Question

In the figure $ABCD$ is a rectangle and $DEC$ is an equilateral. Find $\angle DAE$.

Answer 1

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

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

In $△ADE$ and $△BCE$

$AD=BC$(side of a square)

$DE=CE$(side of an equilateral triangle)

$\angle ADE=\angle BCE$

$\therefore △ADE\cong △BCE$ by SAS rule

$\therefore AE=BE$ by CPCT

In $△ADE$

$AD=DE$ since $AD=DC=DE$

$\Rightarrow \angle 1=\angle 2$

But $\angle 1+\angle 2+\angle ADE={180}^{\circ }$ by angle sum property of $△$

$\angle 1+\angle 2+{150}^{\circ }={180}^{\circ }$

$\Rightarrow \angle 1+\angle 2={180}^{\circ }-{150}^{\circ }={30}^{\circ }$

$\angle 1=\angle 2$

$\Rightarrow 2\angle 1=2\angle 2={30}^{\circ }$

$\Rightarrow \angle 1=\angle 2=\frac{{30}^{\circ }}{2}={15}^{\circ }$

$\therefore \angle DAE={15}^{\circ }$