Cho hình chóp S.ABCD có đáy ABCD là hình thang cân có hai đường chéo AC, BD vuông góc với nhau, \(AD = 2a\sqrt 2 ;BC = a\sqrt 2 \).

Do \(\left( {SAC} \right) \bot \left( {ABCD} \right),\left( {SBD} \right) \bot \left( {ABCD} \right),\left( {SAC} \right) \cap \left( {SBD} \right) = SO \Rightarrow SO \bot \left( {ABCD} \right)\)

Dựng góc giữa \(\left( {SCD} \right),(ABCD)\):

\(\left( {SCD} \right) \cap \left( {ABCD} \right) = DC\). Kẻ \(OK \bot DC \Rightarrow SK \bot DC \Rightarrow \left( {\widehat {\left( {SCD} \right),\left( {ABCD} \right)}} \right) = \widehat {SKO}\)

Kéo dài MO cắt DC tại E

Ta có:

\(\widehat {{A_1}} = \widehat {{D_1}};\widehat {{A_1}} = \widehat {{M_1}};\widehat {{M_1}} = \widehat {{M_2}} = \widehat {{O_1}} \\\Rightarrow \widehat {{D_1}} = \widehat {{O_1}};\widehat {{O_1}} + \widehat {EOD} = {90^0} \\\Rightarrow \widehat E = {90^0} \\ \Rightarrow E \equiv K\)

Ta có:  

\(OK = \frac{{2a.a}}{{a\sqrt 5 }};OM = \frac{{AB}}{2} = \frac{{a\sqrt 5 }}{2};MK = \frac{{9a\sqrt 5 }}{{10}}\)

\(\begin{array}{l} \frac{{d(O,(SCD))}}{{d(M,(SCD))}} = \frac{{OE}}{{ME}} = \frac{9}{4} \Rightarrow d\left( {M,(SCD)} \right)\\ = \frac{9}{4}d\left( {O,(SCD)} \right) = \frac{9}{4}OH\\ OS = OK.\tan {60^0} = \frac{{2a\sqrt {15} }}{5} \end{array}\)

\( \Rightarrow OH = \frac{{OK.OS}}{{\sqrt {O{K^2} + O{S^2}} }} = \frac{{a\sqrt {15} }}{5} \Rightarrow d\left( {M,(SCD)} \right) = \frac{{9a\sqrt {15} }}{{20}}\)