[Example 5.6.3] Divide each side of triangle $ABC$ into $n$ equal parts, and connect the points of division by lines parallel to the sides. Find the total number $f(n)$ of all parallelograms formed within the boundary of the resulting network.
First, we consider parallelograms whose sides are not parallel to $BC$.

Extending the sides of these parallelograms, they intersect with $BC$, so each such parallelogram corresponds to four points or three points on $BC$ (when a vertex of the parallelogram lies on $BC$, it corresponds to three points).

To ensure that extending the sides of the parallelograms corresponds to different points, we can extend $AB$ and $AC$ by $\frac{1}{n}$, obtaining $\triangle AB^{\prime}C^{\prime}$. At this time, $B^{\prime}C^{\prime}$ has $n+2$ division points. Thus, in $\triangle ABC$, each parallelogram with sides not parallel to $BC$ corresponds to four different division points on $B^{\prime}C^{\prime}$.

Conversely, for four different division points on $B^{\prime}C^{\prime}$, draw lines parallel to $AB^{\prime}$ through the two points closer to $B^{\prime}$, and draw lines parallel to $AC^{\prime}$ through the two points closer to $C^{\prime}$. These four parallel lines intersect to form a parallelogram whose sides are not parallel to $BC$.
Thus, we can establish the following correspondence:
\begin{tabular}{|c|c|}
\hline \begin{tabular}{c} 
Parallelograms with sides \\
not parallel to $BC$
\end{tabular} & Extending $AB$ by $\frac{1}{n}$, $AC$ by $\frac{1}{n}$ \\
& and drawing parallel lines to $AB^{\prime}$ and $AC^{\prime}$ \\
\hline
\end{tabular}

In this way, the problem of counting parallelograms in the network within the triangle is transformed into the problem of counting ordered quadruples in $\{1, 2, \cdots, n+2\}$.

Since there are $C_{n+2}^{4}$ ordered quadruples in $\{1, 2, \cdots, n+2\}$, the number of parallelograms with sides not parallel to $BC$ is $C_{n+2}^{4}$. By symmetry, the total number of parallelograms required by the problem is $3 C_{n+2}^{4}$.