(9) $2 \sqrt{2}$ Hint: Take the midpoint $O$ of $B D$, it is easy to prove that $\triangle A C O$ is an equilateral triangle with side length $2 \sqrt{2}$. Therefore, $A C=2 \sqrt{2}$. Let $M$ be the midpoint of $B C$, in $\triangle A C B$
$$
A M=\sqrt{\frac{1}{2}\left(A C^{2}+A B^{2}\right)-\frac{1}{4} B C^{2}}=2 \sqrt{2} .
$$
9 The side length of the square $A B C D$ is 4. When it is folded along $B D$ to form a $60^{\circ}$ dihedral angle, the distance between the midpoint of $B C$ and $A$ is $\qquad$ .