Let \(d=(a,c)\). If \(d|c\), then \(d|(a+b)\), that means:
$$\begin{gather} kd=(a+b) \wedge hd=a \implies \\ (k-h)d=b \implies d|b \end{gather}$$
If \(d\) divides both \(a\) and \(b\), then \(d\) is a multiple of \((a, b)\):
$$\begin{gather} d|(a,b) \\ (a, b) = 1 \implies d=1 \end{gather}$$
This shows that \((a, c) = 1\). Similar proof for \((b, c) = 1\).