Question

In $∆ABC,$$AD$the angle bisector of $\angle A$meets side $BC$in $D.$If $\angle ADC={90}^o,$prove that $∆ABC$is an isosceles triangle.

Answer 1

Step-1: Given the information

$\angle ADC={90}^o,$so $\angle ADB={90}^o$

The angle bisector of $\angle A$meets the side $BC$in $D.$

So,$\angle DAB=\angle DAC$

Step-2: In $∆ADB$and $∆ADC$

$\angle DAB=\angle DAC$(as $AD$is the angle bisector of $\angle A$)

$AD=AD$(given)

$\angle ADB=\angle ADC={90}^o$

So, $∆ADB$$\cong$$∆ADC$(By $ASA$Congruency)

Hence, $AB=AC$(by $CPCT$)

$\angle ABD=\angle ACD$(by $CPCT$)

Step-3: Since, $AB=AC$and $\angle ABD=\angle ACD$ So $∆ABC$is an isosceles triangle.

Hence, $∆ABC$is an isosceles triangle.