Ta có:

Có $\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD}$

$\overrightarrow{CA} = \overrightarrow{BA} - \overrightarrow{BC}$

$\overrightarrow{BD} = \overrightarrow{BC} + \overrightarrow{CD}$

Có:

$\overrightarrow{AB} . \overrightarrow{CD} + \overrightarrow{BC} . \overrightarrow{AD} + \overrightarrow{CA} . \overrightarrow{BD}$

= $\overrightarrow{AB} . \overrightarrow{CD} + \overrightarrow{BC} (\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD}) + (\overrightarrow{BA} - \overrightarrow{BC}) (\overrightarrow{BC} + \overrightarrow{CD})$

$\overrightarrow{AB} . \overrightarrow{CD} + \overrightarrow{BC} . \overrightarrow{AD} + \overrightarrow{CA} . \overrightarrow{BD}$

= $\overrightarrow{AB} . \overrightarrow{CD} + \overrightarrow{BC} . \overrightarrow{AB} + \overrightarrow{BC}^{2} + \overrightarrow{BC} . \overrightarrow{CD} + \overrightarrow{BA} . \overrightarrow{BC} + \overrightarrow{BA} . \overrightarrow{CD} - \overrightarrow{BC}^{2} - \overrightarrow{BC} . \overrightarrow{CD}$

$\overrightarrow{AB} . \overrightarrow{CD} + \overrightarrow{BC} . \overrightarrow{AD} + \overrightarrow{CA} . \overrightarrow{BD}$

= $(\overrightarrow{AB} . \overrightarrow{CD} + \overrightarrow{BA} . \overrightarrow{CD}) + (\overrightarrow{BC} . \overrightarrow{AB}$

+ $\overrightarrow{BA} . \overrightarrow{BC}) +(\overrightarrow{BC}^{2} - \overrightarrow{BC}^{2}) + (\overrightarrow{BC} . \overrightarrow{CD} - \overrightarrow{BC} . \overrightarrow{CD}$)

= 0

$=> \overrightarrow{AB} . \overrightarrow{CD} + \overrightarrow{BC} . \overrightarrow{AD} + \overrightarrow{CA} . \overrightarrow{BD}$ = 0