Show that $\gcd(a,b) = \gcd(a+b,b)$ Show that $\gcd(a,b) = \gcd(a+b,b)$
I understand the proof up till this point:
By definition of $\gcd(a,b)$ it implies that $d|a$ and $d\mid b$ thus, $d\mid a+b$, a linear combination of the two.
By definition of $\gcd(a+b,b)$ it implies that $d\mid a+b$ and $d\mid b$ thus, $d\mid(a+b)-b = a +b - b  =b$
In the second statement, I don't understand where the subtraction comes from in $d\mid(a+b)-b = a +b - b  =b$

In the first statement,it should be like that:
Let $d=gcd(a,b)$
By definition, $d \mid a$, $d \mid b$,so $d$ divides also any linear combination of $a,b$,so it divides also their sum $a+b$.
In the second statement:
Let $d=gcd(a+b,b)$.
By definition, $d \mid a+b \text{ and } d \mid b$,so $d$ divides also any linear combination of $a+b,b$,so it divides also their difference $a+b-b=a$.
