Question

$\text{In Fig.}AD=BC\text{and}BD=CA.$

$\text{Prove that}\angle ADB=\angle BCA\text{and}\angle DAB=\angle CBA$

Answer 1

In triangles ABD and BAC, we have
AD = BC [Given]
BD = CA [Given] and AB = AB [Common]
So, by SSS congruence criterion, we have
$△ABD\cong BAC\Rightarrow \angle DAB=\angle CBA[\because \text{corresponding parts of congruent triangles are equal]}$
$\Rightarrow \angle ADB=\angle BCA$