Question:
If D and E are points on sides AB and AC respectively of a ∆ABC such that DE || BC and BD = CE. Prove that ∆ABC is isosceles.
Solution:
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