ISL 1991 Problem 2

$ABC$ is an acute-angled triangle. $M$ is the midpoint of $BC$ and $P$ is the point on $AM$ such that $MB = MP$ . $H$ is the foot of the perpendicular from $P$ to $BC$ . The lines through $H$ perpendicular to $PB$ , $PC$ meet $AB, AC$ respectively at $Q, R$ . Show that $BC$ is tangent to the circle through $Q, H, R$ at $H$ .

Original Formulation:

For an acute triangle $ABC, M$ is the midpoint of the segment $BC, P$ is a point on the segment $AM$ such that $PM = BM, H$ is the foot of the perpendicular line from $P$ to $BC, Q$ is the point of intersection of segment $AB$ and the line passing through $H$ that is perpendicular to $PB,$ and finally, $R$ is the point of intersection of the segment $AC$ and the line passing through $H$ that is perpendicular to $PC.$ Show that the circumcircle of $QHR$ is tangent to the side $BC$ at point $H.$