Example 4 Let $D$ be a point on the side $AB$ of $\triangle ABC$, point $D$ moves along a direction parallel to $BC$ to point $E$ on side $AC$; then from point $E$ along a direction parallel to $AB$ to point $F$ on side $BC$; then from point $F$ along a direction parallel to $CA$ to point $G$ on side $AB$, $\cdots \cdots$ each move along a side parallel to another side counts as one move. Then, at most $n$ moves, point $D$ can return to its original starting point for the first time. What is the value of $n$?
Analysis: We can estimate through graphical experiments that point $D$ can return to the original starting point after 6 times.
In fact, as shown in

Figure 1, by the intercept theorem of parallel lines,
we get
$$
\begin{array}{l}
\frac{A D}{B D}=\frac{A E}{E C} . \\
=\frac{B F}{F C}=\frac{B G}{A G} \\
=\frac{C H}{A H}=\frac{C K}{B K}=\frac{A M}{B M} .
\end{array}
$$

By the theorem of the sum of ratios, we get
$$
\frac{A D+B D}{B D}=\frac{A M+B M}{B M},
$$

which means
$$
\frac{A B}{B D}=\frac{A B}{B M} \text { . }
$$

Therefore, point $D$ coincides with point $M$, so $n=6$. Of course, if we take the midpoint of any side as the starting point, it is obvious that it only takes 3 times to return to the original starting point for the first time.