1. Given that $ABC$ is an isosceles triangle with $AC = BC$, and $N$ is a point inside the triangle such that $2 \angle ANB = 180^\circ + \angle ACB$. This implies:
$$\angle ANB = 90^\circ + \frac{1}{2} \angle ACB$$
This means that $N$ lies on the circle $\omega$ passing through $A$ and $B$ and tangent to $CA$ and $CB$.

2. Let $P$ be the intersection of the angle bisectors of $\angle CAN$ and $\angle ABN$. Since $P$ is the midpoint of the arc $NA$ of $\omega$, it lies on the perpendicular bisector of $\overline{AN}$.

3. Since $D$ is the intersection of the line $BN$ and the line parallel to $AN$ passing through $C$, we have $CD \parallel AN$. This implies:
$$\angle CDB = \angle AND = 90^\circ - \frac{1}{2} \angle ACB$$
Since $\angle CAB = \angle CDB$, point $D$ lies on the circumcircle of $\triangle ABC$.

4. Now, consider the angles $\angle ADN$ and $\angle ADB$. Since $D$ lies on the circumcircle of $\triangle ABC$, we have:
$$\angle ADB = \angle ACB$$
Therefore:
$$\angle ADN = \angle ACB = 90^\circ - \frac{1}{2} \angle DNA$$
This implies that $\triangle DAN$ is isosceles with $AD = DN$.

5. Since $\triangle DAN$ is isosceles, $DP$ is the perpendicular bisector of $\overline{AN}$. Hence, $DP \perp AN$.

$\blacksquare$