Question 20
In an acute-angled triangle $ABC$ the orthocenter is $H$. $I_H$ is the incenter of $\triangle BHC$. 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)
Prove that $2KF + BC = BH + HC$.
(A. Voidelevich)