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Team Selection Test for IMO

Geometry Difficulty 7.7 National Olympiad, round 2 Prove it Turkey

In a convex quadrilateral $ABCD$ , let $E$ be the intersection of the diagonals. It is given that $\angle EDC = \angle DEC = \angle BAD$ . If $F$ is a point on the line segment $[BC]$ such that $\angle BAF + \angle EBF = \angle BFE$ , then show that $A, B, F, D$ are concyclic.

Solution

It is easy to observe that the point $F$ is unique when $A, B, C, D$ are fixed. Hence it is enough to show that the intersection of the circumcircle of $ABD$ and $[CB]$ satisfies the properties of the point $F$ . Let the intersection be $F'$ .
$\angle DBF' = \angle DAF'$ . Then $\angle BAF' + \angle EBX = \angle DAB$ . On the other hand we have $\angle DEC = \angle DAB = \angle DF'C$ . Thus, $D, E, F', C$ are concyclic. Hence $\angle EF'B = \angle EDC = \angle DAB = \angle BAF' + \angle EBF'$ and we are done.

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