Given: In a circle with center \( O \), the arc \( APB \) subtends the central angle \( \angle AOB \) and an inscribed angle \( \angle ACB \). To Prove: \[ \angle AOB = 2\angle ACB \] Construction: Join \( C, O \) and extend to \( D \). Proof: In \( \triangle AOC \), \( AO = OC \) [radius of the same circle]. \[ \therefore \angle OCA = \angle OAC \] Extending side \( CO \) to point \( D \), the exterior angle: \[ \angle AOD = \angle OAC + \angle OCA \] \[ = 2\angle OCA \quad \text{--------(i)} \quad [\because \angle OAC = \angle OCA] \] Similarly, in \( \triangle BOC \), \( OB = OC \) [radius of the same circle]. \[ \therefore \angle OBC = \angle OCB \] Extending side \( CO \) to point \( D \), the exterior angle: \[ \angle BOD = \angle OCB + \angle OBC \] \[ = 2\angle OCB \quad \text{--------(ii)} \quad [\because \angle OBC = \angle OCB] \] Adding (i) and (ii): \[ \angle AOD + \angle BOD = 2\angle OCA + 2\angle OCB \] \[ \therefore \angle AOB = 2(\angle OCA + \angle OCB) = 2\angle ACB \] \( \text{(Proved)} \)