Question

If $D$ and $E$ are points on sides $AB$ and $AC$ respectively of a $△$ ABC such that DE $\parallel$ BC and $BD=CE.$ Prove that $△$ ABC is isosceles.

Answer 1

If $DE\parallel BC$

Then, $\frac{AD}{BD}=\frac{AE}{CE}$ [ By basic proportionality theorem ]

$\Rightarrow$ $\frac{AD}{BD}+1=\frac{AE}{CE}+1$ [ Adding 1 on both sides ]

$\Rightarrow$ $\frac{AD+BD}{BD}=\frac{AE+CE}{CE}$

$\Rightarrow$ $\frac{AB}{BD}=\frac{AC}{CE}$

$\Rightarrow$ $BD=CE$ [ Given ]

$\Rightarrow$ So, $\frac{AB}{BD}=\frac{AC}{BD}$

$\therefore$ $AB=AC$

As two sides of the triangle are equal

$\therefore$ $ABC$ is an isosceles triangle. $\left[henceproved\right]$