1. **Problem statement:** Given points A, B, C, D on a circle with center O, lines AOBE and DCE are straight, and CO = CE. Given \(\angle AOD = 69^\circ\), find \(x = \angle CEO\). 2. **Key facts and formulas:** - Since O is the center, segments like CO and CE are radii or equal lengths. - \(CO = CE\) implies triangle COE is isosceles with \(\angle CEO = \angle OCE = x\). - The straight line DCE means \(\angle DCE + \angle ECE = 180^\circ\). - \(\angle AOD = 69^\circ\) is a central angle. 3. **Analyze the figure:** - Since AOBE is a straight line, \(\angle AOB = 180^\circ\). - \(\angle AOD = 69^\circ\) is given, so \(\angle DOB = 180^\circ - 69^\circ = 111^\circ\). 4. **Using the isosceles triangle COE:** - Since \(CO = CE\), \(\triangle COE\) is isosceles with base OE. - The angles at C are \(x\) and \(x\), so the vertex angle at O is \(180^\circ - 2x\). 5. **Relate angles at O:** - \(\angle AOD = 69^\circ\) and \(\angle COE = 180^\circ - 2x\). - Since A, O, D, C are points on the circle and lines intersect, \(\angle AOD + \angle COE = 180^\circ\) (they are supplementary). 6. **Set up equation:** $$69^\circ + (180^\circ - 2x) = 180^\circ$$ 7. **Solve for x:** $$69^\circ + 180^\circ - 2x = 180^\circ$$ $$69^\circ = 2x$$ $$x = \frac{69^\circ}{2} = 34.5^\circ$$ **Final answer:** \(x = 34.5^\circ\)