Vì (P)//BD nên [tex](P)\cap (SBD)=IE[/tex]//BD
BD_|_SO , BD_|_AC => BD_|_(SAC)
=> BD_|_AM
=> IE_|_AM
Ta có: [tex]\frac{SM}{MC}.\frac{AC}{AO}.\frac{OK}{KS}=1\Rightarrow 2.2.\frac{OK}{KS}=1\Leftrightarrow \frac{OK}{KS}=\frac{1}{4}[/tex]
[tex]\frac{SM}{MC}.\frac{IE}{BD}.\frac{SK}{SO}=\frac{4}{5}\Rightarrow IE=\frac{4}{5}BD\Rightarrow IE=\frac{4a\sqrt{2}}{5}[/tex]
Vì AC²=SA²+SC² => ∆SAC vuông tại S
[tex]SM=\frac{2a}{3};AM=\sqrt{a^2+\left ( \frac{2a}{3} \right )^2}=\frac{a\sqrt{13}}{3}[/tex]
[tex]S_{AEMI}=\frac{1}{2}EI.AM=\frac{2\sqrt{26}a^2}{15}[/tex]