II. (25 points) As shown in Figure 4, the incircle $\odot I$ of $\triangle ABC$ touches $AB$ and $AC$ at points $D$ and $E$, respectively. Extend $DI$ to $M$ and $EI$ to $N$ such that $IM = IN = 9IE$. Points $B$, $I$, $C$, $M$, and $N$ are concyclic. The ratio of the perimeter of $\triangle ABC$ to side $BC$ is the simplest fraction $\frac{m}{n}$ (where $m$ and $n$ are coprime positive integers). Find the value of $100m + n$.
As shown in the figure,
10, let $A I$ intersect the circumcircle of $\triangle A B C$ at point $P, Q$ be the midpoint of side $B C$, and draw $P F \perp D M$ at point $F$, connecting $P C$. Then
$P Q \perp B C, C Q=\frac{1}{2} B C$,
Rt $\triangle A D I \sim$ Rt $\triangle P F I$,
Rt $\triangle C Q P \sim \mathrm{Rt} \triangle A D I$.
It is easy to prove $P B=P C=P I$.
Therefore, $B, C, I$ lie on the circle $\odot P$ with center $P$ and radius $P B$.
From the problem, $M, N$ are also on $\odot P$.
Thus, $I F=M F=\frac{1}{2} I M$.
Hence, $\triangle C Q P \cong \triangle P F I, P F=C Q=\frac{1}{2} B C$.
Since $\odot I$ is the incircle of $\triangle A B C$, then $I D=I E, A D=A E, B D+C E=B C$.

And $\triangle A D I \sim \triangle P F I$, so,
$$
\frac{P F}{A D}=\frac{F I}{I D}=\frac{\frac{1}{2} I M}{I D}=\frac{\frac{1}{2} \times 9 I E}{I E}=4.5 \text {. }
$$

Therefore, $P F=4.5 A D, B C=9 A D$.
Thus, the perimeter of $\triangle A B C$ is
$$
\begin{array}{l}
A B+A C+B C \\
=(A D+B D)+(A E+C E)+B C \\
=A D+(B D+C E)+A D+B C \\
=2 A D+2 B C=20 A D . \\
\text { Then } \frac{m}{n}=\frac{20 A D}{B C}=\frac{20}{9} .
\end{array}
$$

Since $m, n$ are coprime positive integers, $m=20, n=9$.
Therefore, $100 m+n=2009$.