Example 10 As shown in Figure 7, in $\triangle ABC$, $BC=6$, $AC=4\sqrt{2}$, $\angle C=45^{\circ}$, $P$ is a moving point on side $BC$. A line through point $P$ parallel to $AB$ intersects $AC$ at point $D$. Connect $AP$, the areas of $\triangle ABP$, $\triangle APD$, and $\triangle CDP$ are denoted as $S_{1}$, $S_{2}$, and $S_{3}$, respectively. Let $BP=x$.
(1) Try to express $S_{1}$, $S_{2}$, and $S_{3}$ in terms of $x$;
(2) When point $P$ is at a certain position on $BC$ such that the area of $\triangle APD$ is maximized, what quantitative relationships can you derive between $S_{1}$, $S_{2}$, and $S_{3}$ (require at least 3 relationships)?
(18th Jiangsu Province Mathematics Competition (Junior High School))
Explanation: (1) Draw $P M \perp A C, A N \perp B C$. From the problem, we know that $B P=x(0<x<6)$ and $\frac{A D}{B P}=\frac{A C}{B C}$. Therefore,
$$
\begin{array}{l}
A D=\frac{A C}{B C} \cdot B P=\frac{4 \sqrt{2}}{6} x=\frac{2 \sqrt{2}}{3} x, \\
S_{1}=S_{\triangle A B P}=\frac{1}{2} B P \cdot A N=2 x, \\
S_{2}=S_{\triangle A D P}=\frac{1}{2} A D \cdot P M \\
=\frac{1}{2} \times \frac{2 \sqrt{2}}{3} x \times \frac{\sqrt{2}}{2}(6-x)=2 x-\frac{1}{3} x^{2}, \\
S_{3}=S_{\triangle A B C}-S_{1}-S_{2}=\frac{1}{3}(6-x)^{2} .
\end{array}
$$
(2) Since $S_{2}=2 x-\frac{1}{3} x^{2}=3-\frac{1}{3}(x-3)^{2}$, when $x=3$, $S_{2}$ reaches its maximum value, which is 3. At this time, $S_{1}=6, S_{3}=3$.
Therefore, the quantitative relationships between $S_{1}, S_{2}, S_{3}$ are
$$
S_{1}=S_{2}+S_{3}, S_{2}=S_{3}, S_{1}=2 S_{2}, S_{1}=2 S_{3} \text {. }
$$