Let $AD$ be the bisector of angle $A$ of the triangle $ABC$. One considers the points M, N on the half-lines $(AB$ and $(AC$, respectively, such that $\angle MDA = \angle B$ and $\angle NDA = \angle C$. Let $AD \cap MN=\{P\}$. Prove that: $$AD^3 = AB \cdot AC\cdot AP$$