In the acute triangle $ABC$, the altitude $AH$ is drawn. Using segments $AB,BH,CH$ and $AC$ as diameters circles $\omega_1, \omega_2, \omega_3$ and $\omega_4$ are constructed respectively. Besides the point $H$, the circles $\omega_1$ and $\omega_3$ intersect at the point $P,$ and the circles $\omega_2$ and $\omega_4$ interext at point $Q$. The lines $BQ$ and $CP$ intersect at point $N$. Prove that this point lies on the midline of triangle $ABC$, which is parallel to $BC$.