Question 159
In the triangle $ABC$ the points $P$ and $Q$ are chosen on the side $BC$ such that $\angle BAP = \angle ACB$ and $\angle CAQ = \angle ABC$. Let $N$ be a point on the line $AP$ such that $AP = PN$ and $M$ be a point on the line $AQ$ such that $AQ = QM$. The lines $BN$ and $CM$ intersect at the point $L$ and let $K$ be the midpoint of $BC$. Prove that $\angle BAK = \angle CAL$.