Question

In the given figure, AD is the angle bisector of $\angle A$. If $\angle ADC={90}^0$. Prove that $△ABC$ is isosceles.

Answer 1

$AD$ is angular bisector

$⟹\angle BAD=\angle DACAD=AD\angle ADB=\angle ADC={90}^{\circ }⟹\Delta ADB=\Delta ADC$

So $AB=AC$ since they are corresponding sides

Hence the triangle is isosceles