$N\in (SAD)\cap(BNC)\Rightarrow Nx=(SAD)\cap(BNC)$Ta co:$\left\{ \begin{array}{l} BC\subset (BNC)\\ AD\subset (SAD)\\AD//BC\\(ASD)\cap(BNC)=Nx\end{array} \right.$$\Rightarrow Nx//AD//BC$
Goi $I=Nx\cap SA\Rightarrow NI//AD$ hay $NI//BK (1)$
Ta co :$NI//AD\Rightarrow \frac{NI}{AD}=\frac{SN}{SD}=\frac{1}{3} ma AD=BC\Rightarrow \frac{NI}{BC}=\frac{1}{3}$
Mat khac $\frac{BK}{BC}=\frac{1}{3}$$\Rightarrow BK=NI(2)$
Tu $(1),(2)$ suy ra $ BKNI$ la hinh binh hanh $\Rightarrow NK//BI$ hay $NK//(SAB)$