Question 110
Let $ABCD$ be an isosceles trapezoid with $AB = CD$ and $BC \lt AD$. The bisector of the angle $\angle ABC$ meets the side $AD$ at point $E$ and the line passing through $E$ parallel to the bisector of the angle $\angle CDA$ meets the side $BC$ at point $F$. Prove that $\angle AFE = \angle EFB$.