In quadrilateral $ABCD$, $AC \perp BD$ and $AC \cap BD=O, OC=OD$. Try to find the maximum value of $\frac{OB}{AB}+\frac{OA}{AD}+\frac{OC}{BC}$.
Solution: As shown in Figure 4, let $\angle A B O = a, \angle O A D = \beta, \angle B C O = \gamma$. According to the problem,
$$
\begin{array}{l}
\tan \alpha = \frac{O A}{O B}, \\
\tan \beta = \frac{O D}{O A}, \\
\tan \gamma = \frac{O B}{O C}.
\end{array}
$$

Since $O D = O C$, we have $\tan \alpha \cdot \tan \beta \cdot \tan \gamma = 1$, and $\alpha, \beta, \gamma \in \left(0, \frac{\pi}{2}\right)$.
It is easy to see that $\frac{O B}{A B} = \cos \alpha, \frac{O A}{A D} = \cos \beta, \frac{O C}{B C} = \cos \gamma$.
Thus, the problem is transformed into finding the maximum value of $\cos \alpha + \cos \beta + \cos \gamma$.

From $\tan \alpha \cdot \tan \beta \cdot \tan \gamma = 1$, it follows that at least one of $\tan \alpha, \tan \beta, \tan \gamma$ is not less than 1.
Without loss of generality, assume $\tan \gamma \geqslant 1$, then $\cot \gamma \in (0, 1]$.
And $\tan \alpha \cdot \tan \beta = \cot \gamma$
$$
\begin{array}{l}
\Rightarrow 2 \sin \alpha \cdot \sin \beta = 2 \cot \gamma \cdot \cos \alpha \cdot \cos \beta \\
\Rightarrow -\cos (\alpha + \beta) + \cos (\alpha - \beta) \\
\quad = \cot \gamma [\cos (\alpha + \beta) + \cos (\alpha - \beta)] \\
\Rightarrow \cos (\alpha + \beta) = \frac{1 - \cot \gamma}{1 + \cot \gamma} \cos (\alpha - \beta).
\end{array}
$$

Let $t = \cos \alpha + \cos \beta$, then
$$
\begin{array}{l}
t^{2} = \cos^{2} \alpha + \cos^{2} \beta + 2 \cos \alpha \cdot \cos \beta \\
= 1 + \cos (\alpha + \beta) \cdot \cos (\alpha - \beta) + \\
\quad \cos (\alpha + \beta) + \cos (\alpha - \beta).
\end{array}
$$

Substituting equation (1) into equation (2) gives
$$
t^{2} = 1 + \cos^{2} (\alpha - \beta) \cdot \frac{1 - \cot \gamma}{1 + \cot \gamma} + \frac{2}{1 + \cot \gamma} \cos (\alpha - \beta).
$$

Since $-\frac{\pi}{2} < \alpha - \beta < \frac{\pi}{2}$, we have $0 < \cos (\alpha - \beta) \leqslant 1$.
Thus, $t^{2} \leqslant 1 + \frac{1 - \cot \gamma}{1 + \cot \gamma} + \frac{2}{1 + \cot \gamma} = \frac{4}{1 + \cot \gamma}$.
Therefore, $t \leqslant \frac{2}{\sqrt{1 + \cot \gamma}}$.
Hence, $\cos \alpha + \cos \beta + \cos \gamma$
$= t + \cos \gamma \leqslant \frac{2}{\sqrt{1 + \cot \gamma}} + \frac{\cot \gamma}{\sqrt{1 + \cot^{2} \gamma}}$
$\leqslant \frac{2}{\sqrt{1 + \cot \gamma}} + \frac{\sqrt{2} \cot \gamma}{1 + \cos \gamma}$
$= \sqrt{2} + \frac{2}{\sqrt{1 + \cot \gamma}} - \frac{\sqrt{2}}{1 + \cot \gamma}$.
Let $x = \frac{1}{\sqrt{1 + \cos \gamma}}$, then $x \geqslant \frac{\sqrt{2}}{2}$.
Notice that
$f(x) = \sqrt{2} + 2x - \sqrt{2} x^{2} = -\sqrt{2} \left(x - \frac{\sqrt{2}}{2}\right)^{2} + \frac{3 \sqrt{2}}{2}$,
when and only when $x = \frac{\sqrt{2}}{2}$, $f(x)$ reaches its maximum value $\frac{3 \sqrt{2}}{2}$.
Therefore, $\cos \alpha + \cos \beta + \cos \gamma \leqslant \frac{3 \sqrt{2}}{2}$.
When and only when $\alpha = \beta = \gamma = \frac{\pi}{4}$, $\cos \alpha + \cos \beta + \cos \gamma$ reaches its maximum value, that is, when the quadrilateral $A B C D$ is a square, $\frac{O B}{A B} + \frac{O A}{A D} + \frac{O C}{B C}$ reaches its maximum value $\frac{3 \sqrt{2}}{2}$.
(Kuozhengping, Anhui Normal University High School, 241001)