Question

If the bisector of an angle of a triangle bisects the opposite side, prove that the triangle is isosceles.

Answer 1

Produce AD upto E such that $AD=DE$

In $△ABD$ and $△EDC$

$AD=DE$ [By construction]

$BD=CD$ [Given]

$\angle 1=\angle 2$ [Vertically opposite angles]

$\therefore △ABD\cong △EDC$ [SAS}

$\Rightarrow AB=CE...........\left(1\right)$

and $\angle BAD=\angle CED$

But, $\angle BAD=\angle CAD$ [AD is bisector of $\angle BAC$]

$\therefore \angle CED=\angle CAD$

$\Rightarrow AC=CE...........\left(2\right)$

From (1) and (2)

$AB=AC$

Hence, ABC is an isosceles triangle.