Solution problem 2 of shortlist problem geometry in IMO 2015
Let ABC be a triangle inscribed into a circle with center O. A circle with center A meets the side BC at points D and E such that D lies between B and E. Moreover, let F and G be the common points of and . We assume that F lies on the arc AB of not containing C, and G lies on the arc AC of not containing B. The circumcircles of the triangles BDF and CEG meet the sides AB and AC again at K and L, respectively. Suppose that the lines FK and GL are distinct and intersect at X. Prove that the points A, X, and O are collinear.
It suﬃces to prove that the lines FK and GL are symmetric about AO. Now the segments AF and AG, being chords of with the same length, are clearly symmetric with respect to AO. Hence it is enough to show
Let us denote the circumcircles of BDF and CEG by , respectively. To prove (1), we start from
In view of the circles , this may be rewritten as
Due to the circles , we obtain . Thereby the problem is solved.
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