If the bisector of an angle of a triangle bisects the opposite side, then the triangle is

Cleary AD is bisector of $\angle A$

i.e.,  $\angle BAD=\angle CAD(\therefore \text{given})$

$AD=AD(\therefore \text{common side})$

$BD=CD(\therefore \text{Given})$

By S.S.A congruance  (sometimes acceptable )

$\Delta BAD\cong \Delta CAD$

i. e., $AB=AC\because \left(\text{cpct}\right)$

$\Delta ABC$ is Isosceles