Question

In the given figure, $△CDE$ is an equilateral triangle on a side $CD$ of a square $ABCD$. Show that $△ADE\cong △BCE$.

Answer 1

According to figure,
$\begin{array}{c}\angle ADC=\angle BCD={90}^{\circ }\to square\\ \angle CDE=\angle DCE={60}^{\circ }\to equilateral\_triangle\\ \therefore \angle ADE=\angle BCE={150}^{\circ }\\ In\Delta ADEand\Delta BCE\\ AD=BC\\ DE=CE\\ \angle ADE=\angle BCE\\ \Delta ADE\cong \Delta BCE\end{array}$