Suppose the gcd (a,b) = 1 and c divides a + b. Prove that gcd (a,c) = 1 = gcd (b,c) I am lost. 
So far...
If $\gcd (a,b) = 1$, by Bezout's Formula
$ax + by = 1$
If $c|(a+b)$, then
$cf = a+b$
Then,
$a (x-y) + cfy = 1$ 
$b(yx) + cfx = 1$
Am I on the right track? Any suggestions?

If by contradiction, $(a,c)=(b,c)=m>1$ then $m|a,m|b$ and $m|c$, which means that there exist an $m>1$ that simultaneously divides both $a$ and $b$ which then implies that $gcd(a,b)\geq m$ which contradicts our assumption.
