Example 5 As shown in Figure 2, in trapezoid $ABCD$, there is $AD // BC, AB \perp BC, AD=8 \text{~cm}, BC=16 \text{~cm}$, $AB=6 \text{~cm}$. Moving points $M, N$ start from points $B, C$ simultaneously, moving along $BC, CD$ in the directions of $BC, CD$, with the speeds of points $M, N$ being $2 \text{~cm} / \text{s}, 1 \text{~cm} / \text{s}$, respectively.
(1) After how many seconds do points $M, N$ move such that $MN // BD$?
(2) When point $M$ is moving on side $BC$, let the time of point $M$'s movement be $t(\text{~s})$. Is there a certain moment $t(\text{~s})$ such that the area of $\triangle AMN$ is minimized? If it exists, find the value of $t$; if not, explain the reason.
(2004, Chongqing City Junior High School Mathematics Competition Preliminary)
(1) Suppose when points $M$ and $N$ have moved for $x \, \text{s}$, $M N \parallel B D$. Then $B M = 2x$, $C M = 16 - 2x$, $C N = x$.
From the given information, we have $B D = 10$.
Draw $D H \perp B C$ at $H$, then $B H = A D = 8$.
Thus, $C H = 8$, $C D = B D = 10$.
Therefore, $D N = 10 - x$.
When $M N \parallel B D$, $\frac{C M}{B M} = \frac{C N}{D N}$.
From $\frac{16 - 2x}{2x} = \frac{x}{10 - x}$, solving gives $x = \frac{40}{9}$.
Hence, when points $M$ and $N$ have moved for $\frac{40}{9} \, \text{s}$, $M N \parallel B D$.
(2) It exists.
$$
\begin{array}{l}
S_{\triangle A M N} = S_{\text{trapezoid BCD}} - \left(S_{\triangle A B M} + S_{\triangle C M N} + S_{\triangle A D N}\right) \\
= \frac{1}{2}(8 + 16) \times 6 - \\
\frac{1}{2}\left[6 \times 2t + (16 - 2t) \times \frac{3}{5}t + 8\left(6 - \frac{3}{5}t\right)\right] \\
= 72 - \left(6t + \frac{24}{5}t - \frac{3}{5}t^2 + 24 - \frac{12}{5}t\right) \\
= \frac{3}{5}\left(t^2 - 14t\right) + 48 \\
= \frac{3}{5}(t - 7)^2 + \frac{93}{5} (0 < t < 8). \\
\end{array}
$$

When $t = 7 \, \text{s}$, the area of $\triangle A M N$ is minimized, and the minimum value is $\frac{93}{5} \, \text{cm}^2$.

Note: Clearly, this is a problem of finding the minimum value of a quadratic function.