Question

In $\Delta ABC$ , ray BD bisects $\angle ABC$ and ray CE
bisects $\angle ACB$ If $segAB\cong segAC$ then prove
that $ED\parallel BC$ .

Answer 1

Given $AB=AC$

In $△ABC$

$BD$ is the bisector of $\angle ABC$

$\therefore$ by using angle bisector property we have

$\frac{AD}{DC}=\frac{AB}{BC}$ .....$\left(1\right)$

and $CE$ is the angle bisector of $\angle ACB$

$\therefore$ by using angle bisector property,

$\frac{AE}{EB}=\frac{AD}{BC}$ ....$\left(2\right)$

But segment $AB=$segment $AC$ ......$\left(3\right)$

From $\left(1\right)$,$\left(2\right)$ and $\left(3\right)$ we have

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

$\therefore$segment $ED\parallel$segment $BC$ by converse of basic proportionality theorem.