How to show $((a,b),b)=(a,b)$? I want to prove $((a,b),b)=(a,b)$ for $a$, $b$ integers with some formal mathematical statement like Bezout's identity i.e., $(a,b)=d=am+bn$ or something else. 
Any ideas? 
Here $(a,b)$ means gcd.

My first sight is let $(a,b)=d$ and factors $a=d q_a$, $b=d q_b$ and proceed. 
I want more rigorous proof
 A: Hint $\,\ ((a,b),b) = (a,(b,b)) = (a,b)\ $ by the gcd Associative Law.

Or directly $\,\ d\mid((a,b),b)\iff d\mid(a,b),b\iff d\mid a,b,\, b\iff d\mid a,b\iff d\mid (a,b)\,$ where we used the gcd Universal Property $3$ times.

Or $\ d\mid b\,\Rightarrow\, (d,b) = (d, b\bmod d) = (d,0) = d\,$ by the Euclidean algorithm, or  $\,(d,b) = d(1,b/d) = d\,$ by the gcd Distributive Law.

Or use $\ (a\Bbb Z + b\Bbb Z) + b\Bbb Z = a\Bbb Z + (b\Bbb Z + b\Bbb Z) = a\Bbb Z + b\Bbb Z\ $ and Bezout.
A: Just note that $d=(a,b)$ is a divisor of $b$ and that $(x,y)=x$ whenever $x\mid y$.
A: Hint for proof with Bezout: because $\gcd(a,b)$ is a linear combination of $a$ and $b$, any linear combination of $\gcd(a,b)$ and $b$ is also a linear combination of $a$ and $b$. 
A: $$ \exists x, y \in \mathbb{Z} : (a, b)x+by= ((a, b), b)$$
$$(a, b) \mid (a, b) \land (a, b) \mid b \Longrightarrow (a, b) \mid (a, b)x+by=((a, b), b)$$
$$(a, b) \mid ((a, b), b) \land ((a, b), b) \mid (a, b) \Longrightarrow (a, b) = ((a, b), b)$$
For the last step you need that both $((a, b), b)$ and $(a, b)$ are positive because otherwise you could get that $((a, b), b)= -(a, b)$.
