Question

If a quadrilateral is cyclic, then sum of each pair of opposite angles is ${180}^o$.

Answer 1

Let ABCD be a cyclic quadrilateral as shown in the figure.

Let us join OB and OD.

Then, $\angle BOD=2\angle BCD$

$\angle BCD=\frac{1}{2}\angle BOD$

Similarly, $\angle BAD=\frac{1}{2}\angle DOB$

$\therefore \angle BAD+\angle BCD=\frac{1}{2}\angle BOD+\frac{1}{2}\angle DOB$

$=\frac{1}{2}(\angle BOD+\angle DOB)$

$=\frac{1}{2}\times {360}^{\circ }$

$={180}^{\circ }$

Similarly $\angle B+\angle D={180}^{\circ }$