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FKMO 2017 Problem 5

    Let there be cyclic quadrilateral ABCDABCD with LL as the midpoint of ABAB and MM as the midpoint of CDCD . Let AC∩BD=EAC \cap BD = E , and let rays ABAB and DCDC meet again at FF . Let LM∩DE=PLM \cap DE = P . Let QQ be the foot of the perpendicular from PP to EMEM . If the orthocenter of △FLM\triangle FLM is EE , prove the following equality. EP2EQ=12(BD2DF−BC2CF)\frac{EP^2}{EQ} = \frac{1}{2} \left( \frac{BD^2}{DF} - \frac{BC^2}{CF} \right)

    Dataset text AI-MO AoPS. Quoted with source.

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