Question

Prove that the sum of opposite angles of a cyclic quadrilateral is ${180}^{\circ }$

Answer 1

To prove: 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 $\angle B+\angle D=180°$