In an equilateral (irregular) pentagon $A B C D E$, angle $A B C$ is twice the angle $D B E$. Find the measure of angle $A B C$.

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Since $\angle E B D=\angle A B E+\angle C B D$, we can take a point $P$ on side $E D$ such that $\angle E B P=\angle A B E=\angle A E B$, i.e., $B P \| A E$. Then $\angle P B D=\angle E B D-\angle E B P=\angle C B D=\angle B D C$, i.e., $B P \| C D$. Therefore, $A E \| C D$, and since $A E = C D$, $C D E A$ is a parallelogram. Thus, $A C=E D$, which means triangle $A B C$ is equilateral and $\angle A B C=60^{\circ}$.