a/ \(SD\perp\left(ABCD\right)\Rightarrow SD\perp AB\)
Mà \(AB\perp AD\Rightarrow AB\perp\left(SAD\right)\)
b/ \(SD\perp\left(ABCD\right)\Rightarrow SD\perp BC\)
\(BC\perp CD\Rightarrow BC\perp\left(SCD\right)\)
Mà \(BC\in\left(SBC\right)\Rightarrow\left(SBC\right)\perp\left(SCD\right)\)
c/ Từ D kẻ \(DH\perp SA\Rightarrow DH\perp\left(SAB\right)\)
\(\Rightarrow DH=d\left(D;\left(SAB\right)\right)\)
Áp dụng hệ thức lượng:
\(\frac{1}{DH^2}=\frac{1}{SD^2}+\frac{1}{AD^2}\Rightarrow DH=\frac{SD.AD}{\sqrt{SD^2+AD^2}}=\frac{2a\sqrt{3}}{3}\)
d/ \(OB=\frac{1}{2}DB\Rightarrow d\left(O;\left(SAB\right)\right)=\frac{1}{2}d\left(D;\left(SAB\right)\right)=\frac{a\sqrt{3}}{3}\)
