Proof: By Euclid
(related to Proposition: 7.01: Sufficient Condition for Coprimality)
 For if $AB$ and $CD$ are not prime to one another then some number will measure them.
 Let (some number) measure them, and let it be $E$.
 And let $CD$ measuring $BF$ leave $FA$ less than itself, and let $AF$ measuring $DG$ leave $GC$ less than itself, and let $GC$ measuring $FH$ leave a unit, $HA$.
 In fact, since $E$ measures $CD$, and $CD$ measures $BF$, $E$ thus also measures $BF$.^{1} And ($E$) also measures the whole of $BA$.
 Thus, ($E$) will also measure the remainder $AF$.^{2} And $AF$ measures $DG$.
 Thus, $E$ also measures $DG$.
 And ($E$) also measures the whole of $DC$.
 Thus, ($E$) will also measure the remainder $CG$.
 And $CG$ measures $FH$.
 Thus, $E$ also measures $FH$.
 And ($E$) also measures the whole of $FA$.
 Thus, ($E$) will also measure the remaining unit $AH$, (despite) being a number.
 The very thing is impossible.
 Thus, some number does not measure (both) the numbers $AB$ and $CD$.
 Thus, $AB$ and $CD$ are prime to one another.
 (Which is) the very thing it was required to show.
∎
Thank you to the contributors under CC BYSA 4.0!
 Github:

 nonGithub:
 @Fitzpatrick
References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes