If D and E are points on sides AB and AC respectively

It is given that in $\triangle A B C, D E \| B C$ and $B D=C E$.

We have to prove that ∆ABC is isosceles.

By Thales theorem we have

$\frac{A D}{B D}=\frac{A E}{E C}$

$\Rightarrow A D=A E$

Now $B D=C E$ and $A D=A E$

So $A D+B D=A E+C E$

Hence $A B=A C$

So, $\triangle \mathrm{ABC}$ is isosceles