$\begin{array}{l} \left\{ \begin{array}{l} SA \bot \left( {ABCD} \right) \Rightarrow SA \bot DC\\ DC \bot AD\\ SA \cap AD = \left\{ A \right\} \end{array} \right.\\  \Rightarrow DC \bot \left( {SAD} \right) \Rightarrow DC \bot AK\\ AK \bot SD\left( {gt} \right) \Rightarrow AK \bot \left( {SDC} \right) \Rightarrow AK \bot SC\left( 1 \right)\\ tt:BC \bot \left( {SAB} \right) \Rightarrow AH \bot BC,AH \bot SB \Rightarrow AH \bot \left( {SBC} \right)\\  \Rightarrow AH \bot SC\left( 2 \right)\\ \left( 1 \right),\left( 2 \right) \Rightarrow SC \bot \left( {AHK} \right) \end{array}$  

$\triangle SAD=\triangle SAB$ nên ta được $SK=SH$
$\Rightarrow \dfrac{SK}{SD}=\dfrac{SH}{SB}\Rightarrow KH//BD$

$\begin{array}{l}
\left. \begin{array}{l}
BD \bot AC\\
BD \bot SA
\end{array} \right\} \Rightarrow BD \bot \left( {SAC} \right) \Rightarrow HK \bot \left( {SAC} \right)\\
 \Rightarrow HK \bot AM \Rightarrow M \in \left( {AHK} \right)\\
 \Rightarrow AM \subset \left( {AHK} \right)
\end{array}$