Giải thích các bước giải:
Sửa đề: $MC=\dfrac12MB$
a.Gọi $E$ là trung điểm $CD$
Vì $G$ là trọng tâm $\Delta SCD\to \dfrac{GE}{GA}=\dfrac12$
Gọi $EM\cap AB=F$
Ta có $AB//CD$
$\to \dfrac{ME}{MF}=\dfrac{MC}{MB}=\dfrac12=\dfrac{GE}{GA}$
$\to GM//SF$
b.Gọi $ME\cap AC=H, SH\cap GM=I$
$\to GM\cap (SAC)=I$
c.Ta có $CD//AB\to \dfrac{BF}{CE}=\dfrac{BM}{CM}=2$
$\to BF=2CE=CD$
$\to AF=2CD$
Mà $CE=\dfrac12CD$
$\to \dfrac{HE}{HF}=\dfrac{EC}{AF}=\dfrac14$
$\to \dfrac{HE}{HE+HF}=\dfrac1{4+1}$
$\to\dfrac{HE}{EF}=\dfrac15\to HE=\dfrac15EF$
Mà $\dfrac{ME}{MF}=\dfrac{MC}{MB}=\dfrac12$
$\to \dfrac{ME}{ME+MF}=\dfrac1{1+2}$
$\to\dfrac{ME}{EF}=\dfrac13$
$\to ME=\dfrac13EF$
$\to \dfrac{HE}{ME}=\dfrac35$
$\to \dfrac{HE}{ME-HE}=\dfrac{3}{5-3}$
$\to \dfrac{HE}{HM}=\dfrac32$
Ta có $S, I,H$ thẳng hàng
$\to \dfrac{SE}{SG}.\dfrac{IG}{IM}.\dfrac{HM}{HE}=1$
$\to \dfrac32.\dfrac{IG}{IM}.\dfrac23=1$
$\to \dfrac{IM}{IG}=1$
