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