a) xét tam giác SAC
gọi \(SO\cap MP=\left\{E\right\}\)\(\left(SO,MP\subset\left(SAC\right)\right)\)
ta có \(MP\subset\left(MNP\right)\)
vậy \(SO\cap\left(MNP\right)=\left\{E\right\}\)
b) xét tam giác SAC
ta có \(MC\cap SO=\left\{F\right\}\left(SO,MC\subset\left(SAC\right)\right)\)
mà \(SO\subset\left(SBD\right)\)
=>\(MC\cap\left(SBC\right)=\left\{F\right\}\)