Question

If the bisectors of the angles $A,B,C$ and $D$ of a quadrilateral 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. $ABCD$
$\angle A+\angle B+\angle C+\angle D={360}^{\circ }$
$\Rightarrow \angle A+\angle B={360}^{\circ }-(\angle C+\angle D)$ ...(i)
In $\Delta AOB$, we have
$\angle AOB+\angle OAB+\angle OBA={180}^{\circ }$
$\angle AOB={180}^{\circ }-(\angle OAB+\angle OBA)$
$={180}^{\circ }-\frac{1}{2}(\angle A+\angle B)$ ($\because$ OA bisects $\angle A$ and OB bisects $\angle B$)
Therefore, from (i),
$\angle AOB={180}^{\circ }-\frac{1}{2}\{{360}^{\circ }-(\angle C+\angle D\left)\right\}$
$\angle AOB={180}^{\circ }-{180}^{\circ }+\frac{1}{2}(\angle C+\angle D)$
$\angle AOB=\frac{1}{2}(\angle C+\angle D)$