5. As shown in Figure 1, in the pentagon $A B C D E$ with all sides equal, $\angle A B C$ $=2 \angle D B E$. Then, $\angle A B C$ is ().
(A) $45^{\circ}$
(B) $60^{\circ}$
(C) $90^{\circ}$
(D) $120^{\circ}$
5. (B).

Rotate $\triangle B A E$ clockwise around point $B$ by $\angle A B C$ to get $\triangle B C F$, as shown in Figure 4.
$$
\because \angle D B F=\angle D B C+\angle A B E=\frac{1}{2} \angle A B C=
$$
$\angle D B E$,
$\therefore \triangle D B E \cong$
$\triangle D B F(S A S)$.
Therefore, $\triangle C D F$ is an

equilateral triangle, and point $C$ is
the circumcenter of
$\triangle D B F$. Hence
$$
\begin{array}{l}
\angle A B C=2 \angle D B F \\
=\angle D C F=60^{\circ} .
\end{array}
$$