Giải thích các bước giải:
a. Ta có:
\(BC \perp OK\) (Do \(\Delta OBC\) cân nên OK là trung tuyến đồng thời đường cao
\(\left\{\begin{matrix} BC \perp OK
& & \\ BC \perp SO (SO \perp (ABCD))
& &
\end{matrix}\right.\)
\(\Rightarrow BC \perp (SOK)\)
b. Ta có: \(SB \subset (SBC)\)
Do \(AD//BC\) nên \(AD//(SBC)\)
Vậy \(d(AD,SB)=d(A,(SBC))\)
Kẻ \(OF \perp SK\)
Ta có:
\(\left\{\begin{matrix} OF \perp SK
& & \\ OK \perp BC
\end{matrix}\right.\)
\(\Rightarrow OF \perp (SBC)\)
\(OF=d(O,(SBC))\)
\(OK=\dfrac{1}{2}BC=\dfrac{a}{2}\)
\(SO=\sqrt{a^{2}-(\dfrac{\sqrt{2}}{2}a)^{2}}=\dfrac{\sqrt{2}}{2}\)
Ta có: \(\dfrac{1}{OF^{2}}=\dfrac{1}{SO^{2}}+\dfrac{1}{OK^{2}}\)
\(\Leftrightarrow OF=\dfrac{\sqrt{6}}{6}\)
Ta có: \(AO \bigcap (SBC)=C\)
Ta có: \(\dfrac{AC}{OC}=\dfrac{d(A,(SBC))}{d(O,(SBC))}\)
\(\Leftrightarrow \dfrac{d(A,(SBC))}{\dfrac{\sqrt{6}}{6}}=2\)
\(\Leftrightarrow d(A,(SBC))=\dfrac{\sqrt{6}}{3}\)
