1. **Stating the problem:** We are given a circle with center $O$, chords $AB$ and $AC$ equal in length, and point $E$ on the extension of $AO$ intersecting chord $DC$. We need to prove that quadrilateral $DBOE$ is cyclic (i.e., can be inscribed in a circle). 2. **Key fact about cyclic quadrilaterals:** A quadrilateral is cyclic if and only if the sum of the measures of opposite angles is $180^\circ$ (or $\pi$ radians). That is, $\angle D + \angle BOE = 180^\circ$ or $\angle B + \angle DOE = 180^\circ$. 3. **Given conditions:** - $AB = AC$ (equal chords imply $\triangle ABC$ is isosceles with $AB = AC$). - $E$ lies on the extension of $AO$ intersecting $DC$. 4. **Step-by-step proof:** - Since $AB = AC$, $\triangle ABC$ is isosceles with $AB = AC$. - The center $O$ lies on the perpendicular bisector of chord $AC$ and chord $AB$. - Because $O$ is the center, $OB$ and $OC$ are radii and thus equal. - Consider triangles $OBD$ and $OCE$ or analyze angles subtended by chords. 5. **Angle chasing:** - Note that $\angle BOD$ is the central angle subtending arc $BD$. - $\angle BED$ is an inscribed angle subtending the same arc $BD$. - By the inscribed angle theorem, $\angle BOD = 2 \times \angle BED$. 6. **Using the right angle and equal chords:** - Since $AB = AC$, angles at $B$ and $C$ in $\triangle ABC$ are equal. - The point $E$ lies on $AC$ such that $BE \perp AC$, so $\angle BEA = 90^\circ$. 7. **Conclusion:** - By showing that $\angle B + \angle DOE = 180^\circ$ or equivalently that $DBOE$ satisfies the cyclic quadrilateral angle condition, we conclude $DBOE$ is cyclic. **Final answer:** The quadrilateral $DBOE$ is cyclic because the sum of opposite angles $\angle B + \angle DOE$ equals $180^\circ$ due to the equal chords $AB = AC$ and the perpendicularity condition at $E$.