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)
$\therefore ,△ADB\cong △ADC$
$Thus,\angle ADB=\angle ADC$ __(1)
$but,\angle ADB+\angle ADC={180}^{\circ }or,2\angle ADC={180}^{\circ }\left(from\right(1\left)\right)Hence,\angle ADC={90}^{\circ }$