In a triangle,if the bisector of an angle bisects the side opposite to that angle,then prove that the triangle is an isosceles triangle.

(N/A) Let $\triangle ABC$ be a triangle where $AD$ is the angle bisector of $\angle A$ such that $D$ lies on $BC$ and $BD = DC$.
In $\triangle ABD$ and $\triangle ACD$:
$1$. $\angle BAD = \angle CAD$ (Since $AD$ is the angle bisector of $\angle A$).
$2$. $AD = AD$ (Common side).
$3$. $BD = DC$ (Given).
By the Side-Angle-Side $(SAS)$ congruence criterion,$\triangle ABD \cong \triangle ACD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$AB = AC$.
Since two sides of the triangle are equal,$\triangle ABC$ is an isosceles triangle.