Cho tứ giác $ABCD$. Lấy các điểm $M, N$ theo thứ tự thuộc $AB, CD$ sao cho $\overrightarrow{AM}=k\overrightarrow{AB}  $ và $\overrightarrow{DN}=k\overrightarrow{DC}  $
a) Chứng minh rằng: $\overrightarrow{MN}=(1-k).\overrightarrow{AD}+k.\overrightarrow{BC}   $
b) Gọi các điểm $E, F, I$ theo thứ tự thuộc $AD, BC, MN$ sao cho $\overrightarrow{AE}=l.\overrightarrow{AD},  \overrightarrow{BF}=l.\overrightarrow{BC},   \overrightarrow{MI}=l.\overrightarrow{MN}      $. Chứng minh rằng $E, F, I$ thẳng hàng
a) Từ giả thiết với điểm $O$ bất kỳ, ta được:
$\overrightarrow{AM}=k\overrightarrow{AB}\Leftrightarrow  \overrightarrow{OM}-\overrightarrow{OA}=k(\overrightarrow{OB} -\overrightarrow{OA}  )                   (1)$
$\overrightarrow{DN}=k\overrightarrow{DC}\Leftrightarrow  \overrightarrow{ON}-\overrightarrow{OC}=k(\overrightarrow{OC} -\overrightarrow{OD}  )           (2)$
Trừ theo vế $(2)$ cho $(1)$ ta được:
$(\overrightarrow{ON}-\overrightarrow{OM}  )-(\overrightarrow{OD}-\overrightarrow{OA}  )=k(\overrightarrow{OC}-\overrightarrow{OB}-\overrightarrow{OD}+\overrightarrow{OA}    )$
$\Leftrightarrow  \overrightarrow{MN}-\overrightarrow{AD}=k(\overrightarrow{BC}-\overrightarrow{AD})\Leftrightarrow  \overrightarrow{MN}=(1-k)\overrightarrow{AD}+k\overrightarrow{BC}$       (đpcm)

b) Sử dụng kết quả câu a) ta được:
$\overrightarrow{EF}=(1-l).\overrightarrow{AB}+l.\overrightarrow{DC}          (3)$
$\overrightarrow{EI}=(1-l).\overrightarrow{AM}+1\overrightarrow{DN}=(1-l).k.\overrightarrow{AB}+l.k.\overrightarrow{DC}     $
       $=k[(1-l).\overrightarrow{AB}+l.\overrightarrow{DC}]       (4)$
Từ $(3), (4)\Rightarrow  \overrightarrow{EI}=k.\overrightarrow{EF}  \Leftrightarrow  E, F, I$ thẳng hàng

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