Question

In the figure, $ABC$ is a triangle in which $AD$ is the bisector of $\angle A$. If $AD\perp BC$, show that $△ABC$ is isosceles.

Answer 1

Given:

$AD$ is the bisector of $\angle A$

$\Rightarrow \angle DAB=\angle DAC$ ...$\left(1\right)$

$AD\perp BC$

$\Rightarrow \angle BDA=\angle CDA={90}^o$

To prove:

$△ABC$ is isosceles.

Proof:

In $△DAB$ and $△DAC$,

$\angle BDA=\angle CDA={90}^o$

$DA=DA$ [common]

$\angle DAB=\angle DAC$ [from $\left(1\right)$]

$\therefore △DAB\cong △DAC$ [By ASA congruence property]

$\Rightarrow AB=AC$

Hence, $△ABC$ is isosceles