Question

The sum of pairs of opposite angles of a cyclic quadrilateral is:

• A. ${360}^o$
• B. ${90}^o$
• C. ${180}^o$
• D. ${60}^o$

Answer 1

The correct option is C ${180}^o$

Sum of opposite angles of a cyclic quadrilateral is ${180}^o$

Let $ABCD$ is cyclic quadrilateral.
To prove that $\angle A+\angle C=180°$ and $\angle B+\angle D=180°$

Construction: Join $OB$ and $OD$.

We know that $\angle BOD=2\angle BAD$ ...(The angle at the centre is twice the angle at the circumference )

$⟹$ $\angle BAD=\frac{1}{2}\angle BOD$.

Similarly, we can write $\angle BCD=\frac{1}{2}\angle BOD$.

Now, add above two equations,

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

$⟹$ $\angle BAD+\angle BCD=\frac{1}{2}(\angle BOD+\angle DOB)$ ...(complete angle)

$⟹$$\angle BAD+\angle BCD=\frac{1}{2}\times 360°=180°$.

So we get, $\angle A+\angle C=180°$

And similarly $\angle B+\angle D=180°$.

Hence,option $C$ is correct.