**Given:** ABCD is a cyclic quadrilateral inscribed in a circle with center O. **To Prove:** \[ \angle ABC + \angle ADC = 180^\circ \quad \text{and} \quad \angle BAD + \angle BCD = 180^\circ \] **Construction:** Join A to O and C to O. **Proof:** The central angle ∠AOC is subtended by arc ADC, and the inscribed angle ∠ABC also subtends the same arc. So, \[ \angle AOC = 2\angle ABC \Rightarrow \angle ABC = \frac{1}{2} \angle AOC \quad \text{—— (i)} \] Again, the central angle ∠AOC is also subtended by arc ABC, and the inscribed angle ∠ADC subtends the same arc. So, \[ \angle AOC = 2\angle ADC \Rightarrow \angle ADC = \frac{1}{2} \angle AOC \quad \text{—— (ii)} \] From (i) and (ii), \[ \angle ABC + \angle ADC = \frac{1}{2} \angle AOC + \frac{1}{2} \angle AOC = \frac{1}{2}(2 \angle AOC) = \angle AOC \] Since ∠AOC is a central angle spanning half the circle, \[ \angle ABC + \angle ADC = 180^\circ \] Similarly, by joining B to O and D to O, we can prove: \[ \angle BAD + \angle BCD = 180^\circ \quad \text{[Proved]} \]