Ta có
$\overrightarrow{AD}.\overrightarrow{BC}$
$=\left ( \overrightarrow{AB}+\overrightarrow{BD} \right )\left ( \overrightarrow{BA}+\overrightarrow{AC} \right )$
$=\overrightarrow{AB}\left ( \overrightarrow{BA}+\overrightarrow{AC} \right )+\overrightarrow{BD}.\overrightarrow{BA}+\overrightarrow{BD}.\overrightarrow{AC}$
$=\overrightarrow{AB}.\overrightarrow{BC}+\overrightarrow{BD}.\overrightarrow{BA} \quad (1)$. Do $BD \perp AC$ nên $\overrightarrow{BD}.\overrightarrow{AC}=0.$
Tương tự ta có:
$\overrightarrow{AD}.\overrightarrow{BC}=\left ( \overrightarrow{AB}+\overrightarrow{BD} \right )\left ( \overrightarrow{BD}+\overrightarrow{DC} \right) =\overrightarrow{AB}.\overrightarrow{BD}+\overrightarrow{BD}.\overrightarrow{BC} \quad (2)$
Từ (1) và (2) suy ra
$2\overrightarrow{AD}.\overrightarrow{BC} = \overrightarrow{AB}.\overrightarrow{BC}+\overrightarrow{BD}.\overrightarrow{BC} =\overrightarrow{AD}.\overrightarrow{BC} $
suy ra $\overrightarrow{AD}.\overrightarrow{BC} =0\Rightarrow $ đpcm.