Q.Prove that the central angle subtended by an arc of a circle is twice any inscribed angle subtended by the same arc.

Given: In a circle with center O, the arc APB forms 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
Also, since CO is extended to point D, the exterior angle \( \angle \)AOD = \( \angle \)OAC + \( \angle \)OCA
= 2\( \angle \)OCA ---------(i) [∵ \( \angle \)OAC = \( \angle \)OCA]

Similarly, in \( \triangle \)BOC, OB = OC [Radii of the same circle]
∴ \( \angle \)OBC = \( \angle \)OCB
Since CO is extended to point 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)]
Thus, \( \angle \)AOB = 2(\( \angle \)OCA + \( \angle \)OCB) = 2\( \angle \)ACB
(Proved)
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