a) Ta có:
\(\overrightarrow{DE}=\overrightarrow{AE}-\overrightarrow{AD}=\dfrac{1}{4}\overrightarrow{AB}-\overrightarrow{AD}\) \(\left(1\right)\)
\(\overrightarrow{DF}=\overrightarrow{AF}-\overrightarrow{AD}\)
\(=\dfrac{1}{5}\overrightarrow{AC}-\overrightarrow{AD}\)
\(=\dfrac{1}{5}\left(\overrightarrow{AB}+\overrightarrow{BC}\right)-\overrightarrow{AD}\)
\(=\dfrac{1}{5}\overrightarrow{AB}+\dfrac{1}{5}\overrightarrow{AD}-\overrightarrow{AD}\)
\(=\dfrac{1}{5}\overrightarrow{AB}-\dfrac{4}{5}\overrightarrow{AD}\) \(\left(2\right)\)
b) Từ (1) và (2) ta có: \(\overrightarrow{DE}=\dfrac{5}{4}\overrightarrow{DF}\)
Vậy 3 điểm D, E, F thẳng hàng ( đpcm )
