Question

In the figure, it is given that 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 ${90}^{\circ }$

Given, AB = AC.
Also it is given that D bisects side BC.
So, BD = DC.
And, AD = AD (common side)
$\therefore △$ADB $\cong$ $△$ADC [S.S.S congruence]
Thus, $\angle$ADB = $\angle$ADC (CPCT) $\cdots \left(1\right)$
But, $\angle$ADB + $\angle$ADC = ${180}^{\circ }$ [linear pair of angles]
i.e., $2\angle ADC={180}^{\circ }$ (from (1))
$⟹\angle ADC={90}^{\circ }$