Question

In the figure, side BC of $\Delta ABC$ is peoduced to point D such that bisectors of $\angle ABC$ and $\angle ACD$ meet at a point E. If $\angle BAC={68}^{\circ }$, find $\angle BEC.$

Answer 1

In the figure,

Side BC of $\Delta ABC$ is produced to D such that bisectors of $\angle ABC$ and $\angle ACD$ meet at E

$\angle BAC={68}^{\circ }$

In $\Delta ABC,$

Ext. $\angle ACD=\angle A+\angle B$

$\Rightarrow \frac{1}{2}\angle ACD=\frac{1}{2}\angle A+\frac{1}{2}\angle B$

$\Rightarrow \angle 2=\frac{1}{2}\angle A+\angle 1$ ....(i)

But in $\Delta BCE,$

Ext. $\angle 2=\angle E+\angle 1$

$\Rightarrow \angle E+\angle 1=\angle 2=\frac{1}{2}\angle A+\angle 1$ [From (i)]
$\Rightarrow \angle E=\frac{1}{2}\angle A=\frac{{68}^{\circ }}{2}={34}^{\circ }$