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: 
A Bezout argument can be pushed through, but I think the following argument is more natural. 

Suppose to the contrary that $\gcd(a,c)\gt 1$. Then there exists $d\gt 1$ that divides $a$ and $c$. Since $a+b=cq$ for some $q$, we have $b=cq-a$. Since $d$ divides $c$ and $a$, we have $d$ divides $b$, contradicting the fact that $a$ and $b$ are relatively prime.
**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.