2. In a planar quadrilateral $ABCD$, $AB=\sqrt{3}, AD=DC=CB=1$, the areas of $\triangle ABD$ and $\triangle BCD$ are $S$ and $T$ respectively, then the maximum value of $S^{2}+T^{2}$ is $\qquad$ .
2. $\frac{7}{8}$

Analysis: Let $\angle B A D=\theta$, then $S=\frac{\sqrt{3}}{2} \sin \theta, B D^{2}=4-2 \sqrt{3} \cos \theta$,
$$
T^{2}=\frac{\sqrt{3}}{4} \cos \theta(2-\sqrt{3} \cos \theta)
$$

Thus, $S^{2}+T^{2}=\left(\frac{\sqrt{3}}{2} \sin \theta\right)^{2}+\frac{\sqrt{3}}{4} \cos \theta(2-\sqrt{3} \cos \theta)=-\frac{3}{2} \cos ^{2} \theta+\frac{\sqrt{3}}{2} \cos \theta+\frac{3}{4} \leq \frac{7}{8}$,
when $\cos \theta=\frac{\sqrt{3}}{6}$, $S^{2}+T^{2}=\frac{7}{8}$, hence the maximum value is $\frac{7}{8}$.