SHARYGIN 2014 Problem 7

A parallelogram $ABCD$ is given. The perpendicular from $C$ to $CD$ meets the perpendicular from $A$ to $BD$ at point $F$ , and the perpendicular from $B$ to $AB$ meets the perpendicular bisector to $AC$ at point $E$ . Find the ratio in which side $BC$ divides segment $EF$ .

Official text Sharygin Geometry Olympiad 2014 official paper, page 2. Quoted with source.

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