Question

If D and E are points on sides AB and AC respectively of $\Delta ABC$ such that DE || BC and BD = CE. Prove that $\Delta ABC$ is isoscles.

Answer 1

If $DE\parallel BC$
From the fig.
$\frac{AD}{BD}=\frac{AE}{CE}$ (by Basic Proportionality Theorem)
Adding 1 on both sides
$\Rightarrow \frac{AD}{BD}+1=\frac{AE}{CE}+1$
$\Rightarrow \frac{(AD+BD)}{BD}=\frac{(AE+CE)}{CE}$
$\Rightarrow \frac{AB}{BD}=\frac{AC}{CE}$
As from the given data,
$BD=CE$
so,
$\frac{AB}{BD}=\frac{AC}{BD}$
$\Rightarrow AB=AC$
As two sides of the triangle are equal
$\therefore$ABC is an issosceles triangle.