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In the adjoining figure, $\frac{AD}{BD} = \frac{AE}{EC}$ and $\angle BDE = \angle CED$, prove that $\Delta ABC$ is an isosceles triangle.
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$\frac{AD}{DB} = \frac{AE}{EC} \Rightarrow DE \parallel BC$
$\angle BDE + \angle DBC = \angle CED + \angle BCE$
$\angle DBC = \angle BCE$ or $\angle B = \angle C$
Thus $\Delta ABC$ is an isosceles triangle.
$\angle BDE + \angle DBC = \angle CED + \angle BCE$
$\angle DBC = \angle BCE$ or $\angle B = \angle C$
Thus $\Delta ABC$ is an isosceles triangle.