Question

In the given figure, ABCD is a quadrilateral and $\angle ADC=a^{\circ },\angle BCD=b^{\circ }.$ AO and BO are bisectors of $\angle DAB$ and $\angle ABC$ respectively meeting at O. Find $\angle AOB$ in terms of $a^{\circ }$ and $b^{\circ }$.

• A. $a+b$
• B. $(a+2b)\times 2$
• C. $\frac{a+b}{2}$
• D. $\frac{(a+2b)}{2}$

Answer 1

The correct option is C $\frac{a+b}{2}$

Given, ABCD is a quadrilateral.
Also, OA and OB bisect $\angle A$ and $\angle B$ respectively.

In quadrilateral ABCD,
$\angle BAD+\angle ABC+\angle BCD+\angle CDA={360}^{\circ }$

Since OA bisects $\angle A$ and OB bisects $\angle B$,
$\angle DAO=\angle OAB=\frac{1}{2}\angle A$

and,

$\angle CBO=\angle OBA=\frac{1}{2}\angle B$

$\Rightarrow \angle A=2\angle OAB$ and
$\angle B=2\angle OBA$

Let $\angle OAB=x$ and $\angle OBA=y$

$\angle A=2x$ and $\angle B=2y$

$\therefore 2x+2y+a+b={360}^{\circ }$
$\Rightarrow x+y=\frac{360-(a+b)}{2}$
$\Rightarrow \angle AOB={180}^{\circ }-\left(\frac{360-(a+b)}{2}\right)$ [Angle]
Hence, $\angle AOB=\frac{360-360+(a+b)}{2}=\frac{a+b}{2}$