Gọi \(I = CN \cap DM\)
\(\Delta SAB\) đều \( \Rightarrow SM \bot AB\)
Mà \(\left( {SAB} \right) \bot \left( {ABCD} \right),\left( {SAB} \right) \cap \left( {ABCD} \right) = AB\)
\( \Rightarrow SM \bot \left( {ABCD} \right) \Rightarrow SM \bot CN\)
\(\Delta A{\rm{D}}M = \Delta DCN\left( {c.g.c} \right) \Rightarrow \widehat {AM{\rm{D}}} = \widehat {CN{\rm{D}}}\)
Mà \(\widehat {AM{\rm{D}}} + \widehat {A{\rm{D}}M} = {90^ \circ }\)
\(\widehat {CN{\rm{D}}} + \widehat {A{\rm{D}}M} = {90^ \circ } \Rightarrow \widehat {NI{\rm{D}}} = {180^ \circ } - \left( {\widehat {CN{\rm{D}}} + \widehat {A{\rm{D}}M}} \right) = {90^ \circ } \Rightarrow CN \bot DM\)
\(\left. \begin{array}{l}\left. \begin{array}{l}SM \bot CN\\CN \bot DM\end{array} \right\} \Rightarrow CN \bot \left( {SM{\rm{D}}} \right)\\CN \subset \left( {SNC} \right)\end{array} \right\} \Rightarrow \left( {SNC} \right) \bot \left( {SM{\rm{D}}} \right)\)
b) Kẻ \(MH \bot SI\left( {H \in SI} \right)\)
\(CN \bot \left( {SM{\rm{D}}} \right) \Rightarrow CN \bot MH\)
\( \Rightarrow MH \bot \left( {SNC} \right) \Rightarrow d\left( {M,\left( {SNC} \right)} \right) = MH\)
\(\Delta C{\rm{D}}N\) vuông tại \(D\) có đường cao \(DI\)
\(DN = \frac{1}{2}A{\rm{D}} = \frac{a}{2},CN = \sqrt {C{{\rm{D}}^2} + D{N^2}} = \frac{{a\sqrt 5 }}{2},DI = \frac{{C{\rm{D}}.DN}}{{CN}} = \frac{{a\sqrt 5 }}{5}\)
\(DM = CN = \frac{{a\sqrt 5 }}{2} \Rightarrow MI = DM - DI = \frac{{3a\sqrt 5 }}{{10}}\)
\(\Delta SAB\) đều \( \Rightarrow SM = \frac{{AB\sqrt 3 }}{2} = \frac{{a\sqrt 3 }}{2}\)
\(\Delta SMI\) vuông tại \(M\) có đường cao \(MH\)
\( \Rightarrow MH = \frac{{SM.MI}}{{\sqrt {S{M^2} + M{I^2}} }} = \frac{{3a\sqrt 2 }}{8}\)
Vậy \(d\left( {M,\left( {SNC} \right)} \right) = \frac{{3a\sqrt 2 }}{8}\)
