[Sum of angles in a triangle. Theorem of the exterior angle.] Pentagons

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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Let the angles at the base $BE$ of the isosceles triangle $ABE$ be $\alpha$, and the angles at the base $BD$ of the isosceles triangle $BCD$ be $\beta$. From the condition, it follows that $\alpha + \beta = \angle DBE$. Therefore, $\angle AED + \angle CDE = (\alpha + \angle BED) + (\beta + \angle BDE) = (\alpha + \beta) + (\angle BED + \angle BDE) = \angle DBE + (180^\circ - \angle DBE) = 180^\circ$. Hence, $AE \parallel CD$, and since $AE = CD$, $ACDE$ is a parallelogram, so $AC = DE = AB = BC$. Therefore, triangle $ABC$ is equilateral, and $\angle ABC = 60^\circ$.

## Answer

$60^\circ$.