Question

In the adjoining figure, $AB=AC$ and $BD=DC$. Prove that $\Delta ADB\cong \Delta ADC$ and hence show that
$\left(i\right)\angle ADB=\angle ADC={90}^{\circ }$
$\left(ii\right)\angle BAD=\angle CAD$

Answer 1

Given:

$AB=AC$ and $BD=DC$

To prove:

$\Delta ADB\cong \Delta ADC$

$\left(i\right)\angle ADB=\angle ADC={90}^{\circ }$

$\left(ii\right)\angle BAD=\angle CAD$

Proof:

In $\Delta ADB$ and $\Delta ADC$:

$AB=AC$ (Given)

$BD=DC$ (Given)

$DA=DA$ (Common)

By SSS congruence property,$\Delta ADB\cong \Delta ADC$

$\angle ADB=\angle ADC={90}^{\circ }$ (By CPCT)...(1)

$\angle ADB$ and $\angle ADC$ are on the straight line.

$\because \angle ADB+\angle ADC={180}^{\circ }$ [Linear Pair]

$\Rightarrow \angle ADB+\angle ADB={180}^{\circ }$ [Using (1)]

$\Rightarrow 2\angle ADB=\frac{{180}^{\circ }}{2}$

$\Rightarrow \angle ADB=90°$

From (1),$\angle ADB=\angle ADC=90°$

Since, $\Delta ADB\cong \Delta ADC$

By CPCT

$\angle BAD=\angle CAD$ ---(2)

Hence, proved.