\(\overrightarrow{SA}.\overrightarrow{BC}=\overrightarrow{SA}.(\overrightarrow{SC}-\overrightarrow{SB})\)
\(=\overrightarrow{SA}.\overrightarrow{SC}-\overrightarrow{SA}.\overrightarrow{SB}\)
\(= SA.SC.\cos\widehat{ASC} - SA.SB.\cos\widehat{ASB} = 0\)
Vậy \(SA ⊥ BC\).
\(\overrightarrow{SB}.\overrightarrow{AC}=\overrightarrow{SB}.(\overrightarrow{SC}-\overrightarrow{SA})\)
\(=\overrightarrow{SB}.\overrightarrow{SC}-\overrightarrow{SB}.\overrightarrow{SA}\)
\(= SB.SC.\cos\widehat{BSC} - SB.SA.\cos\widehat{ASB} = 0\)
Vậy \(SB ⊥ AC\).
\(\overrightarrow{SC}.\overrightarrow{AB}=\overrightarrow{SC}.(\overrightarrow{SB}-\overrightarrow{SA})\)
\(=\overrightarrow{SC}.\overrightarrow{SB}-\overrightarrow{SC}.\overrightarrow{SA}\)
\(= SC.SB.\cos\widehat{BSC} - SC.SA.\cos\widehat{ASC} = 0\)
Vậy \(SC ⊥ AB\).
