By assumption $\gcd(a,b)=1$, thus we can use Bezout’s lemma to find integers $x,y$ such that $ax+by=1$. Hence $c\cdot(ax+by)=c$ and $acx+bcy=c$. Since $a\mid a$ and $a\mid bc$ (by hypothesis), we conclude that $a\mid acx+bcy=c$ as claimed. ∎