Lời giải:
a) Ta có:
$\begin{cases}SA\perp BC\quad (SA\perp (ABC))\\BC\perp AB\quad (gt)\\SA\cap AB =\{A\}\end{cases}$
$\Rightarrow BC\perp (SAB)$
$\Rightarrow BC\perp AM$
Khi đó:
$\begin{cases}AM\perp SB\quad (gt)\\BC\perp AM\quad (cmt)\\SB\cap BC=\{B\}\end{cases}$
$\Rightarrow AM\perp (SBC)$
b) Xét $\triangle SBC$ có:
$\dfrac{SM}{SB}=\dfrac{SN}{SC}\quad (gt)$
$\Rightarrow MN//BC$ (theo định lý $Thales$ đảo)
mà $BC\perp SB$
nên $MN\perp SB$
Ta có:
$\begin{cases}AM\perp SB\quad (gt)\\MN\perp SB\quad (cmt)\\AM\cap MN = \{M\}\end{cases}$
$\Rightarrow SB\perp (AMN)$
$\Rightarrow SB\perp AN$
