What I know so far:
We know by the definition of congruence that $n$ divides $ab-ac$. So, there exists an integer $k$ such that $a(b-c)=kn$, and since $d=(a,n)$ we know that $a=ds$ and $n=dt$ from some integers $s$ and $t$. Then substituting for $a$ and $n$, we see that $ds(b-c)=k(dt)$.
Since $\gcd(a,n)=d$, we get $$\gcd\bigg(\frac{a}{d},\frac{n}{d}\bigg)=1.$$ Write $$r=\frac{a}{d}\quad\text{and}\quad s=\frac{n}{d}.$$ Then $$\gcd(r,s)=1\quad\text{and}\quad \frac{a}{n}=\frac{r}{s}.$$ Also, we get $$n\big|(ab-ac).$$ Hence, $$\frac{ab-ac}{n}\in\Bbb Z.$$ But, $$\begin{align} \frac{ab-ac}{n}&=\frac{a(b-c)}{n}\\ &=\frac{r(b-c)}{s}. \end{align}$$ Hence, $$\frac{r(b-c)}{s}\in\Bbb Z.$$ Thus, $$s\big|r(b-c).$$ Since $\gcd(r,s)=1$, using the Euclid's Lemma, we get $$s\big|(b-c).$$ Hence, $$b \equiv c \mod s.$$ Thus, $$b \equiv c \mod{\frac{n}{d}}.$$
Note that since $d=(a,n)$ $$ a(b-c)=kn\implies\frac ad(b-c)=k\frac nd\tag{1} $$ Since $\left(\frac ad,\frac nd\right)=\frac{(a,n)}d=1$, Bezout's Identity says there are $x,y\in\mathbb{Z}$ so that $$ x\frac ad+y\frac nd=1\tag{2} $$ Therefore, applying $(1)$ and $(2)$, we get $$ \begin{align} xk\frac nd &=x\frac ad(b-c)\\ &=\left(1-y\frac nd\right)(b-c)\\ \left(xk+y(b-c)\right)\frac nd\tag{3} &=b-c \end{align} $$ which says that $$ b\equiv c\quad\left(\text{mod}\,{\frac nd}\right)\tag{4} $$
Let $\,x=b\!-\!c.\ $ Then $\ n\mid ax\!\iff\! n\mid ax,nx$ $\iff\! n\mid(ax,nx)=(a,n)x\!\iff\! \dfrac{n}{(a,n)}\mid x$
