(i) $ \triangle A B D \cong \triangle B A C $
(ii) $ B D=A C $
(iii) $\angle ABD=\angle BAC$ "
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ABCD is a quadrilateral in which $ A D=B C $ and $ \angle D A B=\angle C B A $. Prove that.

(i) $ \triangle A B D \cong \triangle B A C $
(ii) $ B D=A C $
(iii) $\angle ABD=\angle BAC$ "

Given:
ABCD is a quatrilateral in which \( A D=B C \) and \( \angle D A B=\angle C B A \).
To do:
We have to prove that,
(i) $\vartriangle ABD \cong\ BAC$
(ii) $BD=AC$
(iii) $\angle ABD=\angle BAC$
Solution:
 In $\vartriangle ABD$ and $\vartriangle BAC$,
$AD=BC$ (Given)
$\angle DAB=\angle CBA$ (Given)
$AB=BA$ (Common)
Therefore,
$\vartriangle ABD \cong\ BAC$ (By SAS congruence rule)
We know that,
Corresponding parts of congruent triangles are equal.
Therefore,
$BD=AC$ (By CPCT)
$\angle ABD=\angle BAC$ (By CPCT)
Hence proved.

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Updated on: 10-Oct-2022

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