Given: The central angle \(\angle\)AOB is subtended by the arc APB in a circle centered at O, and the inscribed angle \(\angle\)ACB is subtended by the same arc.
To prove: \(\angle\)AOB = \(2\angle\)ACB.
Construction: Extend CO to point D.
Proof: In \(∆AOC\), AO = OC [radii of the same circle].
∴ \(\angle\)OCA = \(\angle\)OAC.
Also, in \(∆AOC\), extending CO to D makes the exterior angle \(\angle\)AOD = \(\angle\)OAC + \(\angle\)OCA.
\(\therefore \angle\)AOD = \(2\angle\)OCA ---------(i) [Since \(\angle\)OAC = \(\angle\)OCA].
Similarly, in \(∆BOC\), OB = OC [radii of the same circle].
So, \(\angle\)OBC = \(\angle\)OCB.
Also, extending CO to D makes the exterior angle \(\angle\)BOD = \(\angle\)OCB + \(\angle\)OBC.
\(\therefore \angle\)BOD = \(2\angle\)OCB ---------(ii) [Since \(\angle\)OBC = \(\angle\)OCB].
∴ \(\angle\)AOD + \(\angle\)BOD = \(2\angle\)OCA + \(2\angle\)OCB.
[From (i) and (ii)]
That is, \(\angle\)AOB = \(2(\angle\)OCA + \(\angle\)OCB) = \(2\angle\)ACB.
(Proved).