由 $\overset{\large\rightharpoonup}{AB}$$\cdot$$\overset{\large\rightharpoonup}{AC}$ = $\overset{\large\rightharpoonup}{AB}$$\cdot$$\overset{\large\rightharpoonup}{AD}$ 得 $\overset{\large\rightharpoonup}{AB}$$\cdot\left($$\overset{\large\rightharpoonup}{AC}$$-$$\overset{\large\rightharpoonup}{AD}$$\right)=0$。又 $\overset{\large\rightharpoonup}{AC}$$-$$\overset{\large\rightharpoonup}{AD}$ = $\overset{\large\rightharpoonup}{DC}$，所以 $\overset{\large\rightharpoonup}{AB}$$\cdot$$\overset{\large\rightharpoonup}{DC}$$=0$，等價於 $\overset{\large\rightharpoonup}{AB}$$\cdot$$\overset{\large\rightharpoonup}{CD}$$=0$。