$7 \cdot 44$ As shown in the right figure, in $\triangle ABC$, $\angle A = 45^{\circ}, \angle B = 30^{\circ}$. Line $DE$ divides $\triangle ABC$ into two parts of equal area, where $D$ is on $AB$ and $\angle ADE = 60^{\circ}$ (Note: The figure may not be accurate, and $E$ may also be on $CB$). Then the ratio $\frac{AD}{AB}$ is
(A) $\frac{1}{\sqrt{2}}$.
(B) $\frac{2}{2+\sqrt{2}}$.
(C) $\frac{1}{\sqrt{3}}$.
(D) $\frac{1}{\sqrt[3]{6}}$.
(E) $\frac{1}{\sqrt[4]{12}}$.
(38th American High School Mathematics Examination, 1987)
[Solution] First, we prove that the line $DE$ dividing $\triangle ABC$ into two equal areas must intersect $AC$.
If not, draw $CD$ from $C$ such that $\angle CDA = 60^\circ$.
Then $\angle DCB = 180^\circ - 120^\circ - 30^\circ = 30^\circ$,
which implies $CD = DB$.
Also, $\angle ACD = 180^\circ - (45^\circ + 60^\circ) = 75^\circ > \angle CAD$, so $AD > CD$.
By $S_{\triangle ACD} = \frac{1}{2} CD \cdot AD \sin 60^\circ$,
$$
S_{\triangle CDB} = \frac{1}{2} CD \cdot DB \sin (180^\circ - 60^\circ),
$$

then $S_{\triangle ACD} > S_{\triangle CDB}$. This is a contradiction.
As shown in the figure, let $AB = 1$, $AD = \lambda$, then
$S_{\triangle ADE} = \frac{\sin 45^\circ \sin 60^\circ}{2 \sin 75^\circ} \lambda^2$, and $S_{\triangle ABC} = \frac{\sin 45^\circ \sin 30^\circ}{2 \sin 105^\circ} \cdot 1^2$,
and $\frac{S_{\triangle ADE}}{S_{\triangle ABC}} = \frac{1}{2}$, then $\frac{\sin 45^\circ \sin 60^\circ}{2 \sin 75^\circ} \cdot \frac{2 \sin 105^\circ}{\sin 45^\circ \sin 30^\circ} \lambda^2 = \frac{1}{2}$.
Thus, $\lambda^2 = \frac{1}{2 \sqrt{3}} = \frac{1}{\sqrt{12}}$, so $\lambda = \frac{1}{\sqrt[4]{12}}$.
Therefore, the answer is $(E)$.