Giải thích các bước giải:
a.Ta có :
$SM=\dfrac13SD\to\dfrac{DM}{SM}=2$
Mà $AD//BC\to\dfrac{OD}{OB}=\dfrac{AD}{BC}=2$
$\to\dfrac{DM}{SM}=\dfrac{OD}{OB}\to OM//SB$
$\to OM//(SAB)$
b.Ta có M, H, O thẳng hàng
$\to \dfrac{MA}{MI}.\dfrac{HI}{HC}.\dfrac{OC}{OA}=1$
$\to 2.\dfrac{HI}{HC}.\dfrac{1}{2}=1$
$\to HI=HC$
$\to S_{MHC}=\dfrac12S_{MIC}=\dfrac12.\dfrac12S_{MAC}=\dfrac14S_{MAC}$
$\to \dfrac{S_{MAC}}{S_{MHC}}=\dfrac{S_1}{S_2}=4$