In the figure, it is given that $AB = CD$ and $AD = BC$. Prove that $\triangle ADC \cong \triangle CBA$.
"
Given:
$AB = CD$ and $AD = BC$.
To do:
We have to prove that $\triangle ADC \cong \triangle CBA$.
Solution:
From the figure,
In $\triangle ADC$ and $\triangle CBA$,
$CD = AB$ (Given)
$AD = BC$ (Given)
$CA = CA$ (Common)
Therefore, by SSS axiom,
$\triangle ADC \cong \triangle CBA$.
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