Let $k=b-c$. We want to show that $m$ divides $ak$ if and only if $\frac{m}{d}$ divides $k$.
Let $a=a_1d$ and let $m=m_1 d$. Note that $\gcd(a_1,m_1)=1$.
One direction: We show that if $\frac{m}{d}$ divides $k$, then $m$ divides $ak$. By assumption we have $k=\frac{m}{d}q=m_1q$ for some $q$. Thus $ak=(a_1 d)(m_1q)=(a_1q)m$, so $m$ divides $ak$.
The other direction: We show that if $m$ divides $ak$, the $\frac{m}{d}$ divides $k$.
We have $m_1d$ divides $(a_1d)k$, so $m_1$ divides $a_1k$. But since $d=\gcd(a,m)$, we have $\gcd(a_1,m_1)=1$. Then from $m_1$ divides $a_1k$ we can conclude that $m_1$ divides $k$.
Remark: We have used without proof the fact that if $u$ divides $vw$ and $u$ and $v$ are relatively prime, then $u$ divides $w$. That's because it has probably been proved in your course. For a proof, we can use the Bezout argument you began. There are integers $x$ and $y$ such that $ux+vy=1$. Multiply through by $w$. We get $u(xw)+(vw)y=w$. By assumption $u$ divides $vw$, so $u$ divides $u(xw)+(vw)y$, that is, $u$ divides $w$.
