Let $a \mid (b + c)$ and $a \mid b$. Show that $a \mid c$.
We're given
$$b+c = am$$
$$b = an$$
for some integers $m,n$.
Substituting for $b$ in the first equation we have
$$(ab) + c = am$$
Re-arranging,
$$c = a(m-b)$$
That is, $a \mid b$.