If $D$ and $E$ are points on sides $AB$ and $AC$ respectively of a $\triangle ABC$ such that $DE \parallel BC$ and $BD = CE$. Prove that $\triangle ABC$ is isosceles.
Given:
$D$ and $E$ are points on sides $AB$ and $AC$ respectively of a $\triangle ABC$. $DE \parallel BC$ and $BD = CE$.
To do:
We have to prove that $\triangle ABC$ is isosceles.
Solution:
From the figure,
$DE \parallel BC$
By basic proportionality theorem,
$\frac{AD}{DB}=\frac{AE}{EC}$
$\Rightarrow \frac{AD}{DB}=\frac{AE}{DB}$ (Since $BD=CE$)
$\Rightarrow AD=AE$
Adding $DB$ on both sides, we get,
$\Rightarrow AD+DB=AE+DB$
$\Rightarrow AD+DB=AE+EC$ (Since $BD=CE$)
$\Rightarrow AB=AC$
$\Rightarrow \triangle ABC$ is isosceles.
Hence proved.
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