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