Question

If the bisectors of the angles $A$ and $B$ of a quadrilateral $ABCD$ meet at $O$, then $\angle AOB$ is equal to:

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

Answer 1

Answer: (B)

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

In quad $ABCD$

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

$\Rightarrow$ $\angle A+\angle B={360}^o-(\angle C+\angle D)$

$\Rightarrow$ $\frac{\angle A}{2}+\frac{\angle B}{2}={180}^o-\frac{1}{2}(\angle C+\angle D)$

$\Rightarrow$ $\angle OAB+\angle OBA={180}^o-\frac{1}{2}(\angle C+\angle D)$

($\therefore$ $OA$ bisects $\angle A$ and $OB$ bisects $\angle B$)

In $△AOB$, $\angle AOB={180}^o-(\angle OAB+\angle OBA)$

$={180}^o-\left[\right({180}^o-\frac{1}{2}(\angle C+\angle D)]$

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