Question

In a quadrilateral ABCD. AO and BO are bisectors of angle A and angle B respectively. Prove that $\angle$AOB$=\frac{1}{2}\{\angle C+\angle D\}$.

Answer 1

ABCD is a quadrilateral

To prove : $\angle AOB=\frac{1}{2}(\angle C+\angle D)$

$AO$ and $BO$ is bisector of $AandB$

$\angle 1=\angle 2\angle 3=\angle 4...\left(1\right)$

$\angle A+\angle B+\angle C+\angle D=360$

(Angle sum property)

$\frac{1}{2}(\angle A+\angle B+\angle C+\angle D)=\mathrm{180...}\left(2\right)$

In $△AOB$

$\angle 1+\angle 3+\angle 5=\frac{1}{2}(\angle A+\angle B+\angle C+\angle D)$

$\angle 1+\angle 3+\angle 5=\angle 1+\angle 3+\frac{1}{2}(\angle C+\angle D)$

$\angle AOB=\frac{1}{2}(\angle C+\angle D)$

Hence,Proved.