Given: The central angle \( \angle AOB \) and the inscribed angle \( \angle ACB \) subtended by the arc \( APB \) in a circle centered at \( O \). To Prove: \( \angle AOB = 2\angle ACB \). Construction: Extend \( CO \) to \( D \). Proof: In \( \triangle AOC \), \( AO = OC \) [Radii of the same circle]. ∴ \( \angle OCA = \angle OAC \). Extending side \( CO \) to \( D \), the exterior angle is \[ \angle AOD = \angle OAC + \angle OCA = 2\angle OCA \quad \text{(i)} \quad [\because \angle OAC = \angle OCA] \]
Similarly, in \( \triangle BOC \), \( OB = OC \) [Radii of the same circle]. So, \( \angle OBC = \angle OCB \). Extending side \( CO \) to \( D \), the exterior angle is \[ \angle BOD = \angle OCB + \angle OBC = 2\angle OCB \quad \text{(ii)} \]
From (i) and (ii), \[ \angle AOD + \angle BOD = 2\angle OCA + 2\angle OCB \] \[ \Rightarrow \angle AOB = 2(\angle OCA + \angle OCB) = 2\angle ACB \] (Proved).