\(\left\{{}\begin{matrix}BC\perp AB\\BC\perp SA\end{matrix}\right.\) \(\Rightarrow BC\perp\left(SAB\right)\Rightarrow BC\perp AH\Rightarrow AH\perp\left(SBC\right)\Rightarrow AH\perp SC\)
Tương tự ta có \(AK\perp\left(SCD\right)\Rightarrow AK\perp SC\)
Lại có \(AI\perp SC\) mà \(AI,AH,AK\) đồng quy tại A
\(\Rightarrow A,H,I,K\) đồng phẳng
b/ \(AC=a\sqrt{2}\)
Từ câu a ta chứng minh được \(\left\{{}\begin{matrix}AH\perp\left(SBC\right)\Rightarrow AH\perp IH\\AK\perp\left(SCD\right)\Rightarrow AK\perp IK\end{matrix}\right.\)
\(\frac{1}{AH^2}=\frac{1}{SA^2}+\frac{1}{AB^2}\Rightarrow AH=\frac{SA.AB}{\sqrt{SA^2+AB^2}}=\frac{a\sqrt{3}}{2}\)
\(AI=\frac{SA.AC}{\sqrt{SA^2+AC^2}}=\frac{a\sqrt{30}}{5}\) ; \(AK=\frac{SA.AD}{\sqrt{SA^2+AD^2}}=\frac{a\sqrt{3}}{2}\)
\(IH=\sqrt{AI^2-AH^2}=\frac{3a\sqrt{5}}{10}\) ; \(IK=\sqrt{AI^2-AK^2}=\frac{3a\sqrt{5}}{10}\)
\(\Rightarrow S_{AHIK}=S_{AIH}+S_{AIK}=\frac{1}{2}.AH.IH+\frac{1}{2}AK.IK=\frac{3a^2\sqrt{15}}{20}\)
