2. For a square $A B C D$ with side length $a$ and a square $A B E F$ on two half-planes forming a dihedral angle of $120^{\circ}$, let $M, N$ be moving points on the diagonals $A C$ and $B F$ respectively, and $A M=F N$. Then the range of $M N$ is
A. $\left(\frac{\sqrt{2}}{2} a, a\right]$
B. $\left(\frac{\sqrt{3}}{2} a, \sqrt{2} a\right)$
C. $\left[\frac{\sqrt{2}}{2} a, \sqrt{2} a\right)$
D. $\left[\frac{\sqrt{3}}{2} a, a\right]$
Solution:
As shown in the figure, draw $N P / / A F$ intersecting $A B$ at point $P$, and connect $P M$. Then $\frac{A M}{M C}=\frac{F N}{N B}=\frac{A P}{P B} \Rightarrow P M / / B C$,
thus $\angle N P M=120^{\circ}$. Let $A P=\lambda a(0 \leqslant \lambda \leqslant 1)$, then $P M=\lambda a, P N=(1-\lambda) a$.
Therefore, $M N^{2}=\left[\lambda^{2}+(1-\lambda)^{2}+\lambda(1-\lambda)\right] a^{2}$
$$
=\left(\lambda^{2}-\lambda+1\right) a^{2} \in\left[\frac{3}{4} a^{2}, a^{2}\right] \Rightarrow M N \in\left[\frac{\sqrt{3}}{2} a, a\right] \text {, }
$$

Therefore, the answer is $D$.