In the given figure, $$ABCD$$ is a quadrilateral in which $$AD=BC$$ and $$\angle DAB=CBA$$. Prove that $$\triangle ABD \cong \triangle BAC$$
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The figure $$ABCD$$ is a quadrilateral In which $$AD=BC$$ $$\angle DAB =\angle CBA$$ To prove: (i) $$\triangle ABD =\triangle BAC$$ (ii) $$\angle ABD =\angle BAC$$ Proof: In $$\triangle ABD $$ and $$\triangle ABC$$ $$AB=AB$$ (common) $$\angle DAB =\angle CBA$$ (Given) $$AD=BC$$ (i) $$\triangle ABD\cong \triangle ABC$$ ($$SAS$$ axiom) (ii) $$BD=AC$$ (iii) $$\angle ABD=\angle BAC$$
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