# Greatest common divisor addition of numbers [duplicate]

Show that $\text{gcd}(m+n,m) = \text{gcd}(m,n)$

Here is what I have so far, using the definition of $\text{GCD}$ for some integer d we have

$d = x(m+n) + y(m)=(x)(n) + (x+y) m$ Now let $x' = x+y$ and $y'=y$,

$d'= x'(m) + y'(n)=(x+y)m+y(n)$ where $\text{GCD}(m,n)=d'=d$

Is this reasoning sufficient?