Question

In triangle $ABC$, the bisector of angle $BAC$ meets opposite side $BC$ at point $D$. If $BD=CD$, prove that $\Delta ABC$ is isosceles.

Answer 1

Given $AD$ is angle bisector and $BD=CD.$

Construction:

Draw a line from $C$ parallel to $AB$ with length $AC.$

Now, in $△ABD$ and $△FCD,$

$\angle ABD=\angle FCD$ [Corresponding Angles]

$\angle BAD=\angle CFD$ [Corresponding Angles]

So, by AAA criteria of similarity,

$△ABD\sim △FCD$

$\Rightarrow \frac{AB}{BD}=\frac{CF}{CD}$

But given $BD=CD$ and we took $CF=AC$

$\Rightarrow AB=AC$

$\therefore △ABC$ is isosceles.