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Proving ∠BHF=90

Source: IGO 2021 Advanced P1

December 30, 2021
geometrycircumcircleIGO

Problem Statement

Acute-angled triangle ABCABC with circumcircle ω\omega is given. Let DD be the midpoint of ACAC, EE be the foot of altitude from AA to BCBC, and FF be the intersection point of ABAB and DEDE. Point HH lies on the arc BCBC of ω\omega (the one that does not contain AA) such that ∠BHE=∠ABC\angle BHE=\angle ABC. Prove that ∠BHF=90∘\angle BHF=90^\circ.
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