Let $AD$ be the bisector of triangle $ABC$. Circle $\omega$ passes through the vertex $A$ and touches the side $BC$ at point $D$. This circle intersects the sides $AC$ and $AB$ for the second time at points $M$ and $N$ respectively. Lines $BM$ and $CN$ intersect the circle for the second time $\omega$ at points $P$ and $Q$, respectively. Lines $AP$ and $AQ$ intersect side $BC$ at points $K$ and $L$, respectively. Prove that $KL=\frac12 BC$