2. Let $D$ be any point on the side $A B$ of $\triangle A B C$ (excluding points $A$ and $B$). $D$ moves parallel to $B C$ to point $E$ on side $A C$, then from $E$ moves parallel to $A B$ to point $F$ on side $B C$; then from $F$ moves parallel to $C A$ to point $G$ on side $A B$; and so on. A movement along a line parallel to one side to another side is called a "point translation transformation." When $D$ returns to its original starting point, the maximum number of point translation transformations is denoted as $m$, and the minimum number is denoted as $n$. Then the value of $m+n$ is ( ).
(A) 3
(B) 6
(C) 9
(D) Uncertain
2. C.

According to the problem:
(1) When point $D$ starts from the midpoint of side $AB$, points $E$ and $F$ are the midpoints of $AC$ and $BC$, respectively. Therefore, such a point translation transformation requires only three times (minimum).
So, $n=3$.
(2) As shown in Figure 4, when point $D$ starts from any point other than the midpoint of side $AB$, by the theorem of parallel lines dividing segments proportionally, we have
$$
\begin{array}{l}
\frac{AD}{BD}=\frac{AE}{EC}=\frac{BF}{FC} \\
=\frac{BG}{GA}=\frac{CH}{HA}=\frac{CK}{KB}=\frac{AM}{MB}. \\
\text{Then } \frac{AD+BD}{BD}=\frac{AM+MB}{MB}, \text{ i.e., } \frac{AB}{BD}=\frac{AB}{BM}.
\end{array}
$$

Thus, $BD=BM$. This indicates that point $D$ coincides with point $M$. Therefore, such a point translation transformation requires six times (maximum).
So, $m=6$.
Therefore, $m+n=9$.