9. A square $A B C D$ with side length 4 is folded along $B D$ to form a $60^{\circ}$ dihedral angle. Then the distance between the midpoint of $B C$ and $A$ is $\qquad$ .
9. $2 \sqrt{2}$.

Take the midpoint $O$ of $B D$. It is easy to know that $\triangle A C O$ is an equilateral triangle with side length $2 \sqrt{2}$. Therefore, $A C=2 \sqrt{2}$.
Let the midpoint of $B C$ be $M$.
In $\triangle A C B$, we have
$$
A M=\sqrt{\frac{1}{2}\left(A C^{2}+A B^{2}\right)-\frac{1}{4} B C^{2}}=2 \sqrt{2} .
$$