Chapter: Circle· Mathematics Part 2 (Geometry)
Answer: \(\angle B + \angle D = 180^\circ\) and \(\angle A + \angle C = 180^\circ\) (proved below)
[Figure: \(\square ABCD\) inscribed in a circle with centre O; arc ADC and arc ABC are marked.]
Given: \(\square ABCD\) is cyclic. To prove: \(\angle B + \angle D = 180^\circ\) and \(\angle A + \angle C = 180^\circ\).
Proof: Arc ABC is intercepted by inscribed \(\angle ADC\). \(\therefore \angle D = \dfrac{1}{2} m(\text{arc ABC})\) ...... (I) (inscribed angle theorem)
Arc ADC is intercepted by inscribed \(\angle ABC\). \(\therefore \angle B = \dfrac{1}{2} m(\text{arc ADC})\) ...... (II)
Adding (I) and (II): \(\angle B + \angle D = \dfrac{1}{2}\left[m(\text{arc ABC}) + m(\text{arc ADC})\right] = \dfrac{1}{2} \times 360^\circ = 180^\circ\) (the two arcs together form the whole circle).
The angle sum of a quadrilateral is \(360^\circ\), so \(\angle A + \angle C = 360^\circ - 180^\circ = 180^\circ\). Hence proved.