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Geometry Difficulty 7.0 National olympiad, round 2 Prove it China

Given an acute triangle $ABC$ with $AB < BC < CA$ . Let $D$ be a moving point on side $BC$ , and $E$ be a moving point on the minor arc $\widehat{BC}$ of the circumcircle of $ABC$ , such that $\angle BAD = \angle BED$ .
Let $F$ be the intersection point of the line through $D$ perpendicular to $AB$ with the extension of $AC$ . Prove that $\angle BEF$ is constant.

Solution

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. $\square$

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