1) Ta có:
$\begin{cases}(SAB)\perp (ABCD)\quad (gt)\\(SBC)\perp (ABCD)\quad (gt)\\(SAB)\cap (SBC) = SB\end{cases}$
$\to SB\perp (ABCD)$
$\to SB\perp BC$
mà $BC\perp AB$
nên $BC\perp (SAB)$
$\to (SBC)\perp (SAB)$
Mặt khác:
$SB\perp (ABCD)\quad (cmt)$
$\to SB\perp CD$
mà $CD\perp BC$
nên $CD\perp (SBC)$
$\to (SCD)\perp (SBC)$
2) Ta có:
$\begin{cases}AM\perp BD\quad (gt)\\AM\perp SB\quad (SB\perp (ABCD))\\SB\cap BD =\{B\}\end{cases}$
$\to AM\perp (SBD)$
$\to (SAM)\perp (SBD)$
3) Ta có:
$SB\perp (ABCD)$ (câu 1)
$\to SB\perp AD$
mà $AD\perp AB$
nên $AD\perp (SAB)$
$\to AD\perp BH$
mà $BH\perp SA\quad (gt)$
nên $BH\perp (SAD)$
$\to BH\perp SD$
Mặt khác:
$CD\perp (SBC)$ (câu 1)
$\to CD\perp BK$
mà $BK\perp SC\quad (gt)$
nên $BK\perp (SCD)$
$\to BK\perp SD$
4) Ta có:
$\begin{cases}SD\perp BH\quad \text{(câu 3)}\\SD\perp BK\quad \text{(câu 3)}\\BH\cap BK=\{B\}\end{cases}$
$\to SD\perp (BHK)$
$\to (SBD)\perp (BHK)$
