elementary number theory - Given $d mid a$ and $d mid b$ show that $d mid gcd(a, b-a)$

I can't seem to make this work.

I understand that $d \mid a \wedge d\mid b \Rightarrow d\mid b-a$, and I see that both $d$ and $\gcd(a, b-a)$ will divide $b-a$, but I can't seem to prove that $d \mid \gcd(a, b-a)$.

I've looked at examples, and it seems to be true, but as always, formulating a proof is harder. There has to be something elementary I'm overlooking.

Any help appreciated!