Proof about divisibility and greatest common divisor that doesn't use Bezout's formula

Question

Suppose the $\gcd(a,b) = 1$ and $c$ divides $a + b$. Prove that $\gcd (a,c) = 1 = \gcd (b,c)$.

This is relatively easy using Bezout's formula. See here. But is there a way to do it without Bezout?

Answer 1

Assume that $(a,c) \not = 1$ and let $p$ be a prime divisor of $(a,c)$. Then $p \mid c \mid a+b$, but as $p \mid a$ we have that $p \mid b$. But this contradicts the fact that $(a,b) = 1$. Therfore $(a,c) = 1$. Similarly $(b,c) = 1$