Given a triangle $ ABC$ with $ \angle A = 90^{\circ}$ and $ AB \neq AC$. The points $ D$, $ E$, $ F$ lie on the sides $ BC$, $ CA$, $ AB$, respectively, in such a way that $ AFDE$ is a square. Prove that the line $ BC$, the line $ FE$ and the line tangent at the point $ A$ to the circumcircle of the triangle $ ABC$ intersect in one point.
Problem
Source: Baltic Way 2000 Problem 3
Tags: geometry, circumcircle, trigonometry, geometry unsolved