Cho tứ giác $ABCD$.Lấy các điểm $M,N$ theo thứ tự thuộc $AB$ và $CD$ 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 $AD,BC$ và $MN$ sao cho: $\overrightarrow {AE}=l.\overrightarrow {AD},\overrightarrow {BF}=l.\overrightarrow {BC}$ và $\overrightarrow {MI}=l.\overrightarrow {MN}$.
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)
$\overrightarrow {DN}=k.\overrightarrow {DC} \Leftrightarrow \overrightarrow {ON}-\overrightarrow {OD}=k(\overrightarrow {OC}-\overrightarrow {OD})$ (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 trong câu a),ta được:
$\overrightarrow {EF}=(1-l)\overrightarrow {AB}+l.\overrightarrow {DC}$ (3)
$\overrightarrow {EF}=(1-l)\overrightarrow {AM}+l.\overrightarrow {DN}=(1-l).k.\overrightarrow {AB}+l.k.\overrightarrow {DC}$
$=k[(1-l).\overrightarrow {AB}+l.\overrightarrow {DC}]$ (4)
Từ (3) và (4),suy ra:
$\overrightarrow {EI}=k.\overrightarrow {EF} \Leftrightarrow E,F,I$ thẳng hàng.

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