Problem

Source: IGO 2021 Advanced P1

Tags: geometry, circumcircle, IGO



Acute-angled triangle $ABC$ with circumcircle $\omega$ is given. Let $D$ be the midpoint of $AC$, $E$ be the foot of altitude from $A$ to $BC$, and $F$ be the intersection point of $AB$ and $DE$. Point $H$ lies on the arc $BC$ of $\omega$ (the one that does not contain $A$) such that $\angle BHE=\angle ABC$. Prove that $\angle BHF=90^\circ$.