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Geometry Difficulty 7.7 National olympiad, round 2 Prove it Turkey

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

Solution

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

Figure 1

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