First; parmenides51 wrote: In an acute-angled triangle $ABC$, point $I$ is the incenter, $H$ is the orthocenter, $O$ is the center of the circumscribed circle, $\color{red}T$ and $ \color {red}K $ are the touchpoints of the $A$-excircl and incircle with side $BC$ respectively. It turned out that the segment $TI$ is passing through the point $O$. Prove that $HK$ is the angle bisector of $\angle BHC$. (Matvii Kurskyi) by angle computation it suffices to prove that $ HK\parallel AI$ Let $M$ midpoint of $BC$ , since $OM\parallel KI$ then $O$ is midpoint of IT so $\overrightarrow {AH}=\overrightarrow {2OM }=\overrightarrow { IK}$ hence $AHKI$ is parallelogram hence $HK\parallel AI$.