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Question

In the given figure, $$ABCD$$ is a quadrilateral in which $$AD=BC$$ and $$\angle DAB=CBA$$. Prove that
$$\triangle ABD \cong \triangle BAC$$

Solution
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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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