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

(N/A) We are given a point $D$ on side $BC$ of a $\Delta ABC$ such that $\angle BAD = \angle CAD$ and $BD = CD$. We are to prove that $AB = AC$.
Produce $AD$ to a point $E$ such that $AD = DE$ and then join $CE$.
Now,in $\Delta ABD$ and $\Delta ECD$,we have:
$BD = CD$ (Given)
$AD = DE$ (By construction)
$\angle ADB = \angle EDC$ (Vertically opposite angles)
Therefore,$\Delta ABD \cong \Delta ECD$ ($SAS$ congruence rule).
So,$AB = EC$ and $\angle BAD = \angle CED$ $(CPCT)$ ... $(1)$
Also,$\angle BAD = \angle CAD$ (Given) ... $(2)$
From $(1)$ and $(2)$,we get $\angle CAD = \angle CED$.
In $\Delta ACE$,since $\angle CAD = \angle CED$,the sides opposite to these angles must be equal.
Therefore,$AC = EC$ ... $(3)$
From $(1)$ and $(3)$,we have $AB = AC$.
Hence,$\Delta ABC$ is an isosceles triangle.