1. Let's create a problem similar to the ones described involving angles in a circle with intersecting chords. 2. **Problem:** In circle $O$, chords $AB$ and $CD$ intersect at point $E$ inside the circle. Given that angle $AEB = 50^\circ$ and angle $CED = 70^\circ$, find the measures of angles $AEC$ and $BED$. 3. **Formula:** When two chords intersect inside a circle, the measure of each angle formed is half the sum of the measures of the arcs intercepted by the angle and its vertical angle. Mathematically, if two chords intersect at $E$, then: $$\angle AEC = \frac{1}{2}(\text{arc } AC + \text{arc } BD)$$ and $$\angle BED = \frac{1}{2}(\text{arc } AD + \text{arc } BC)$$ 4. **Important rule:** Vertical angles formed by intersecting chords are equal. 5. Since $\angle AEB = 50^\circ$ and $\angle CED = 70^\circ$ are vertical angles, they correspond to the pairs of arcs as per the formula. 6. Using the property of vertical angles and the given angles, we find: $$\angle AEC = \angle CED = 70^\circ$$ and $$\angle BED = \angle AEB = 50^\circ$$ 7. **Answer:** $$\boxed{\angle AEC = 70^\circ, \quad \angle BED = 50^\circ}$$