RMO 2016 B5

Let $ABC$ be a right angled triangle with $\angle B=90^{\circ}$ . Let $AD$ be the bisector of angle $A$ with $D$ on $BC$ . Let the circumcircle of triangle $ACD$ intersect $AB$ again at $E$ ; and let the circumcircle of triangle $ABD$ intersect $AC$ again at $F$ . Let $K$ be the reflection of $E$ in the line $BC$ . Prove that $FK = BC$ .