If c|ab And gcd(c, a) = 1, Then c|b
Since gcd(c, a) = 1:
$$\displaylines{1=cx+ay\\b=b(cx)+b(ay)}$$
Since c|bc and c|ab, then c|(bcm + abn), where m and n can be any integers, including x and y:
$$\begin{align} c|(bcm + abn) & \implies c|(bcx + bay)\\ & \implies c|(b) \end{align}$$