Question 37
Let $ABC$ be a triangle with $\hat{A} \ne 90^\circ$. Points $E, F$ are the foot of altitudes from $B, C$ to $AC, AB$ respectively. The angle bisector of $\hat{A}$ intersects $EF, BC$ at $M, N$ respectively. Perpendicular lines to $EF$ at $M$ and to $BC$ at $N$ intersect each other at $P$. Prove that $P$ lies on the median of vertex $A$.