Let $ABC$ be an acute triangle. Let $O$ be the circumcenter and $H$ be the orthocenter of triangle $ABC$. Let $D$ in the minor arc $BC$. and $K$ such that $DK$ is a diameter of the circumcircle. Let $I$ be the foot of the perpendicular dropped from $K$ to $AC$ ($I$ in $AC$), and $KP$ be the perpendicular of $BC$ ($P$ in $BC$). Let $F$ the intersection of $IP$ and $HK$. Prove that $HF$ = $FK$.
(If you love geometry, trying to solve this problem is the good choice, use all your knowledge about geometry such as Cyclic quadrilateral, Similar triangles, parallel, tangent of the circle, orthocenter,Parallelogram, Midline of the triangles, Ceva theorem or Menelaus theorem, Euler circle ... . it isn't the homework, it justs the quzzle for you who really good at Geometry :D)
