Xét \(\left(ABCD\right)\) có AB//CD
Theo Thales có \(\dfrac{IA}{IC}=\dfrac{AB}{CD}=\dfrac{a}{2a}=\dfrac{1}{2}\)
\(\Rightarrow IC=2IA\) \(\Rightarrow\dfrac{IC}{AC}=\dfrac{2}{3}\)
Vì IM//CD//AB nên theo Thales ta có :
\(\dfrac{CM}{BM}=\dfrac{IC}{CA}=\dfrac{2}{3}\)
Mà \(\dfrac{CN}{SC}=\dfrac{2}{3}\) do \(CN=2NS\left(gt\right)\)
\(\Rightarrow\dfrac{CN}{SC}=\dfrac{CM}{BM}=\dfrac{2}{3}\) (mặt phẳng SBC)
\(\Rightarrow NM//SB\left(1\right)\)
Lại có \(IM//CD//AB\left(2\right)\)
mà \(\left\{{}\begin{matrix}MN\subset\left(IMN\right)\\\left(AB\right)\subset\left(SAB\right)\end{matrix}\right.\) (3)
\(\left(1\right)\left(2\right)\left(3\right)\Rightarrow\left(IMN\right)//\left(SAB\right)\) => A đúng
