If $a$ divides $bc$ and $\gcd(a,b) = d$ then $\frac a d$ divides c

I'm trying to prove that if $a$ divides $bc$ and $\gcd(a,b) = d$ then $\frac a d$ divides c. I tried using Bezout identity but couldn't get anywhere.

Let $a=a'd$ and $b=b'd$. Note that $a'$ and $b'$ are relatively prime. We want to show that $a'$ divides $c$. Since $a'd$ divides $b'dc$, it follows that $a'$ divides $b'c$.

By the Bezout Identity there are integers $x$ and $y$ such that $a'x+b'y=1$. Multiply through by $c$. Note that $a'$ divides $a'xc$ and $a'$ divides $b'cy$. The result follows.

$a\mid ac,bc\,\Rightarrow\, a\mid(ac,bc)\overset{\rm\color{#C00}{DL}}=(a,b)c\,\Rightarrow\,a/(a,b)\mid c\ $ by the GCD Distributive Law ($\rm\color{#C00}{DL}$)