In the given figure, $A B = A C$ and $A P = A Q$ . Prove that
$△ A P C = △ A Q B$

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Solution

In $△ A P C$ and $△ A Q B$
$A B = A C$ and $A P = A Q$
From the given figure, $∠ A = ∠ A$ [common in both the triangles
Therefore, using $S A S$ axiom we have $△ A P C ≅ △ A Q B$

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