2024 CGMO

Proof. Let $H$ be the orthocenter of triangle $ABD$. Since $DF\perp AB$, the points $H,D,F$ are colinear, as shown:

Case 1: When $\angle ADB<90^{\circ }$: • $H$ lies inside triangle $ABD$ • By orthocenter properties, $\angle BAD$ and $\angle BHD$ are supplementary • Given $\angle BED=\angle BAD$, we have $\angle BED=180^{\circ }-\angle BHD$ • Thus, $B,H,D,E$ are concyclic This implies: • $\angle EHF=\angle EBD=\angle EBC=\angle EAF$

Therefore, A, H, E, F are concyclic Consequently, $\angle AEF=\angle AHF=180^{\circ }-\angle ABD=180^{\circ }-\angle ABC$

Other Cases: When $\angle ADB>90^{\circ }$, D lies inside triangle ABH When $\angle ADB=90^{\circ }$, D coincides with H In all cases (using directed angles when necessary), we conclude: $$\angle BEF=\angle BEA+\angle AEF=\angle BCA+(180^{\circ }-\angle ABC)$$ Since $\angle BCA$ and $\angle ABC$ are fixed angles of triangle $ABC$, $\angle BEF$ is constant. $\square$