Q.Prove that the angle subtended at the center of any circle by an arc is double the angle subtended by the same arc at any point on the remaining part of the circle.

Thanks for the clarification! Here's the English translation of your previous post with no additional lines added: Given: A central angle ∠AOB and an inscribed angle ∠ACB are subtended by arc APB of a circle with center O. To prove: ∠AOB = 2∠ACB Construction: Join C and O, and extend it to point D. Proof:In triangle ∆AOC, AO = OC [radii of the same circle] ∴ ∠OCA = ∠OAC Since side CO is extended to point D, the exterior angle ∠AOD = ∠OAC + ∠OCA = 2∠OCA ——— (i) In triangle ∆BOC, OB = OC [radii of the same circle] So, ∠OBC = ∠OCB And the exterior angle formed by extending CO to D is ∠BOD = ∠OCB + ∠OBC = 2∠OCB ——— (ii) ∴ ∠AOD + ∠BOD = 2∠OCA + 2∠OCB = 2(∠OCA + ∠OCB) So, ∠AOB = 2∠ACB (Proved)
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