1. **Problem statement:** Find the measure of angle $\angle ACB$ given a circle with points $D, E, B, C$ on the circumference and point $A$ inside the circle, with $\angle DAE = 125^\circ$ and $\angle ADE = 50^\circ$. 2. **Key concepts:** - The sum of angles in triangle $ADE$ is $180^\circ$. - The measure of an inscribed angle in a circle is half the measure of its intercepted arc. 3. **Step 1: Find $\angle DEA$ in triangle $ADE$** $$\angle DEA = 180^\circ - \angle DAE - \angle ADE = 180^\circ - 125^\circ - 50^\circ = \cancel{180^\circ} - 125^\circ - 50^\circ = 5^\circ$$ 4. **Step 2: Use the fact that $\angle ACB$ intercepts the same arc as $\angle DEA$** Since $D, E, B, C$ lie on the circle, $\angle ACB$ and $\angle DEA$ intercept the same arc $DE$. 5. **Step 3: Calculate $\angle ACB$** The inscribed angle $\angle ACB$ is half the measure of the intercepted arc, so $$\angle ACB = \frac{1}{2} \times \angle DEA = \frac{1}{2} \times 5^\circ = 2.5^\circ$$ **Final answer:** $$\boxed{2.5^\circ}$$