In the given figure $A B = C D$ and $A D = B C$ . Prove that $∠ B A C = ∠ A C D$ .

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Solution

In $Δ A B C$ and $Δ C D A$ :
$A C = A C$ (common)
$A B = C D$ (Given)
$A D = B C$ (Given)
By $S S S$ property:
$Δ$ ABC $≅ Δ$ CDA
Thus, $∠$ BAC = $∠$ ACD
Hence, proved.

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