The quadrilateral $ABCD$ has $AD = DC = CB < AB$ and $AB \parallel CD$. Points $E$ and $F$ lie on the sides $CD$ and $BC$ such that $\angle ADE = \angle AEF$. Prove that: (a) $4CF \le CB$. (b) If $4CF = CB$, then $AE$ is the angle bisector of $\angle DAF$.