VN-TST 2006 Problem 1

Given an acute angles triangle $ABC$ , and $H$ is its orthocentre. The external bisector of the angle $\angle BHC$ meets the sides $AB$ and $AC$ at the points $D$ and $E$ respectively. The internal bisector of the angle $\angle BAC$ meets the circumcircle of the triangle $ADE$ again at the point $K$ . Prove that $HK$ is through the midpoint of the side $BC$ .