Question

CDE is an equilateral triangle formed on a side a CD of a square ABCD. Show that $\Delta ADE\cong \Delta BCE.$

Answer 1

Given : An equilateral $\Delta CDE$ is formed on the side of square ABCD. AE and BE are joined

To prove : $\Delta ADE\cong \Delta BCE$

Proof : In $\Delta ADEand\Delta BCE,$

AD = BC (Sides of a square)

DE = CE (Sides of equilateral triangle)

$\angle ADE=\angle BCE(Each={90}^{\circ }+{60}^{\circ }={150}^{\circ })$

$\therefore \Delta ADE\cong \Delta BCE\text{(SAS axiom)}$