Question

In the figure, it is given that $\angle A={90}^{\circ },AB=AC,$ and point $D$ bisects the side $BC.$ Find $\angle ADC.$

• A. ${30}^{\circ }$
• B. ${60}^{\circ }$
• C. ${90}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

The correct option is C

${90}^{\circ }$

We have, $AB=AC$ [given]
also, it is given that $D$ bisects side $BC,$ so,
$BD=DC$
and, $AD=AD$ [common side]

Therefore, $△ADB\cong △ADC$ [SSS congruence rule]
thus, $\angle ADB=\angle ADC$ ...(i) [CPCTE]
but, $\angle ADB+\angle ADC={180}^{\circ }$
or, $2\angle ADC={180}^{\circ }$ [from i]
Hence, $\angle ADC={90}^{\circ }$