Question

In the figure, $segAB\cong segAC$,ray $CE$ bisect $\angle ACB$, ray $BD$ bisect $\angle ABC$. then prove that ray $ED||BC$ ?

Answer 1

$CE$ is angle bisects of $ACB$

From vertical angle bisects theorem.

$\frac{AE}{EB}=\frac{AC}{BC}----\left(1\right)$

$BD$ is angle bisects of $ABC$

From vertical angle bisects theorem,

$\frac{AD}{DC}=\frac{AB}{BC}=\frac{AC}{BC}(\because AB=AC)----\left(2\right)$

From $\left(1\right)$ & $\left(2\right)$

$\frac{AE}{EB}=\frac{AD}{DC}$

So, $ED$ is parallel to $BC$.