Question 166
Let $ABC$ be an acute triangle with $AB \gt AC$. Consider point $N$ inside the triangle such that $\angle ANB = 180^\circ - \angle BAC$, $\angle NBA = 180^\circ - 2\angle ACB$. The angle bisector of $\angle NBA$ intersects the angle bisector of $\angle NAC$ at point $P$. The line passing through $C$ and parallel to line $AN$ intersects line $NB$ at point $D$. Prove that $\angle DPN = 2\angle ABC$.