18. As shown in Figure 5, in $\triangle ABC$, $BC=6$, $AC=4\sqrt{2}$, $\angle C=45^{\circ}$, $P$ is a moving point on side $BC$. A line through $P$ 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 $S_{3}$ (at least 3 relationships are required)?
18. (1) From the problem, we know that $B P=x, 0<x<6$, and $\frac{A D}{B P}=\frac{A C}{B C}$,

so $A D=\frac{A C}{B C} \cdot B P=\frac{4 \sqrt{2}}{6} x=\frac{2 \sqrt{2}}{3} x$.
$S_{2}=S_{\triangle A P D}=\frac{1}{2} A D \cdot P M$
$=\frac{1}{2} \times \frac{2 \sqrt{2}}{3} x \times \frac{\sqrt{2}}{2}(6-x)=2 x-\frac{1}{3} x^{2}$.
$S_{1}=S_{\triangle B P}=\frac{1}{2} B P \cdot A N=2 x$,
$S_{3}=S_{\triangle C D P}=S_{\triangle B C}-S_{1}-S_{2}=\frac{1}{3}(6-x)^{2}$.
(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. $S_{2}=3$, at this time $S_{1}=6, S_{3}=3$. Therefore, the quantitative relationships between $S_{1} 、 S_{2} 、 S_{3}$ are
$$
S_{1}=S_{2}+S_{3}, S_{2}=S_{3}, S_{1}=2 S_{2}, S_{1}=2 S_{3} .
$$