Prove that $a\mid c$; given that $a\mid bc$ and $\gcd(a,b) = 1$. Please help answer the number theory problem. The proof I came across goes like this:
Let $d = \gcd(a,b)$ then $d = ma + nb$.
$ma + nb = 1$.
then
$mac + nbc = c$.
$a\mid mac$, $a\mid nbc$, therefore $a\mid c$.
Not sure how last step unfolded. Thanks. 
(This is my personal question I have not posted it anywhere else, and  the context of my question is my own) 

We know $\gcd(a,b)=1\iff au+bv=1$ and $a\mid bc\iff bc=ak$ for some $k\in\Bbb Z$. Then $$au+bv=1\implies auc+bvc=c\implies auc+akv=c\implies a(uc+kv)=c\implies a\mid c$$
Note, after the second implication we introduced $bc=ak$.
