Question 109
Let $ABC$ be an acute triangle, $D$ be the foot of the perpendicular from $A$ to $BC$, and $M$ be the midpoint of $AC$. Let $P$ be a point on the segment $BM$ such that $\angle PAM = \angle MBA$. Given $\angle BAP = 41^\circ$ and $\angle PDB = 115^\circ$, find the value of $\angle BAC$.