Solution:
Let $\omega$ be a circle passing through the points $A$, $N$, $D$ and let $M$ be the point where $\omega$ intersects $BD$ for the second time. The quadrilateral $ANDM$ is cyclic and it follows that
$$\angle NDM + \angle NAM = \angle NDM + \angle BDC = 180^\circ$$
and

Figure 1
$$\angle NAM = \angle BDC$$
Now we have
$$\angle BDC = \angle ACD = \angle NAC$$
and
$$\angle NAM = \angle NAC$$
So the points $A$, $M$, $C$ are collinear and $M \equiv E$.

In this solution we do not need the circle passing through the points $A$, $N$ and $D$.
Because of the given symmetry we have
$$\angle ANE = \angle ABD$$
and from the equality $AD = AB$ the triangle $ABD$ is isosceles with
$$\angle ABD = \angle ADE$$
From (1) and (2) we get that $\angle ANE = \angle ADE$, which means that the quadrilateral $ANDE$ is cyclic.