1. **Stating the problem:** We are given a circle with points A, B, C, D, E on the circumference. We know $m\angle ACB = 66^\circ$ and the length $mAB = 132$. We need to find $m\angle ADB$, $m\angle AEB$, and confirm $m\angle ACB$. 2. **Important circle theorem:** The measure of an inscribed angle is half the measure of its intercepted arc. That is, for an inscribed angle $\angle XYZ$, $m\angle XYZ = \frac{1}{2} m\overset{\frown}{XZ}$. 3. **Using the given $m\angle ACB = 66^\circ$:** Since $\angle ACB$ is an inscribed angle intercepting arc $AB$, the measure of arc $AB$ is: $$m\overset{\frown}{AB} = 2 \times 66^\circ = 132^\circ$$ 4. **Finding $m\angle ADB$:** $\angle ADB$ is also an inscribed angle intercepting the same arc $AB$ (since points A, D, B lie on the circle). Therefore, $$m\angle ADB = \frac{1}{2} m\overset{\frown}{AB} = \frac{1}{2} \times 132^\circ = 66^\circ$$ 5. **Finding $m\angle AEB$:** $\angle AEB$ is an inscribed angle intercepting arc $AB$ as well (points A, E, B on the circle). So, $$m\angle AEB = \frac{1}{2} m\overset{\frown}{AB} = 66^\circ$$ 6. **Summary:** - $m\angle ADB = 66^\circ$ - $m\angle AEB = 66^\circ$ - $m\angle ACB = 66^\circ$ (given) All three angles intercept the same arc $AB$ of measure $132^\circ$. Final answers: $$m\angle ADB = 66^\circ$$ $$m\angle AEB = 66^\circ$$ $$m\angle ACB = 66^\circ$$