A circumscribed quadrilateral $ABCD$ is given. It is known that $\angle{ACB} = \angle{ACD}$. On the angle bisector of $\angle{C}$, a point $E$ is marked such that $AE \bot BD$. Point $F$ is the foot of the perpendicular line from point $E$ to the side $BC$. Prove that $AB = BF$.