Question 32
The triangle $ABC$ has $\angle ABC = 90^\circ$ and $\angle BCA = 30^\circ$. Let $AD$ be the bisector of the angle $\angle BAC$, $D \in BC$, and $BE \perp AC$, $E \in AC$. Denote $M$ the intersection of the lines $AD$ and $BE$, and $P$ the midpoint of the segment $CM$. Prove that $AC = 4 \cdot DP$.