Question

In the given figure, $△DA$ and $△CAB$ are on the same base $AB$. Prove that $△DAB\equiv △CAB$

Answer 1

Consider $△DAB$ and $△CAB$
$\angle DAB={35}^o+{20}^o={55}^o=\angle CBA$ (given)
$\angle DBA=\angle CAB={20}^o$
$AB$ is common to both the triangles.
$\therefore △DBA\equiv △CAB$ (by ASA axiom)