\(BC=2BM\Rightarrow d\left(B;\left(SAC\right)\right)=2d\left(M;\left(SAC\right)\right)\)
\(MA=2HA\Rightarrow d\left(M;\left(SAC\right)\right)=2d\left(H;\left(SAC\right)\right)\)
\(\Rightarrow d\left(B;\left(SAC\right)\right)=4d\left(H;\left(SAC\right)\right)\)
Gọi N là trung điểm AC \(\Rightarrow\left\{{}\begin{matrix}NH//CM\\NH=\frac{1}{2}CM=\frac{1}{4}BC=a\sqrt{2}\end{matrix}\right.\) (đường trung bình)
\(\Rightarrow NH\perp AC\)
Trong mp (SNH), từ H kẻ \(HK\perp SN\Rightarrow HK\perp\left(SAC\right)\Rightarrow HK=d\left(H;\left(SAC\right)\right)\)
Từ H kẻ \(HP\perp BC\Rightarrow P\) là trung điểm CM (đường trung bình)
\(HP=\frac{1}{2}AC=\frac{8a}{2\sqrt{2}}=2a\sqrt{2}\)
\(BP=\frac{3}{4}BC=\frac{3}{4}.\frac{8a}{\sqrt{2}}=3a\sqrt{2}\)
\(\Rightarrow BH=\sqrt{HP^2+BP^2}=a\sqrt{26}\)
\(SH=\sqrt{SB^2-BH^2}=\frac{a\sqrt{521}}{2}\)
\(d\left(B;\left(SAC\right)\right)=4HK=4.\frac{SH.NH}{\sqrt{SH^2+NH^2}}=...\)
