ and $CD$.
 Let it be $G$.
 Therefore, since $G$ measures $AB$, but $AB$ measures $ED$, $G$ will thus also measure $ED$.
 And it also measures the whole of $CD$.
 Thus, $G$ will also measure the remainder $CE$.
 But $CE$ measures $FB$.
 Thus, $G$ will also measure $FB$.
 And it also measures the whole (of) $AB$.
 And (so) it will measure the remainder $AF$, the greater (measuring) the lesser.
 The very thing is impossible.
 Thus, some magnitude greater than $AF$ cannot measure (both) $AB$ and $CD$.
 Thus, $AF$ is the greatest common measure of $AB$ and $CD$.
 Thus, the greatest common measure of two given commensurable magnitudes, $AB$ and $CD$, has been found.
 (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"