Question

State true or false.

In the adjoining figure, AB = AC and BD = DC.
If $\Delta ADB\cong \Delta ADC$ then $\angle ADB+\angle ADC={90}^{\circ }$.

• A. True
• B. False

Answer 1

The correct option is B False

In $△ADB$ and $△ADC$

$AB=AC$(given)

$BD=DC$(given)

$AD=AD$(common)

$\Rightarrow △ADB\cong △ADC$(SSS)

$\Rightarrow \angle ADB=\angle ADC$ by CPCT

Now,$\angle ADB+\angle ADC={180}^{\circ }$

$\Rightarrow \angle ADB+\angle ADB={180}^{\circ }$

Hence $={180}^{\circ }=2\angle ADC={90}^{\circ }\to \angle ADC={90}^{\circ }$

$\therefore$ the given statement is false.