Proof: By Euclid
(related to Proposition: 7.02: Greatest Common Divisor of Two Numbers  Euclidean Algorithm)
 In fact, if $CD$ measures $AB$, $CD$ is thus a common measure of $CD$ and $AB$, (since $CD$) also measures itself.
 And (it is) manifest that (it is) also the greatest (common measure).
 For nothing greater than $CD$ can measure $CD$.
 But if $CD$ does not measure $AB$ then some number will remain from $AB$ and $CD$, the lesser being continually subtracted, in turn, from the greater, which will measure the (number) preceding it.
 For a unit will not be left.
 But if not, $AB$ and $CD$ will be prime to one another [Prop. 7.1].
 The very opposite thing was assumed.
 Thus, some number will remain which will measure the (number) preceding it.
 And let $CD$ measuring $BE$ leave $EA$ less than itself, and let $EA$ measuring $DF$ leave $FC$ less than itself, and let $CF$ measure $AE$.
 Therefore, since $CF$ measures $AE$, and $AE$ measures $DF$, $CF$ will thus also measure $DF$.
 And it also measures itself.
 Thus, it will also measure the whole of $CD$.
 And $CD$ measures $BE$.
 Thus, $CF$ also measures $BE$.
 And it also measures $EA$.
 Thus, it will also measure the whole of $BA$.
 And it also measures $CD$.
 Thus, $CF$ measures (both) $AB$ and $CD$.
 Thus, $CF$ is a common measure of $AB$ and $CD$.
 So I say that (it is) also the greatest (common measure).
 For if $CF$ is not the greatest common measure of $AB$ and $CD$ then some number which is greater than $CF$ will measure the numbers $AB$ and $CD$.
 Let it (so) measure ($AB$ and $CD$), and let it be $G$.
 And since $G$ measures $CD$, and $CD$ measures $BE$, $G$ thus also measures $BE$.
 And it also measures the whole of $BA$.
 Thus, it will also measure the remainder $AE$.
 And $AE$ measures $DF$.
 Thus, $G$ will also measure $DF$.
 And it also measures the whole of $DC$.
 Thus, it will also measure the remainder $CF$, the greater (measuring) the lesser.
 The very thing is impossible.
 Thus, some number which is greater than $CF$ cannot measure the numbers $AB$ and $CD$.
 Thus, $CF$ is the greatest common measure of $AB$ and $CD$.
 (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"