9. (14 points) In $\square A B C D$, it is known that $A B=x$, $B C=1$, the diagonals $A C$ and $B D$ intersect at point $O$, and $\angle B O C = 45^{\circ}$. Let the distance between the lines $A B$ and $C D$ be $h(x)$. Find the expression for $h(x)$ and state the range of $x$.

In $\square A B C D$, it is known that $A B=x$, $B C=1$, the diagonals $A C$ and $B D$ intersect at point $O$, and $\angle B O C = 45^{\circ}$. Let the distance between the lines $A B$ and $C D$ be $h(x)$. Find the expression for $h(x)$ and state the range of $x$.
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9. From the property that the sum of the squares of the diagonals of a parallelogram equals the sum of the squares of its four sides, we have
$$
\begin{array}{l}
O B^{2}+O C^{2}=\frac{1}{2}\left(A B^{2}+B C^{2}\right) \\
=\frac{1}{2}\left(x^{2}+1\right) .
\end{array}
$$

In $\triangle O B C$, by the cosine rule, we get
$$
O B^{2}+O C^{2}-\sqrt{2} O B \cdot O C=1 \text {. }
$$

Combining equations (1) and (2), we have
$$
O B \cdot O C=\frac{x^{2}-1}{2 \sqrt{2}} \text {. }
$$

Thus, $S_{\triangle A B C D}=4 S_{\triangle O B C}=2 O B \cdot O C \sin \angle B O C$
$$
\begin{array}{l}
\Rightarrow A B \cdot h(x)=\sqrt{2} O B \cdot O C=\frac{x^{2}-1}{2} \\
\Rightarrow h(x)=\frac{x^{2}-1}{2 x} .
\end{array}
$$
$$
\begin{array}{l}
\text { From } O B^{2}+O C^{2} \geqslant 2 O B \cdot O C>0 \\
\Rightarrow \frac{1}{2}\left(x^{2}+1\right) \geqslant 2 \times \frac{x^{2}-1}{2 \sqrt{2}}>0 \\
\Rightarrow 1<x \leqslant \sqrt{2}+1 .
\end{array}
$$
```