5. On the sides $AB$ and $BC$ of the square $ABCD$, take points $P$ and $Q$ respectively, and connect $DP$, $DQ$, and $PQ$. Let the areas of $\triangle DPQ$, $\triangle DAP$, $\triangle DQC$, and $\triangle PBQ$ be $S_{1}$, $S_{2}$, $S_{3}$, and $S_{4}$ respectively. When do the points $P$ and $Q$ achieve the minimum value of $S_{1}^{2}+S_{2}^{2}+S_{3}^{2}+S_{4}^{2}$?
Let's assume the side length of the square is 1, and establish a coordinate system as shown. Let $P(0, b), Q(a, 0)$, then,
$$
\begin{array}{l}
S_{2}=\frac{1}{2}(1-b), \\
S_{3}=\frac{1}{2}(1-a), \\
S_{4}=\frac{1}{2} a b, \\
S_{1}=1-S_{2}-S_{3}-S_{4} . \\
\quad=\frac{1}{2}(a+b-a b) .
\end{array}
$$

Thus,
$$
\begin{aligned}
S_{1}^{2} & +S_{2}^{2}+S_{3}^{2}+S_{4}^{2} \\
= & \frac{1}{4}\left(1-2 b+b^{2}+1-2 a+a^{2}\right. \\
& +a^{2} b^{2}+a^{2}+b^{2}+a^{2} b^{2}+2 a b \\
& \left.-2 a^{2} b-2 a b^{2}\right) \\
= & \frac{1}{2}\left(a^{2}+b^{2}-a-b+1+a b-a^{2} b\right. \\
& \left.-a b^{2}+a^{2} b^{2}\right) \\
= & \frac{1}{2}\left\{\left(b^{2}-b+1\right) a^{2}-\left(b^{2}-b\right.\right. \\
& \left.+1) a+\left(b^{2}-b+1\right)\right\} \\
= & \frac{1}{2}\left(a^{2}-a+1\right)\left(b^{2}-b+1\right) \\
= & \frac{1}{2}\left[\left(a-\frac{1}{2}\right)^{2}+\frac{3}{4}\right] \\
& \cdot\left[\left(b-\frac{1}{2}\right)^{2}+\frac{3}{4}\right] . \text { (1) }
\end{aligned}
$$

Clearly, when $a=\frac{1}{2}, b=\frac{1}{2}$, the value of (1) is minimized. When $b C$ is divided into 4 points, $S_{1}^{2}+S_{2}^{2}+S_{3}^{2}+S_{4}^{2}$ achieves its minimum value $\frac{9}{32}$.