Three, (Full marks 15 points) For the rectangle $\triangle B C(1)$, $A B=20$ cm, $B C=10$ cm. If points $M, N$ are taken on $A C$ and $A B$ (as shown in Figure 2), to make the value of $B M+M N$ the smallest, find this minimum value.
Three, as shown in Figure 1, construct the symmetric point $B'$ of $B$ with respect to $AC$, and connect $AB'$. Then the symmetric point of $N$ with respect to $AC$ is the point $N'$ on $AB$. At this time, the minimum value of $B$ and $M$ to $V$ is equal to the minimum value of $B \rightarrow M \rightarrow N'$, which is equal to the distance from $B$ to $AB'$, $B'H'$, and the minimum value of $BM + MN$ is $B'H'$. Now, we need to find $B'H'$. Connect $B$ with the intersection point $P$ of $AB'$ and $IC$. Then the area of $\triangle ABP$ is $\frac{1}{2} \times 20 \times 10 = 100$ square centimeters. Note that $PA = PC$ $(\because \angle PAC \stackrel{\text{symmetry}}{=} \angle BAC \stackrel{\text{alternate}}{=} \angle PAC)$, let $AP = x$. Then $P^2 - x, DP = 20 - x$.
So, $x^2 = (20 - x)^2 + 10^2, x = 12.5$.
$B'H' = \frac{100 \times 2}{12.5} = 16$ (centimeters).
Answer: The minimum value of $BM + MN$ is 16 centimeters.