State and prove the inscribed angle theorem.

Step 1 ：\(\text{Inscribed Angle Theorem}: \angle APB = \frac{1}{2}\angle AOB\) where A, B, and P are points on the circle, and O is the center

Step 2 ：\(\text{Proof} \)

Step 3 ：\(\text{Case 1: }\angle AOB \text{ is a minor arc}\)

Step 4 ：\(\text{Extend }AP, BP \text{ to meet at }D \text{. Let }\angle AOB = x \text{, } OD \text{ bisects } \angle AOB \text{.}\)

Step 5 ：\(\text{Since }P \in OD \text{ bisects } \angle AOB , \angle AOP = \angle POB = \frac{x}{2}\)

Step 6 ：\(\text{In } \triangle AOP, AP = OP \text{, so } \triangle AOP \text{ is isosceles, then } \angle APO = \angle AOP = \frac{x}{2}\)

Step 7 ：\(\text{In } \triangle BOP, BP = OP \text{, so } \triangle BOP \text{ is isosceles, then } \angle BPO = \angle BOP = \frac{x}{2}\)

Step 8 ：\(\angle APB = \angle APO + \angle BPO = \frac{x}{2} + \frac{x}{2} = \frac{x}{1} = x \text{, thus } \angle APB = \frac{1}{2}\angle AOB \)

Step 9 ：\(\text{Case 2: }\angle AOB \text{ is a major arc}\)

Step 10 ：\(\text{Let }\angle AOB = x \text{, } OD \text{ bisects } \angle AOB \)

Step 11 ：\(\text{Since }P \in OD \text{ bisects } \angle AOB , \angle AOP = \angle POB = \frac{360^\circ - x}{2}\)

Step 12 ：\(\text{In } \triangle AOP, AP = OP \text{, so } \triangle AOP \text{ is isosceles, then } \angle APO = \angle AOP = \frac{360^\circ - x}{2}\)

Step 13 ：\(\text{In } \triangle BOP, BP = OP \text{, so } \triangle BOP \text{ is isosceles, then } \angle BPO = \angle BOP = \frac{360^\circ - x}{2}\)

Step 14 ：\(\angle APB = \angle APO + \angle BPO = \frac{360^\circ - x}{2} + \frac{360^\circ - x}{2} = 360^\circ - x \text{, thus } \angle APB = \frac{1}{2}\angle AOB \)