5. As shown in Figure 2, in pentagon $A B C D E$, $\angle A=$ $120^{\circ}, \angle B=\angle E=$ $90^{\circ}, A B=B C=1, A E$ $=D E=2$. Find points $M$ and $N$ on $B C$ and $D E$ respectively, such that the perimeter of $\triangle A M N$ is minimized. Then the minimum perimeter of $\triangle A M N$ is ( ).
(A) $2 \sqrt{6}$
(B) $2 \sqrt{7}$
(C) $4 \sqrt{2}$
(D) 5
5.B.

As shown in Figure 5, extend $AB$ to $P$ such that $PB = AB$; extend $AE$ to $Q$ such that $EQ = AE$. Connect $PQ$ to intersect $BC$ and $DE$ at points $M$ and $N$, respectively. Then the perimeter of $\triangle AMN$ is minimized, and the minimum perimeter is the length of segment $PQ$. Draw $PF \perp AE$ at $F$.
It is easy to know that $\angle PAF = 60^{\circ}$, $PF = \frac{\sqrt{3}}{2} AP = \sqrt{3}$,
$$
AF = \frac{1}{2} AP = 1 \text{. }
$$

Also, $AQ = 2AE = 4$, then in the right triangle $\triangle PQF$, by the Pythagorean theorem we get
$$
PQ = \sqrt{PF^2 + FQ^2} = 2\sqrt{7}.
$$

In $\triangle APQ$, by the cosine rule we get
$$
\cos \angle APQ = \frac{AP^2 + PQ^2 - AQ^2}{2 \cdot AP \cdot PQ} = \frac{2\sqrt{7}}{7} \text{. }
$$

Therefore, $BM < BC$.
So, point $M$ lies on segment $BC$.
Similarly, point $N$ lies on segment $DE$.