Question

In a quadrilateral ABCD, AO and BO are bisectors of angle A and angle B respectively.

Show that :

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

Answer 1

Given : AO and BO are the bisectors of $\angle Aand\angle B$ respectively.

$\therefore \angle 1=\angle 4and\angle 3=\angle 5........\left(i\right)$

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

Proof : In quadrilateral ABCD $\angle A+\angle B+\angle C+\angle D={360}^o\frac{1}{2}(\angle A+\angle B+\angle C+\angle D)={180}^o.....\left(ii\right)\text{Now in}\Delta AOB\angle 1+\angle 2+\angle 3={180}^o.......\left(iii\right)\text{Equating equation (ii) and equation (iii), we get}\angle 1+\angle 2+\angle 3=\angle A+\angle B+\frac{1}{2}(\angle C+\angle D)\angle 1+\angle 2+\angle 3=\angle 1+\angle 3+\frac{1}{2}(\angle C+\angle D)\angle 2=\frac{1}{2}(\angle C+\angle D)\therefore \angle AOB=\frac{1}{2}\angle C+\angle D$

Hence proved.