smc

First we show that $E$ is on $AC$, as in the given figure, by demonstrating that if $E=C$, then $△ADE$ has more than half the area, so $DE$ is too far to the right. Specifically, assume $E=C$ and drop an altitude from $C$ to $AB$ that meets $AB$ at $F$. Without loss of generality, assume that $CF=1$, so that $\frac{\text{Area}△EAD}{\text{Area}△EAB}=\frac{AD}{AB}$ (as they have the same height, and area is $\frac{1}{2}$ times the product of base and height), which is $\frac{1+\frac{1}{\sqrt{3}}}{1+\sqrt{3}}=\frac{1}{\sqrt{3}}>\frac{1}{2}$, as required. Thus we must move $DE$ to the left, scaling $△EAD$ by a factor of $k$ such that $\text{Area}△EAD=\frac{1}{2}\text{Area}△CAB⟹\frac{1}{2}{k}^2(1+\frac{1}{\sqrt{3}})=\frac{1}{4}(1+\sqrt{3})⟹k^2=\frac{\sqrt{3}}{2}⟹k=(\frac{3}{4}{)}^{\frac{1}{4}}$. Thus $\frac{AD}{AB}=\frac{k(1+\frac{1}{\sqrt{3}})}{1+\sqrt{3}}=\frac{k}{\sqrt{3}}=(\frac{3}{4}{)}^{\frac{1}{4}}\times (\frac{1}{9}{)}^{\frac{1}{4}}=(\frac{1}{12}{)}^{\frac{1}{4}}=\frac{1}{\sqrt[4]{12}}$, which is answer $\boxed{\frac{1}{\sqrt[4]{12}}}$.