Question

In the below given figure of parallelogram, AF and BE are angle bisectors of $\angle A$ and $\angle B$ respectively. Find the measure of $x$.

$\left[4Marks\right]$

Answer 1

Given:
ABCD is a parallelogram
AF bisects $\angle BAD$
BE bisects $\angle ABC$

In the parallelogram ABCD,
$\angle A=\angle C={60}^{\circ }$ [Opposite angles of a parallelogram]
and $\angle B=\angle D$
$\left[1Mark\right]$

Also, in a parallelogram, the adjacent angles are supplementary.

$\therefore \angle C+\angle D={180}^{\circ }$

$\Rightarrow \angle D={180}^{\circ }-\angle C={120}^{\circ }$

AF and BE bisect $\angle BAD$ and $\angle ABC$ respectively

$\therefore \angle BAF=\angle FAD=\frac{\angle BAD}{2}$

$=\frac{{60}^{\circ }}{2}$

$={30}^{\circ }$
$\left[1Mark\right]$

And,
$\angle ABE=\angle EBC=\frac{\angle ABC}{2}$

$=\frac{{120}^{\circ }}{2}$

$={60}^{\circ }$

And,
$\angle EOF=\angle AOB=x^{\circ }$ [Vertically opposite angles]

In $\Delta OAB$,
$\angle OAB+\angle OBA+\angle AOB={180}^{\circ }$ [Angle sum property]
$\left[1Mark\right]$

$\therefore \angle AOB={180}^{\circ }-(\angle OAB+\angle OBA)$

$={180}^{\circ }-({30}^{\circ }+{60}^{\circ })$

$\angle AOB={180}^{\circ }-{90}^{\circ }={90}^{\circ }$

$\therefore x=\angle AOB={90}^{\circ }$
[Vertically opposite angles]
$\left[1Mark\right]$