Find the angle x in the shape

In $\triangle {ADE}$ , we have $\dfrac{|\overline {AE}|}{|\overline {DE}|}=\dfrac{\sin {75\degree}}{\sin x}$ . In $\triangle {AEC}$ , we have $\dfrac{|\overline {AE}|}{|\overline {EC}|}=$ $\dfrac{\sin 30\degree}{\sin (45\degree -x)}$ . Since $|\overline {DE}|=|\overline {EC}|$ , therefore $\sin {30\degree}\times \sin x=\sin {75\degree}\times {\sin {(45\degree -x) }}$ . This implies $\tan x=\dfrac{1}{√3}$ , or $x=\boxed {30\degree}$

Let $AF$ be the altitude from $A$ to $BC$ ; then $\angle DAF = 15^\circ$ . Let $AC=2$ ; then $AF = BC \times \sin 30^\circ = 1$ and $FC = \sqrt 3$ . We note that $FD = AF \times \tan \angle DAF = \tan 15^\circ$ and $DE = \dfrac {\sqrt 3-\tan 15^\circ}2$

$\begin{aligned} \implies FE & = \tan 15^\circ + \frac {\sqrt 3 - \tan 15^\circ}2 \\ & = \frac {\sqrt 3 + \tan 15^\circ}2 & \small \blue{\text{Since }\tan 15^\circ = 2-\sqrt 3} \\ & = 1 \end{aligned}$

Since $\tan \angle FAE = \dfrac {FE}{AF} = 1 \implies \angle FAE = 45^\circ$ and $\angle DAE = \angle FAE - \angle DAF = 45^\circ - 15^\circ = \boxed{30}^\circ$ .

Here is a construction. All green segments are equal to AC.