In the figure AD = BC and BD = CA. Prove that $∠ A D B = ∠ B C A$ and $∠ D A B = ∠ C B A$ .

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Solution

In triangles $A B D$ and $B A C$ , we have
$A D = B C$ (given);
$A B = A B$ (common);
$A C = B D$ (given).
We can use SSS condition to conclude that $△ A B D ≅ △ B A C$ . From this we conclude that
$∠ A D B = ∠ B C A$ and $∠ D A B = ∠ C B A$ .

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