Question

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

Answer 1

(Refer to Image)

Given $ABCD$ is a square. $CDE$ is an equilateral $△$ .

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

Proof: $AD=BC$ (sides of a square)

$DE=EC$ (sides of equilateral $△$)

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

Hence, $\Delta ADE\cong \Delta BCE$

By $SAS$ congruency.