5. In $\triangle A B C$, it is known that $\angle C A B=60^{\circ}, D, E$ are points on sides $A B, A C$ respectively, and $\angle A E D=60^{\circ}$, $E D+D B=C E, \angle C D B=2 \angle C D E$. Then $\angle D C B$ $=(\quad)$.
(A) $15^{\circ}$
(B) $20^{\circ}$
(C) $25^{\circ}$
(D) $30^{\circ}$