Question

Cho hình chóp tam giác S.ABC có SA = SB = SC và có \(\widehat{\:ASB}=\widehat{BSC}=\widehat{CSA}\). Chứng minh rằng \(SA\perp BC;SB\perp AC;SC\perp AB\) ?

Answer 1

(h.3.19)

= SA.SC.cos - SA.SB.cos = 0.

Vậy SA ⊥ BC. 
\(\overrightarrow{SB}.\overrightarrow{AC}=\overrightarrow{SB}\left(\overrightarrow{SC}-\overrightarrow{SA}\right)=\overrightarrow{SB}.\overrightarrow{SC}-\overrightarrow{SB}.\overrightarrow{SA}\)
\(=SB.SC.cos\widehat{BSC}-SB.SA.cos\widehat{BSA}=0\).
Vậy \(SB\perp AC\).
\(\overrightarrow{SC}.\overrightarrow{AB}=\overrightarrow{SC}.\left(\overrightarrow{SB}-\overrightarrow{SA}\right)=\overrightarrow{SC}.\overrightarrow{SB}-\overrightarrow{SC}.\overrightarrow{SA}\)
\(=SC.SB.cos\widehat{BSC}-SC.SA.cos\widehat{CSA}=0\).
Vậy \(SC\perp AB\).