Example 1: On a leftover piece of an acute triangle, process it into a square part so that the four vertices of the square are all on the sides of the triangle. If the three sides of the triangle are $a$, $b$, and $c$, and $a > b > c$. Which side should two vertices of the square be placed on to maximize the area of the square part?
(1997, Mathematics Competition for Junior High School in Some Areas of Anhui Province)
Solution: As shown in Figure 2, from equation (1) we get $\frac{A E}{A D}=\frac{P S}{B C}$, which means
$$
P S=\frac{A D \cdot B C}{A D+B C}
$$

As shown in Figure 3, from equation (1) we get $M N=\frac{A B \cdot C G}{A B+C G}$.
Let the area of $\triangle A B C$ be $S$. Note that
$$
A B \cdot C G=B C \cdot A D=2 S \text {, }
$$

Thus, $M N-P S=2 S\left(\frac{1}{A B+C G}-\frac{1}{B C+\overline{A D}}\right)$
$$
\begin{array}{l}
=2 S \cdot \frac{B C+A D-A B-C G}{(A B+C G)(B C+A D)} \\
=2 S \cdot \frac{(a-c)(c a-2 S)}{(c+C G)(a+A D) c a} .
\end{array}
$$

Since $a-c>0, c a-2 S>0$, therefore, $M N>$ $P S$, which means the area of the inscribed square $P Q R S$ on side $A B$ is the largest.