Question

$D$ and $E$ are points on sides $AB$ and $AC$ respectively such that $BD=CE$. If $\angle B=\angle C$, show that $DE||BC$.

Answer 1

In $△ABC$, we have

$\angle B=\angle C$.

$\Rightarrow AC=AB$ [Sides opposite to equal angles are equal]

$\Rightarrow AE+EC=AD=+DB$

$\Rightarrow AE+CE=AD+BD$

$\Rightarrow AE+CE=AD+CE$ [because BD=CE]

$\Rightarrow AE=AD$

Thus, we have

$AD=AE$ and $BD=CE$

$\therefore \frac{AD}{BD}=\frac{AE}{CE}$

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

$\Rightarrow DE||BC$ [By the converse of Basic Proportionality Theorem] $\left[henceproved\right]$