In an acute-angled triangle $ABC$, the orthocenter is $H$. $I_H$ is the incenter of $\vartriangle BHC$. The bisector of $\angle BAC$ intersects the perpendicular from $I_H$ to the side $BC$ at point $K$. Let $F$ be the foot of the perpendicular from $K$ to $AB$. Prove that $2KF+BC=BH +HC$ A. Voidelevich