**Given:** Let ∠ACB and ∠ADB be any two angles subtended by the arc AB of a circle with center O. **To Prove:** All angles subtended by the arc AB at the circumference (i.e., ∠ACB and ∠ADB) are equal. **Construction:** Join points O and A, and O and B with straight lines. **Proof:** ∠AOB is the central angle formed by the arc AB, and ∠ACB and ∠ADB are angles at the circumference. ∴ ∠AOB = 2∠ACB and ∠AOB = 2∠ADB Therefore, 2∠ACB = 2∠ADB ∴ ∠ACB = ∠ADB [Proved]