Given: ABCD is a cyclic quadrilateral inscribed in a circle with center O. To Prove: ∠ABC + ∠ADC = 180° and ∠BAD + ∠BCD = 180° Construction: Join A to O and C to O. Proof: The central angle ∠AOC is subtended by arc ADC, and the inscribed angle ∠ABC is also subtended by the same arc. ∴ ∠AOC = 2∠ABC ⇒ ∠ABC = \(\frac{1}{2}\) ∠AOC ——(i) Similarly, ∠AOC is also subtended by arc ABC, and the inscribed angle ∠ADC is subtended by the same arc. ∴ ∠AOC = 2∠ADC ⇒ ∠ADC = \(\frac{1}{2}\) ∠AOC ——(ii) From (i) and (ii), we get: ∠ABC + ∠ADC = \(\frac{1}{2}\) ∠AOC + \(\frac{1}{2}\) ∠AOC = \(\frac{1}{2}\)(∠AOC + ∠AOC) = \(\frac{1}{2}\) × 360° (since ∠AOC + ∠COA = full angle around center) = 180° Similarly, by joining B to O and D to O, we can prove that ∠BAD + ∠BCD = 180° (Proved)