Question

In the given figure, $BE$ is the bisector of $\angle B$ and $CE$ is the bisector of $\angle ACD.$ Prove that $\angle BEC=\frac{1}{2}\angle A.$

Answer 1

Given: $BE$ is the bisector of angle $\angle B$ and $CE$ is the bisector of $\angle ACD$ ​

Side $BC$ of triangle $ABC$ is produced to $D.$

$\therefore \angle ACD=\angle B+\angle A$ [Exterior angle property]

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

$\Rightarrow \angle ECD=\frac{1}{2}\angle B+\frac{1}{2}\angle A\dots \dots \left(i\right)$

Also, side $BC$ of triangle $EBC$ is produced to $D.$

$\therefore \angle ECD=\angle CBE+\angle BEC$ [Exterior angle property]

$\Rightarrow \angle ECD=\frac{1}{2}\angle B+\angle BEC\dots \dots \left(ii\right)$

From $\left(i\right)$ and $\left(ii\right),$ we get

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

$\therefore \angle BEC=\frac{1}{2}\angle A$