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)