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 \(D\).
Proof: In \(∆AOC\), we have \(AO = OC\) [Radius of the same circle].
∴ \(\angle OCA = \angle OAC\)
Also, since \(CO\) is extended to \(D\), the exterior angle \(\angle AOD = \angle OAC + \angle OCA\)
\(= 2\angle OCA\) ---------(i) [∵ \(\angle OAC = \angle OCA\)]
Similarly, in \(∆BOC\), we have \(OB = OC\) [Radius of the same circle].
Thus, \(\angle OBC = \angle OCB\)
Since \(CO\) is extended to \(D\), the exterior angle \(\angle BOD = \angle OCB + \angle OBC\)
\(= 2\angle OCB\) -------- (ii) [∵ \(\angle OBC = \angle OCB\)]
∴ \(\angle AOD + \angle BOD = 2\angle OCA + 2\angle OCB\) [From (i) and (ii)]
That is, \(\angle AOB = 2(\angle OCA + \angle OCB) = 2\angle ACB\)
(Proved)