Selection and Training Session

First, easy counting of angles shows that $AK\parallel H{I}_H$. By condition, $K{I}_H\parallel AH$, so we conclude that $AKI{H}_H$ is a parallelogram. Hence $AK=H{I}_H$. Let $T$ be the tangency point of the incircle of $△BHC$ with the side $BH$, then $I_{H}T\perp HT$; thus from the above $△EKA=△TH{I}_H$. So $EK=TH$. By the well-known formula $TH=0.5(BH+HC-BC)$ whence the required equality follows.