Proof

By hypothesis, $c > 0$ is a common divisor of two integers $a$ and $b$ and the integers $\frac ac$ and $\frac bc$ are co-prime.
This means that $\gcd\left(\frac ac,\frac bc\right)=1$.
Therefore, at least one of the integers $\frac ac$ and $\frac bc$, and so at least one of the integers $a$ and $b$ must be different from zero.
Set $d:=\gcd(a,b)$, and we have $d\neq 0.$
From the definition of the greatest common divisor we have that $c\mid d$. Thus, $\frac dc$ is an integer.
Now, from the multiplication of rational numbers it follows that \[\frac dc \frac ad=\frac ac,\quad\frac dc \frac bd=\frac bc,\] which means that \[\frac dc\mid \frac ac,\quad \frac dc\mid \frac bc.\]
Because $\gcd\left(\frac ac,\frac bc\right)=1$, $d > 0$, $c > 0$, any divisor of $\gcd\left(\frac ac,\frac bc\right)$ must divide $1$.
This is only true for $\frac dc=1,$ which means that $c=d=\gcd(a,b)$.
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Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927