Question: If $\gcd (a,b)=1$ and $c\mid a+b$, then $\gcd (a,c)=\gcd(b,c)=1$ 

so far I have:
$ax+by=1$

$(a+b)/c=q$ for some integer $q$ so $a+b=cq$

$au+cv=1$

$bl+ck=1$

$au+cv=bl+ck$

Where should I go from here? step by step explanation please!

Answer: 
As you've started, using Bezout's Lemma we have:

$$ax + by = 1$$

has a solution in $\mathbb{Z}$. Now from the condition we have: $a + b = cq \implies b = cq - a$. Substituting we have:

$$ax + (cq - a)y = 1$$
$$a(x-y) + c(qy) = 1$$

Since $x+y$ and $qy$ are integers $gcd(a,c) = 1$. Simularly we get $gcd(b,c) = 1$
**Hint** $\ a\,$ is coprime to $\,a+b\,$ since $\,(a,a+b) = (a,b) = 1,\ $ using $\ (a,n) = (a,\,n\ {\rm mod}\ a).\,$  

Thus $\ a\,$ is coprime to every factor of $\,a+b,\,$ including $\,c.\,$ Ditto for $\,b\,$ by $\, a\leftrightarrow b\,$ symmetry.