$19 \cdot 2$ As shown in the figure, 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) $60^{\circ}$.
(B) $120^{\circ}$.
(C) $90^{\circ}$.
(D) $45^{\circ}$.
(China Heilongjiang Province Harbin Junior High School Mathematics Competition, 1992)
[Solution] Given that all sides of the pentagon are equal, without loss of generality, considering the symmetric case, as shown in the figure, we have
$$
\left\{\begin{array}{l}
2 \alpha+\gamma=180^{\circ}, \\
2 \alpha+\beta=2 \beta, \\
\beta+2 \delta=180^{\circ}, \\
4 \alpha+\beta+2 \gamma+2 \delta=540^{\circ} .
\end{array}\right.
$$

Solving these equations, we get $\alpha=15^{\circ}, \beta=30^{\circ}, \gamma=150^{\circ}$,
$$
\delta=75^{\circ} \text {. }
$$

Therefore, $\angle A B C=2 \alpha+\beta=60^{\circ}$.
Hence, the correct choice is $(A)$.