Question 46
Let $H$ be the orthocenter of an acute triangle $ABC$, and $M$ be the midpoint of the side $BC$. Let $P$ be the point of intersection of the line $AM$ and the line through $H$ and perpendicular to the line $AM$. Prove that $AM \cdot PM = BM^2$ holds. Here for a line segment $XY$ its length is also denoted by $XY$.