RTP: If gcd(a,b)=1, then gcd(a, (a+b))=1, where a,b are positive integers

Proof
Note 1=as+bt
Assume gcd(a, (a+b))=k
Then
am+(a+b)n=a(m+n)+bn=k*1
a(m+n)+bn=aks+bkt
Thus
m+n=ks,n=kt
So
m+kt=ks,m=ks-kt
So
a(ks-kt)+bkt=aks+bkt
-akt=0
This is impossible, so our assumption is wrong
Thus our statement is true

Questions
1. Is this proof valid?
2. Can I generalize to :If gcd(a,b)=k, then gcd(a, (a+b))=k
3. Is there a simpler non-contradiction, non-FTA proof?