Question

Question 3
In the given figure, $\Delta \text{CDE}$ is an equilateral triangle formed on a side CD of a square ABCD. Show that $\Delta \text{ADE}\cong \Delta \text{BCE}$

Answer 1

Given in figure $\Delta CDE$ is an equilateral triangle formed on a side CD of a square ABCD to show $\Delta ADE\cong \Delta BCE$
Proof in $\Delta ADE$ and $\Delta BCE$. DE =CE [sides of an equilatera 1 trianlges]
$\angle ADE=\angle BCE$
$\left[\begin{array}{cc}\because \angle ACE=\angle BCD={90}^{\circ }and& \angle EDC=\angle ECD={60}^{\circ }\\ \therefore \angle ADE={90}^{\circ }+{60}^{\circ }+{150}^{\circ }and& \angle BCE={90}^{\circ }+{60}^{\circ }={150}^{\circ }\end{array}\right]$
and AD =BC [sides of a square]
$\therefore \Delta ADE=\Delta BCE$ [by SAS congruence rule]