$\displaystyle m\left( {\angle EAD} \right) = \frac{\pi }{6}$. WHY?
$\displaystyle m\left( {\angle ADE} \right) = m\left( {\angle DEA} \right)$ WHY?
$\displaystyle m\left( {\angle ADE} \right) + m\left( {\angle DEA} \right) + m\left( {\angle EAD} \right) = ?$
Well that hurts my feelings. Did you think about it at all?
$\displaystyle \Delta EAB$ is equilateral so that tells you $\displaystyle m\left( {\angle EAB} \right)$.
So that tells you $\displaystyle m\left( {\angle EAD} \right)$.
The $\displaystyle \Delta EAD$ is isosceles. That gives you $\displaystyle x$.
if you will ONLY read well the post of Mr. PLATO, you almost there.
observe that, angle DAE = angle DAB - angle EAB = 90 - 60 = 30 degrees
but angle DAE is isoceles triangle.
Sum up the angles of triangle DAE, here we go
180 = (angle DAE) + (angle ADE) + (angle AED),
Therefore ....