Question
CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that $\triangle A D E \cong \triangle B C E$.

Answer

In square  ABCD,  AD=BC and in equilateral $\triangle C D E, C E=D E ;$ also $\angle A D E=\angle B C E=60^{\circ}$. So by SAS criterion,
$\triangle A D E \cong \triangle B C E$

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