Answer: E

Small Hint:

Scale so $AB=1$ and place $A=(0,0),\ B=(1,0)$

Big Hint:

Find the height of $C$ and the height of $E$ in terms of $d=AD$, then equate half the total area to the area of $\triangle ADE$

Solution:

Set $AB=1,\ A=(0,0),$ and $B=(1,0).$ Since $\angle A=45^\circ,$ side $AC$ lies on $y=x.$ The line through $B$ making angle $30^\circ$ with $BA$ meets it at height $$h_C=\frac1{1+\sqrt3}.$$ Let $d=AD.$ The ray $DE$ has slope $-\sqrt3,$ so its intersection with $y=x$ has height $$h_E=\frac{\sqrt3\,d}{1+\sqrt3}.$$ Equal areas require $$\frac12d h_E=\frac12\left(\frac12h_C\right).$$ Therefore $$\begin{aligned} \frac{\sqrt3\,d^2}{2(1+\sqrt3)} &=\frac1{4(1+\sqrt3)},\\ d^2&=\frac1{2\sqrt3} =\frac1{\sqrt{12}}. \end{aligned}$$ Hence $\frac{AD}{AB}=d=\frac{1}{\sqrt[4]{12}}.$

Thus the correct answer is E.