Given: The central angle \(\angle\)AOB is formed by the arc APB in a circle with center O, and an inscribed angle \(\angle\)ACB is formed by the same arc.
To Prove: \(\angle\)AOB = 2\(\angle\)ACB
Construction: Extend line CO to point D.
Proof: In \(∆AOC\), AO = OC [Radii of the same circle]
∴ \(\angle\)OCA = \(\angle\)OAC
Also, by extending CO to point D, the exterior angle \(\angle\)AOD = \(\angle\)OAC + \(\angle\)OCA
= 2\(\angle\)OCA ---------(i) [Since \(\angle\)OAC = \(\angle\)OCA]
Similarly, in \(∆BOC\), OB = OC [Radii of the same circle]
So, \(\angle\)OBC = \(\angle\)OCB
Again, extending CO to point D gives the exterior angle \(\angle\)BOD = \(\angle\)OCB + \(\angle\)OBC
= 2\(\angle\)OCB -------- (ii) [Since \(\angle\)OBC = \(\angle\)OCB]
Thus, \(\angle\)AOD + \(\angle\)BOD = 2\(\angle\)OCA + 2\(\angle\)OCB [from (i) and (ii)]
Which means, \(\angle\)AOB = 2(\(\angle\)OCA + \(\angle\)OCB) = 2\(\angle\)ACB
(Proved).