AB and CD are chords of a circle with center O. When extended, they intersect at point P. To prove: \(\angle AOC - \angle BOD = 2\angle BPC\) Construction:Join A and D. **Proof:** Angle \(\angle AOC\) is a central angle subtended by arc AC, and \(\angle ADC\) is the corresponding inscribed angle. ∴ \(\angle AOC = 2\angle ADC\) Similarly, Angle \(\angle BOD\) is a central angle subtended by arc BD, and \(\angle BAD\) is the corresponding inscribed angle. ∴ \(\angle BOD = 2\angle BAD\) In triangle ADP, the exterior angle \(\angle ADC = \angle PAD + \angle APD\) ⇒ \(\angle ADC - \angle PAD = \angle APD\) Multiply both sides by 2: \(2(\angle ADC - \angle PAD) = 2\angle APD\) ⇒ \(2\angle ADC - 2\angle PAD = 2\angle APD\) ⇒ \(2\angle ADC - 2\angle BAD = 2\angle BPC\) ⇒ \(\angle AOC - \angle BOD = 2\angle BPC\) (Proved)