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BALTIC-WAY 2000 Problem 3

    Given a triangle ABCA B C with ∠A=90∘\angle A=90^{\circ} and ∣AB∣≠∣AC∣|A B| \neq|A C|. The points D,E,FD, E, F lie on the sides BC,CA,ABB C, C A, A B, respectively, in such a way that AFDEA F D E is a square. Prove that the line BCB C, the line FEF E and the line tangent at the point AA to the circumcircle of the triangle ABCA B C intersect in one point.

    Dataset text NuminaMath olympiads. Quoted with source.

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