Giải thích các bước giải:
Kẻ $AH\bot SB=H$
Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
BC \bot SA\left( {do:SA \bot \left( {ABC} \right)} \right)\\
BC \bot AB\\
SA \cap AB = A
\end{array} \right.\\
\Rightarrow BC \bot \left( {SAB} \right)\\
\Rightarrow BC \bot AH
\end{array}$
Khi đó:
$\begin{array}{l}
\left\{ \begin{array}{l}
AH \bot BC\\
AH \bot SB\\
BC \cap SB = B
\end{array} \right.\\
\Rightarrow AH \bot \left( {SBC} \right)\\
\Rightarrow d\left( {A,\left( {SBC} \right)} \right) = AH
\end{array}$
Lại có:
$\begin{array}{l}
\Delta SAB;\widehat A = {90^0};SA = AB = a;AH \bot SB = H\\
\Rightarrow AH = \sqrt {\dfrac{1}{{\dfrac{1}{{S{A^2}}} + \dfrac{1}{{A{B^2}}}}}} = \dfrac{a}{{\sqrt 2 }}\\
\Rightarrow d\left( {A,\left( {SBC} \right)} \right) = \dfrac{a}{{\sqrt 2 }}
\end{array}$
Vậy $d\left( {A,\left( {SBC} \right)} \right) = \dfrac{a}{{\sqrt 2 }}$
