\(\left\{{}\begin{matrix}\left(SAB\right)\perp\left(ABCD\right)\\\left(SAC\right)\perp\left(ABCD\right)\\\left(SAB\right)\cap\left(SAC\right)=SA\end{matrix}\right.\) \(\Rightarrow SA\perp\left(ABCD\right)\)
Ta có: DM cắt (SBC) tại S, mà \(MS=\frac{1}{2}DS\Rightarrow d\left(M;\left(SBC\right)\right)=\frac{1}{2}d\left(D;\left(SBC\right)\right)\)
Lại có \(AD//BC\Rightarrow AD//\left(SBC\right)\Rightarrow d\left(D;\left(SBC\right)\right)=d\left(A;\left(SBC\right)\right)\)
\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp BC\\BC\perp AB\end{matrix}\right.\) \(\Rightarrow BC\perp\left(SAB\right)\)
Trong mặt phẳng (SAB), từ A kẻ \(AH\perp SB\Rightarrow AH\perp\left(SBC\right)\Rightarrow AH=d\left(A;\left(SBC\right)\right)\)
\(\frac{1}{AH^2}=\frac{1}{SA^2}+\frac{1}{AB^2}=\frac{2}{a^2}\Rightarrow AH=\frac{a\sqrt{2}}{2}\)
\(\Rightarrow d\left(M;\left(SBC\right)\right)=\frac{1}{2}AH=\frac{a\sqrt{2}}{4}\)
