If $a|(b+c)$ and $\gcd(b,c)=1$, prove that $\gcd(a,b)=1$ and $\gcd(a,c)=1$. If $a|(b+c)$ and $\gcd(b,c)=1$, prove that $\gcd(a,b)=1$ and $\gcd(a,c)=1$.

I started with:
Suppose $a|(b+c)$ and $\gcd(b,c)=1$.
This means that $ak=b+c$, for some integer $k$.
And $1|b$ and $1|c$.
I know I can solve this using the theorem that if $\gcd(a,b)=1$ then there exists integers $u$ and $v$ such that $au+bv=1$.
But I was wondering how else I can solve this, without using this theorem.
The fact that $\gcd(b,c)=1$ implies there are $k,l \in \Bbb Z$ such that $kb + lc= 1$ (Bézout identity). $a|(b+c)$ tells us $ma = b+c$ for some $m \in \Bbb Z$.
Can you find a Bezout identity for the pairs $(a,b)$ and $(a,c)$ from that?
One thing I like to do is figure if $k$ is a common (positive) divisor of $a$ and $b$, and $a|b+c$ then $k|b+c$.  But we also have $k|b$ so $k| (b+c)-b = c$.  So $k$ is a common (positive) divisor of $b,c$.  But $\gcd(b,c) = 1$ so the only common (positive) divisor of $b,c$ is one so $k=1$.  So the only common divisor of $a$ and $b$ is $1$.  So $\gcd(a,b)=1$.
And doing similar to prove $\gcd(a,c) =1$ is nearly entirely the same.  If $d$ is a common divisor of $a$ and $c$ then $d|b+c$ and $(b+c)-c=b$ and so $d$ is a common divisor of $b$ and $c$ which are relatively prime.
....
Or maybe even easier: $a|b+c$.  So $\gcd(a,b)|a$ so $\gcd(a,b)|b+c$ but $\gcd(a,b)|b$ so $\gcd(a,b)|(b+c)-b=c$.  So $\gcd(a,b)|b$ and $\gcd(a,b)|c$ so $\gcd(a,b)|\gcd(b,c)=1$.  So $\gcd(a,b) = 1$.