Vẫn dùng kĩ thuật cũ:
\(\overrightarrow{AD}-2\overrightarrow{BC}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{AS}+\overrightarrow{SD}-2\overrightarrow{BS}-2\overrightarrow{SC}=0\)
\(\Leftrightarrow\overrightarrow{SA}=2\overrightarrow{SB}-2\overrightarrow{SC}+\overrightarrow{SD}\) (1)
Đặt \(\overrightarrow{SC}=x.\overrightarrow{SN}\)
Giả thiết suy ra \(\overrightarrow{SD}=3\overrightarrow{SM}\)
Thế vào (1): \(\overrightarrow{SA}=2\overrightarrow{SB}-2x.\overrightarrow{SN}+3\overrightarrow{SM}\)
Do A, B, N, M đồng phẳng
\(\Rightarrow2-2x+3=1\)
\(\Rightarrow x=2\Rightarrow SC=2SN\Rightarrow SN=\dfrac{1}{2}SC\)