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SHARYGIN 2010 Problem 5

    A point EE lies on the altitude BDBD of a triangle ABCABC, and ∠AEC=90∘\angle AEC = 90^\circ. Points O1O_1 and O2O_2 are the circumcenters of the triangles AEBAEB and CEBCEB; points F,LF, L are the midpoints of the segments ACAC and O1O2O_1O_2. Prove that the points L,E,FL, E, F are collinear.

    Official text Sharygin Geometry Olympiad 2010 official paper, page 2. Quoted with source.

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