Given: In a circle with center \(O\), arc \(APB\) subtends a central angle \(\angle AOB\) and an inscribed angle \(\angle ACB\). To prove: \[ \angle AOB = 2\angle ACB \] Construction: Join \(C\) and \(O\), and extend it to point \(D\). Proof: In triangle \(\triangle AOC\), \(AO = OC\) [radii of the same circle] \[ \Rightarrow \angle OCA = \angle OAC \] Now, since side \(OC\) is extended to point \(D\), by the exterior angle theorem: \[ \angle AOD = \angle OAC + \angle OCA = 2\angle OCA \tag{i} \] Similarly, in triangle \(\triangle BOC\), \(OB = OC\) [radii of the same circle] \[ \Rightarrow \angle OBC = \angle OCB \] Extend side \(OC\) to point \(D\): \[ \angle BOD = \angle OCB + \angle OBC = 2\angle OCB \tag{ii} \] Adding equations (i) and (ii): \[ \angle AOD + \angle BOD = 2\angle OCA + 2\angle OCB \Rightarrow \angle AOB = 2(\angle OCA + \angle OCB) = 2\angle ACB \] Hence proved.