5. In rectangle $ABCD$, the side lengths are $AD=3, AB=2, E$ is the midpoint of $AB$, $F$ is on segment $BC$, and $\frac{BF}{FC}=\frac{1}{2}, AF$ intersects $DE$ and $DB$ at points $M$ and $N$. Then $MN=(\quad$.
(A) $\frac{3 \sqrt{5}}{7}$
(B) $\frac{5 \sqrt{5}}{14}$
(C) $\frac{9 \sqrt{5}}{28}$
(D) $\frac{11 \sqrt{5}}{28}$
5. C.

It is easy to know that $\triangle B N F \backsim \triangle D N A$
$$
\Rightarrow \frac{F N}{A N}=\frac{B N}{N D}=\frac{B F}{A D}=\frac{1}{3} \text {. }
$$

Thus, $F N=\frac{1}{3} A N=\frac{1}{4} A F$.
As shown in Figure 4, extend $D E$ and $C B$ to intersect at point $G$.
Then $\triangle A M D \backsim \triangle F M G$
$$
\Rightarrow \frac{A M}{F M}=\frac{A D}{F G}=\frac{3}{4} \text {. }
$$

Thus, $A M=\frac{3}{4} F M=\frac{3}{7} A F$.
Therefore, $M N=A F-A M-F N$
$$
=\frac{9}{28} A F=\frac{9}{28} \sqrt{A B^{2}+B F^{2}}=\frac{9 \sqrt{5}}{28} .
$$