Giải thích các bước giải:
a) $\eqalign{ & qua\,S\,kẻ\,đường\,thẳng\,d\,song\,song\,với\,AB \cr & Vi\,d//AB\,và\,AB//CD \cr & \Rightarrow \,d//CD \cr & Ta\,co:d\, \in \,(SAB)\,do\,d\,đi\,qua\,S\,va\,d//AB \cr & d\, \in \,(SDC)\,do\,d\,di\,qua\,S\,va\,d//CD \cr & \Rightarrow Giao\,diem\,cua\,(SAB)\,va\,(SCD)\,la\,đường\,thẳng\,d \cr & \cr} $
b) $\eqalign{ & Vi\,AD//BC\,n\^e n\,theo\,định\,lí\,Talet\,ta\,co\, \cr & \frac{{BN}}{{DN}} = \frac{{BK}}{{AD}} \cr & M\`a \,BN = \frac{1}{3}DB \cr & \Rightarrow BN = \frac{1}{2}DN \cr & \Rightarrow \frac{{BN}}{{DN}} = \frac{1}{2} \cr & \Rightarrow \frac{{BK}}{{AD}} = \frac{1}{2}.M\`a \,AD = BC \cr & \Rightarrow BK = \frac{1}{2}BC \cr & \Rightarrow K\,l\`a \,trung\,diem\,cua\,BC \cr & Vì \,M\,là \,trung\,diem\,cua\,SC \cr & \Rightarrow MK\,la \,duong\,trung\,binh\,cua\,\vartriangle SBC \cr & \Rightarrow MK//SB \cr & \Rightarrow MK//(SBD)(dpcm) \cr} $
