Question

In a quadrilateral $ABCD,CO$ and $DO$ are the bisectors of $\angle C$ and $\angle D$ respectively. Prove that $\angle COD=\frac{1}{2}(\angle A+\angle B)$

Answer 1

$\Rightarrow \angle COD=108-[\frac{1}{2}\angle C+\frac{1}{2}\angle D]$

[ $\because OC$ and $OD$ are bisectors of $\angle C$ and $\angle D$ represents ]

$\Rightarrow \angle COD=180-\frac{1}{2}(\angle C+\angle D).......\left(1\right)$

In quadrilateral $ABCD$

$\angle A+\angle B+\angle C+\angle D={360}^{\circ }$

$\angle C+\angle D=360-\angle A+\angle B.........\left(2\right)$

[Angle sum property of quadrilateral ]

Substituting (ii) in (i)

$\Rightarrow \angle COD=180-\frac{1}{2}(360-(\angle A+\angle B\left)\right)$

$\Rightarrow \angle COD=180-180+\frac{1}{2}(\angle A+\angle B)$

$\Rightarrow \angle COD=\frac{1}{2}(\angle A+\angle B)$