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(N/A) Let $	riangle ABC$ be a triangle where $AD$ is the angle bisector of $\angle A$ such that $D$ lies on $BC$ and $BD = DC$.In $	riangle ABD$ and

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(N/A) Let $	riangle ABC$ be a triangle where $AD$ is the angle bisector of $\angle A$ such that $D$ lies on $BC$ and $BD = DC$.In $	riangle ABD$ and $	riangle 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,$	riangle ABD \cong 	riangle ACD$.Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.Therefore,$AB = AC$.Since two sides of the triangle are equal,$	riangle ABC$ is an isosceles triangle.

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.

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