Proof: As shown in the figure, let the midpoints of $AB, BC, CD, DA, AC$, and $BD$ be $E, F, G, H, K, L$ respectively, and $AC \perp BD, AB \perp CD, AD \perp BC$.

Since $EF \parallel AC, GH \parallel AC$, we have $EF \parallel GH$.
Similarly,
$EH \parallel \tilde{F} G$.
Therefore, $EFGH$ is a parallelogram.
$$\begin{array}{l} \text { Since } \because AC \perp BD, \\ \therefore EF \perp EH . \end{array}$$

Thus, $EFGH$ is a rectangle. The intersection point $O$ of its diagonals $EG$ and $FH$ satisfies
$$O E=O F=O G=O H .$$

Similarly, $ELGK$ is also a rectangle, and the intersection point of its diagonals $EG$ and $KL$ is the midpoint $O$ of $EG$, hence
$$O E=O K=O G=O L \text {. }$$

Therefore,
$$O E=O F=O G=O H =O K=O L \text {. }$$

This also indicates that $E, F, G, H, K, L$ lie on a sphere with $O$ as the center and $OE$ as the radius.