Question

In $△$ABC, $\angle B=\angle C$, D and E are the points on AB and AC such that BD = CE, prove that DE || BC.

Answer 1

Given is a $△ABC$ such that $\angle B=\angle CandBD=CE$.

To prove: $DE\parallel BC$

We know that, sides opposite to equal angles of a triangle are equal.

$AB=AC[\because \angle B=\angle C]$

$\therefore AD+DB=AE+EC$

Also, $BD=CE$

$\therefore AD=AE$

Hence, $\frac{AD}{DB}=\frac{AE}{EC}$

According to the converse of basic proportionality theorem, if a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

Therefore, by using the converse of the basic proportionality theorem, we get $DE||BC$. $\left[Henceproved\right]$