Answer: Proof — each pair of opposite angles of a cyclic quadrilateral adds up to \(180^\circ\).
Given: \(\square ABCD\) is a cyclic quadrilateral inscribed in a circle with centre O.
To prove: \(\angle A + \angle C = 180^\circ\) and \(\angle B + \angle D = 180^\circ\).
Proof: arc BCD and arc BAD together make the complete circle.
\(\therefore m(\text{arc } BCD) + m(\text{arc } BAD) = 360^\circ\)
\(\angle A\) is an inscribed angle standing on arc BCD and \(\angle C\) is an inscribed angle standing on arc BAD. By the inscribed angle theorem,
\(\angle A = \dfrac{1}{2}\,m(\text{arc } BCD)\) and \(\angle C = \dfrac{1}{2}\,m(\text{arc } BAD)\).
Adding, \(\angle A + \angle C = \dfrac{1}{2}\left[m(\text{arc } BCD) + m(\text{arc } BAD)\right] = \dfrac{1}{2} \times 360^\circ = 180^\circ\).
Similarly, \(\angle B + \angle D = 180^\circ\).
Hence the opposite angles of a cyclic quadrilateral are supplementary.