Do \(AB||CD\Rightarrow AB||\left(SCD\right)\Rightarrow d\left(AB;SC\right)=d\left(AB;\left(SCD\right)\right)=d\left(A;\left(SCD\right)\right)\)
Trong tam giác SAD, kẻ \(AH\perp SD\) \(\Rightarrow AH\perp\left(SCD\right)\)
\(\Rightarrow AH=d\left(A;\left(SCD\right)\right)\)
Tam giác SAD vuông cân tại A \(\Rightarrow AH=\dfrac{AD}{\sqrt{2}}=\dfrac{a\sqrt{2}}{2}\)
\(\Rightarrow d\left(SC;AB\right)=\dfrac{a\sqrt{2}}{2}\)