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 $EF$.
Problem
Source: Sharygin Geometry Olympiad 2014 - Problem 7
Tags: geometry, parallelogram, ratio, perpendicular bisector, geometry unsolved