IOM 2017 Problem 1

Let $ABCD$ be a parallelogram in which angle at $B$ is obtuse and $AD>AB$ . Points $K$ and $L$ on $AC$ such that $\angle ADL=\angle KBA$ (the points $A, K, C, L$ are all different, with $K$ between $A$ and $L$ ). The line $BK$ intersects the circumcircle $\omega$ of $ABC$ at points $B$ and $E$ , and the line $EL$ intersects $\omega$ at points $E$ and $F$ . Prove that $BF||AC$ .