9. If the sum of the squares of two pairs of opposite sides of a spatial quadrilateral are equal, then the angle formed by its two diagonals is

保留了源文本的换行和格式。
9. $90^{\circ}$ As shown in the figure, $A B C D$ is a spatial quadrilateral, with $A C, B D$ as its diagonals. The midpoints of $A B, B C, C D, D A, B D, A C$ are taken as $M, N, G, P, K, H$ respectively. Using the property that the sum of the squares of the four sides of a parallelogram equals the sum of the squares of its two diagonals, in $\square P H N K$, we have $P N^{2}+K H^{2}=2\left(P K^{2}+P H^{2}\right)=\frac{1}{2}\left(A B^{2}+C D^{2}\right)$. In $\square K M H G$, we have $M G^{2}+K H^{2}=2\left(M K^{2}+M H^{2}\right)=\frac{1}{2}\left(A D^{2}+B C^{2}\right)$. Since $A B^{2}+C D^{2}=A D^{2}+B C^{2}$,

it follows that $P N=M G$.
Thus, $\square M N G P$ is a rectangle, so $M N \perp N G$, hence $A C \perp B D$.