Let us consider a circle with center O, where arc APB subtends a central angle ∠AOB and an inscribed angle ∠ACB. To prove: ∠AOB = 2∠ACB Construction: Join points C and O, and extend line CO to point D. Proof: In triangle △AOC, OA = OC [radii of the same circle] ∴ ∠OCA = ∠OAC Now, since side CO is extended to point D, Exterior angle ∠AOD = ∠OAC + ∠OCA = 2∠OCA ——— (i) [because ∠OAC = ∠OCA] Similarly, in triangle △BOC, OB = OC [radii of the same circle] ∴ ∠OBC = ∠OCB Again, extending side CO to point D, Exterior angle ∠BOD = ∠OCB + ∠OBC = 2∠OCB ——— (ii) [because ∠OBC = ∠OCB] From equations (i) and (ii), we get: ∠AOD + ∠BOD = 2∠OCA + 2∠OCB ∴ ∠AOB = 2(∠OCA + ∠OCB) = 2∠ACB Hence proved: The central angle ∠AOB is twice the inscribed angle ∠ACB subtended by the same arc.