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Bài 5. Cho hình chóp tam giác \(S.ABC\) có \(SA = SB = SC\) và có \(\widehat{ABC}= \widehat{BSC}=\widehat{CSA}.\) Chứng minh rằng \(SA ⊥ BC, SB ⊥ AC, SC ⊥ AB\).
(h.3.19)
\(\overrightarrow{SA}.\overrightarrow{BC}=\overrightarrow{SA}.(\overrightarrow{SC}-\overrightarrow{SB})\)
\(=\overrightarrow{SA}.\overrightarrow{SC}-\overrightarrow{SA}.\overrightarrow{SB}\)
\(= SA.SC.\cos\widehat{ASC} – SA.SB.\cos\widehat{ASB} = 0\).
Vậy \(SA ⊥ BC\).
\(\overrightarrow{SB}.\overrightarrow{AC}=\overrightarrow{SB}.(\overrightarrow{SC}-\overrightarrow{SA})\)
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\(=\overrightarrow{SB}.\overrightarrow{SC}-\overrightarrow{SB}.\overrightarrow{SA}\)
\(= SB.SC.\cos\widehat{BSC} – SB.SA.\cos\widehat{ASB} = 0\).
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 \(SC ⊥ AB\).