1. **Problem statement:** Given a circle with center O, points A, B, C, and D on the circumference, and $\angle BOC = 120^\circ$, find the value of $x$, where $x$ is an angle at the circumference subtended by the arc AD. 2. **Key theorem:** The angle subtended by an arc at the center of the circle is twice the angle subtended by the same arc at any point on the circumference. Mathematically, if $\theta$ is the central angle and $\alpha$ is the inscribed angle subtending the same arc, then: $$\theta = 2\alpha$$ 3. **Identify arcs and angles:** Here, $\angle BOC = 120^\circ$ is the central angle subtending arc BC. 4. **Find the angle subtended by arc BC at the circumference:** Using the theorem, $$120^\circ = 2 \times \angle BAC \implies \angle BAC = \frac{120^\circ}{2} = 60^\circ$$ 5. **Relate $x$ to the given angles:** Since points A, D, B, C lie on the circle, and $x$ is the angle subtended by arc AD at the circumference, and $\angle BOC$ subtends arc BC, the arcs AD and BC together make the full circle of $360^\circ$. 6. **Calculate arc AD:** $$\text{Arc } AD = 360^\circ - 120^\circ = 240^\circ$$ 7. **Calculate angle $x$ at the circumference subtended by arc AD:** $$x = \frac{1}{2} \times 240^\circ = 120^\circ$$ **Final answer:** $$x = 120^\circ$$