In the given figure,$\angle ADE = \angle AED$ and $\frac{AD}{DB} = \frac{AE}{EC}$. Prove that $\triangle ABC$ is an isosceles triangle.

(N/A) Given: $\frac{AD}{DB} = \frac{AE}{EC}$ and $\angle ADE = \angle AED$.
According to the converse of the Basic Proportionality Theorem (Thales Theorem),if a line divides any two sides of a triangle in the same ratio,then the line is parallel to the third side.
Therefore,$DE \parallel BC$.
Since $DE \parallel BC$ and $AB$ is a transversal,the corresponding angles are equal:
$\angle ADE = \angle ABC$ ....... $(1)$
Similarly,since $DE \parallel BC$ and $AC$ is a transversal:
$\angle AED = \angle ACB$ ....... $(2)$
We are given that $\angle ADE = \angle AED$.
Substituting the values from $(1)$ and $(2)$ into this equation,we get:
$\angle ABC = \angle ACB$.
In $\triangle ABC$,since the base angles are equal $(\angle B = \angle C)$,the sides opposite to these angles must also be equal.
Therefore,$AB = AC$.
Since two sides of $\triangle ABC$ are equal,$\triangle ABC$ is an isosceles triangle.