In Figure, AD = BC and BD = CA.

Prove that $∠ A D B = ∠ B C A a n d ∠ D A B = ∠ C B A .$

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Solution

In triangles ABD and BAC, we have
AD = BC [Given]
BD = CA [Given]
and AB = AB [Common]
So, by SSS congruence criterion, we have
$Δ A B D ≅ Δ B A C \Rightarrow ∠ D A B = ∠ C B A$
[Corresponding parts of Congruent triangles are equal]
$\Rightarrow ∠ A D B = ∠ B C A$

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