Given: In a circle with center \(O\), the arc \(APB\) subtends a central angle \(\angle AOB\) and an inscribed angle \(\angle ACB\).
To Prove: \(\angle AOB = 2\angle ACB\).
Construction: Extend \(CO\) to point \(D\).
Proof: In \( \triangle AOC \), \( AO = OC \) [Radii of the same circle].
∴ \(\angle OCA = \angle OAC\).
Again, since \( CO \) is extended to \( D \), the exterior angle \( \angle AOD \) is given by:
\( \angle AOD = \angle OAC + \angle OCA \).
\( = 2\angle OCA \) ---------(i) [∵ \( \angle OAC = \angle OCA \)]
Similarly, in \( \triangle BOC \), \( OB = OC \) [Radii of the same circle].
Thus, \( \angle OBC = \angle OCB \).
Again, since \( CO \) is extended to \( D \), the exterior angle \( \angle BOD \) is given by:
\( \angle BOD = \angle OCB + \angle OBC \).
\( = 2\angle OCB \) -------- (ii)
∴ \( \angle AOD + \angle BOD = 2\angle OCA + 2\angle OCB \) [(From (i) and (ii))].
Thus, \( \angle AOB = 2(\angle OCA + \angle OCB) = 2\angle ACB \).
(Proof Complete)