Let
\(
ABCD\) be a square. If an equilateral triangle is formed externally on side
\(CD\). Establish that \(\triangle ADE \cong \triangle BCE\).
Proof:
\(\angle\)\(=\) \(\angle ADC + \angle EDC\)
\(\angle\) \(=\) \(\angle BCD + \angle ECD\)
\( =\) \(= 150^{\circ}\)
\(\triangle ADE \cong \triangle BCE\) [By Congruence rule]
Hence Proved