Qua E kẻ EM//CF
Xét $\Delta CDF$ có: $DE=EC; ME//CF \Longrightarrow DM=MF=\dfrac{1}{2} DF$
Ta có: $OD=\dfrac{2}{3}OF \Longrightarrow \dfrac{OD}{OF}=\dfrac{2}{3} \Longrightarrow \dfrac{OF}{DF}=\dfrac{3}{5}$
$ \Longrightarrow \dfrac{OM}{DF}=\dfrac{OF}{DF}-\dfrac{MF}{DF}=\dfrac{3}{5}-\dfrac{1}{2}=\dfrac{1}{10} \Longrightarrow \dfrac{DF}{OM}=10$
$\Longrightarrow \dfrac{OF}{DF}.\dfrac{DF}{OM}=\dfrac{3}{5}.10=6$
$\Longrightarrow \dfrac{OF}{OM}=6$
$\Longrightarrow \dfrac{OF}{OM}.\dfrac{OD}{OF}=\dfrac{OD}{OM}=6. \dfrac{2}{3}=4$
Mặt khác, từ $OA = 4OE \Longrightarrow \dfrac{OA}{OE}=4 \Longrightarrow \dfrac{OA}{OE}=\dfrac{OD}{OM} \Longrightarrow EM//AD$( Ta lét đảo) $
\Longrightarrow AD//CF \Longrightarrow AD//CB$ (1)
Xét $\Delta CDF$ có: $DE=CE; DM=MF \Longrightarrow ME=\dfrac{1}{2}CF=\dfrac{1}{4}CB$
Từ $ME//AD \Longrightarrow \dfrac{ME}{AD}=\dfrac{OE}{OA}=\dfrac{1}{4} \Longrightarrow ME=\dfrac{1}{4}AD$
$\dfrac{1}{4}CB=\dfrac{1}{4}AD \Longrightarrow BC=AD$(2)
Từ (1) và (2) $\Longrightarrow ABCD$ là hình bình hành
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