Question

In the adjoining figure, $ABCD$ is a square and $△EDC$ is an equilateral triangle. Prove that
$AE=BE$.

Answer 1

In $△ADE$ & $△BCE$

$AD=BC$ (given )

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

$DE=CE$ (given )

$\therefore △ADE\cong △BCE$ (by SAS congruency rule)

$AE=BE\left(CPCT\right)$