5. In the plane, there are four points $A, B, C, D$, where $A, B$ are fixed points, and $|A B|=\sqrt{3}, C, D$ are moving points, H. $|A D|=|D C|=|B C|=1$, let $S_{\triangle A B D}=S$ be the area of $\triangle A B D$, $S_{\triangle B C D}=T$ be the area of $\triangle B C D$. Then the range of $S^{2}+T^{2}$ is $\qquad$.
5.
5. $\frac{2 \sqrt{3}-3}{4} \leqslant S^{2}+T^{2} \leqslant \frac{7}{8}$.

From $B D^{2}=1^{2}+\left(\sqrt{3}^{2}\right)-2$
$$
\begin{aligned}
& \cdot 1 \cdot \sqrt{3} \cos \alpha \\
= & 1^{2}+1^{2}-2 \cdot 1 \\
\cdot & 1 \cos \beta
\end{aligned}
$$

we get $\cos \beta=\sqrt{3} \cos \alpha-1$.
$$
\text { Also, } \begin{aligned}
S^{2}+T^{2} & =\frac{3}{4} \sin ^{2} \alpha+\frac{1}{4} \sin ^{2} \beta \\
& =-\frac{3}{2}\left(\cos \alpha-\frac{1}{2 \sqrt{3}}\right)^{2}+\frac{7}{8} .
\end{aligned}
$$

It can be proven that $0 \leqslant \alpha \leqslant \frac{\pi}{2}$. Therefore, $S^{2}+T^{2}$ reaches its maximum value of $\frac{7}{8}$ when $\cos \alpha=$ $\frac{1}{2 \sqrt{3}}$, and its minimum value of
$$
\frac{2 \sqrt{3}-3}{4}
$$
when $\cos \alpha=1$.