Question

In given figure, the bisectors of the angles B and C of a triangle ABC, meet the opposite sides in D and E respectively. If DE||BC, prove that the triangle is isosceles.

Answer 1

Given A $△ABC$ in which the bisectors of $\angle B$ and $\angle C$ meet the sides AC and AB at D and E respectively.
To prove $AB=AC$

Construction Join DE

Proof In $△ABC$, BD is the bisector of $\angle B$.

$\therefore$ $\frac{AB}{BC}=\frac{AD}{DC}$...........(i)

In $△ABC$, CE is the bisector of $\angle C$.

$\therefore$ $\frac{AC}{BC}=\frac{AE}{BE}$.......(ii)

Now, $DE||BC$

$\Rightarrow$ $\frac{AE}{BE}=\frac{AD}{DC}$ [By Thale's Theorem]......(iii)

From (iii), we find the RHS of (i) and (ii) are equal. Therefore, their LHS are also equal i.e.

$\frac{AB}{BC}=\frac{AC}{BC}$

$\Rightarrow$ $AB=AC$

Hence, $△ABC$ is isosceles.