空间有 $A,B,C,D$ 四个点，满足 $AC\perp BD$，空间中还有 $A',B',C',D'$，满足 $A'B'=AB$，$A'D'=AD$，$B'C'=BC$，$C'D'=CD$，求证：$A'C'\perp B'D'$．
解析 根据题意，有\[\overrightarrow{AC}\perp \overrightarrow {BD}\iff \overrightarrow{AC}\cdot \left(\overrightarrow {AD}-\overrightarrow {AB}\right)=0\iff \overrightarrow {AC}\cdot \overrightarrow{AD}=\overrightarrow{AC}\cdot \overrightarrow{AB},\]根据余弦定理，有\[AC^2+AD^2-CD^2=AB^2+AC^2-BC^2\iff AB^2+CD^2=AD^2+BC^2,\]从而\[A'B'^2+C'D'^2=A'D'^2+B'C'^2\iff \overrightarrow{A'C'}\perp \overrightarrow{B'D' },\]命题得证．