\(AC=10\sqrt{2}\Rightarrow SA=\sqrt{SC^2-AC^2}=10\sqrt{3}\)
Gọi E là trung điểm BC \(\Rightarrow BD//NE\) (đường trung bình)
\(\Rightarrow BD//\left(MNE\right)\Rightarrow d\left(BD;MN\right)=d\left(BD;\left(MNE\right)\right)=d\left(O;\left(MNE\right)\right)\)
Gọi H là giao điểm AC và NE \(\Rightarrow OH=\frac{1}{2}OC=\frac{1}{2}OA\Rightarrow OH=\frac{1}{3}AH\)
\(\Rightarrow d\left(O;\left(MNE\right)\right)=\frac{1}{3}d\left(A;\left(MNE\right)\right)\)
\(BD\perp AC;BD\perp SA\Rightarrow BD\perp\left(MAH\right)\Rightarrow NE\perp\left(MAH\right)\)
Từ A kẻ \(AK\perp MH\Rightarrow AK\perp\left(MNE\right)\Rightarrow AK=d\left(A;\left(MNE\right)\right)\)
\(MA=\frac{1}{2}SA=5\sqrt{3}\) ; \(AH=\frac{3}{4}AC=\frac{15\sqrt{2}}{2}\)
\(\frac{1}{AK^2}=\frac{1}{MA^2}+\frac{1}{AH^2}\Rightarrow AK=\frac{MA.AH}{\sqrt{MA^2+AH^2}}=3\sqrt{5}\)
\(\Rightarrow d\left(BD;MN\right)=\frac{1}{3}.3\sqrt{5}=\sqrt{5}\)
