\(SA\perp\left(ABCD\right)\Rightarrow SA\perp\left(BC\right)\Rightarrow BC\perp\left(SAB\right)\)
\(\Rightarrow SB\) là hình chiếu vuông góc của SC lên (SAB) \(\Rightarrow\widehat{BSC}=30^0\)
\(BC=AD=\sqrt{AC^2-AB^2}=a\sqrt{3}\)\(\Rightarrow SB=\frac{BC}{tan30^0}=3a\)
\(\Rightarrow SA=\sqrt{SB^2-AB^2}=2a\sqrt{2}\)
Kéo dài MC cắt AD tại N, áp dụng đl Talet:
\(\frac{AN}{DN}=\frac{AM}{CD}=\frac{1}{4}\Rightarrow\frac{AN}{AN+a\sqrt{3}}=\frac{1}{4}\Rightarrow AN=\frac{a\sqrt{3}}{3}\)
Từ A kẻ \(AH\perp MN\Rightarrow\frac{1}{AH^2}=\frac{1}{AN^2}+\frac{1}{AM^2}\Rightarrow AH=\frac{AM.AN}{\sqrt{AM^2+AN^2}}=\frac{a\sqrt{19}}{19}\)
Từ A kẻ \(AK\perp SH\Rightarrow AK\perp\left(SCM\right)\Rightarrow AK=d\left(A;\left(SCM\right)\right)\)
\(\frac{1}{AK^2}=\frac{1}{AH^2}+\frac{1}{SA^2}\Rightarrow AK=\frac{AH.SA}{\sqrt{AH^2+SA^2}}=\frac{2a\sqrt{34}}{51}\)
