Point $H$ is the orthocenter of the acute triangle $ABC$, and $AD$ is its altitude. Tangents are drawn from points $B$ and $C$ to the circle with center $A$ and radius $AD$, which do not coincide with the line $BC$. These tangents intersect at point $P$. Prove that the radius of the incircle of $\triangle BCP$ is equal to $HD$. Proposed by Danylo Khilko