Q.3 (B)(1) Prove that, ‘Opposite angles of a cyclic quadrilateral are supplementary’.
[Figure: a circle with centre O; points A, B, C, D on the circle in order, forming cyclic quadrilateral ABCD; diagonals AC and BD are drawn.]
Best answer
Answer: \(\angle A + \angle C = 180^\circ\) and \(\angle B + \angle D = 180^\circ\).
Given: \(\square ABCD\) is cyclic, i.e. all four vertices lie on a circle. To prove: \(\angle A + \angle C = 180^\circ\) and \(\angle B + \angle D = 180^\circ\).
Proof: The arcs BCD and BAD together make the complete circle, so \(m(\text{arc } BCD) + m(\text{arc } BAD) = 360^\circ\).
\(\angle A = \dfrac{1}{2}m(\text{arc } BCD)\) and \(\angle C = \dfrac{1}{2}m(\text{arc } BAD)\) (inscribed angle theorem).
\(\therefore \angle A + \angle C = \dfrac{1}{2}\left[m(\text{arc } BCD) + m(\text{arc } BAD)\right] = \dfrac{1}{2}\times 360^\circ = 180^\circ\).
Similarly, \(\angle B + \angle D = 180^\circ\). Hence opposite angles of a cyclic quadrilateral are supplementary.
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