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) $30^{\circ}$
5. B.

As shown in Figure 2, extend $A B$ to point $F$ such that $B F = A D$, and connect $C F$. It is easy to see that $\triangle A D E$ is an equilateral triangle. Then
$$
\begin{array}{l}
\angle E D B = 120^{\circ}. \\
\text{Also, } C E = E D + D B \\
= A D + D B \\
= D B + B F \\
= D F,
\end{array}
$$

Thus, $\triangle A C F$ is also an equilateral triangle.
From $\angle E D B = 120^{\circ}$ and $\angle C D B = 2 \angle C D E$, we know $\angle C D B = 80^{\circ}$.
In the equilateral $\triangle A C F$, since $A D = B F$, we have $C D = C B$.
Therefore, $\angle D C B = 180^{\circ} - 2 \angle C D B = 20^{\circ}$.