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a: \(\overrightarrow{BK}=\overrightarrow{BA}+\overrightarrow{AK}\)
\(=\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{AC}\)
\(=\overrightarrow{BA}-\dfrac{1}{3}\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{BC}\)
\(=\dfrac{2}{3}\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{BC}\)
\(\overrightarrow{AB}+\overrightarrow{AC}=2\overrightarrow{AD}\)(D là trung điểm của BC) (1)
\(\overrightarrow{AM}+\overrightarrow{AN}=2\overrightarrow{AK}\)(K là trung điểm của MN) (2)
Lấy (1) trừ (2) có: \(\left(\overrightarrow{AB}+\overrightarrow{AC}\right)-\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=2\left(\overrightarrow{AD}-\overrightarrow{AK}\right)\)
⇔\(\dfrac{\left(\overrightarrow{AB}+\overrightarrow{AC}\right)-\left(\overrightarrow{AM}+\overrightarrow{AN}\right)}{2}\)=\(\overrightarrow{KD}\)
⇔\(\dfrac{\left(\overrightarrow{AB}+\overrightarrow{AC}\right)-\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\right)}{2}\)=\(\overrightarrow{KD}\)
⇔\(\dfrac{\overrightarrow{AB}+\overrightarrow{AC}-\dfrac{1}{2}\overrightarrow{AB}-\dfrac{1}{3}\overrightarrow{AC}}{2}\)=\(\overrightarrow{KD}\)
⇔\(\dfrac{\dfrac{1}{2}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}}{2}\)=\(\overrightarrow{KD}\)
⇔\(\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\)=\(\overrightarrow{KD}\)
\(\overrightarrow{KA}=-\overrightarrow{AK}=-\frac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=-\frac{1}{2}\left(\frac{1}{2}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\right)\)
\(=-\frac{1}{4}\overrightarrow{AB}-\frac{1}{6}\overrightarrow{AC}\)
\(\overrightarrow{KD}=\overrightarrow{AD}-\overrightarrow{AK}=\overrightarrow{AD}+\overrightarrow{KA}=\frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)-\frac{1}{4}\overrightarrow{AB}-\frac{1}{6}\overrightarrow{AC}\)
\(=\frac{1}{4}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\)