In a convex quadrilateral $ABCD$, $\angle ABD=$
$$
\begin{array}{l}
\angle CBD=\angle ADC=45^{\circ}, AB=a, BC=b, CD=c, DA \\
=d(a \neq b). \text{ Determine the range of } \frac{a^{2}-b^{2}}{c^{2}-d^{2}}.
\end{array}
$$
Solution: As shown in Figure 3, connect $AC$, take the circumcenter $O$ of $\triangle ACD$, and connect $OA$, $OB$, $OC$.

Since $\angle AOC = 2 \angle ADC = 90^{\circ}$,

and $\angle ABC = \angle ABD + \angle CBD = 90^{\circ}$, then
$$
\begin{array}{l}
\angle AOC + \angle ABC \\
= 180^{\circ}.
\end{array}
$$

Thus, $O$, $A$, $B$, $C$ are concyclic.
In this circle, since $OA = OC$, we have $\overparen{OA} = \overparen{OC}$.
Therefore, $\angle ABO = \angle CBO$.
Hence, $OB$ bisects $\angle ABC$, which means point $O$ must lie on $BD$.
Let $OA = 1$, then $AC = \sqrt{2}$.
Let $\angle BDC = \alpha$, $\angle BDA = \beta$. Then
$\angle BAC = \angle BOC = 2\alpha$, $\angle BCA = \angle BOA = 2\beta$.
It is easy to see that $c = CD = 2 \cos \alpha$, $d = DA = 2 \cos \beta$. Thus, $c^2 - d^2 = 4(\cos^2 \alpha - \cos^2 \beta)$.
Also, $\frac{a}{\sqrt{2}} = \frac{AB}{AC} = \cos 2\alpha = 2 \cos^2 \alpha - 1$, so
$4 \cos^2 \alpha = 2 + \sqrt{2} a$.
Similarly, $4 \cos^2 \beta = 2 + \sqrt{2} b$.
Thus, $4(\cos^2 \alpha - \cos^2 \beta) = \sqrt{2}(a - b)$.
Therefore, $c^2 - d^2 = \sqrt{2}(a - b)$.
So, $\frac{a^2 - b^2}{c^2 - d^2} = \frac{a^2 - b^2}{\sqrt{2}(a - b)} = \frac{a + b}{\sqrt{2}}$.
In the right triangle $\triangle ABC$, $\angle ABC = 90^{\circ}$, it is easy to see that
$$
AC < AB + BC \leq \sqrt{2} AC,
$$

i.e., $\sqrt{2} < a + b \leq \sqrt{2} \times \sqrt{2} = 2$.
Since $a \neq b$, we have $\sqrt{2} < a + b < 2$.
Substituting into equation (1), we get
$$
1 < \frac{a^2 - b^2}{c^2 + d^2} < \sqrt{2}.
$$