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 triangle $abc$ is isosceles.

Answer 1

By basic proportionality property,

$\Rightarrow$ $\frac{AD}{DB}=\frac{AE}{EC}$

By adding $1$ on both sides,

$\Rightarrow$ $\frac{AD}{DB}+1=\frac{AE}{EC}+1$

$\Rightarrow$ $\frac{AD+DB}{DB}=\frac{AE+EC}{EC}$

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

It is given that, $DE\parallel BC$ and $BD=EC$

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

$\Rightarrow$ $AB=AC$

$\therefore$ $△ABC$ is isosceles triangle.