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\(\overrightarrow{AF}=2\overrightarrow{FC}\Rightarrow\overrightarrow{AF}=\frac{2}{3}\overrightarrow{AC}\)
\(\overrightarrow{AM}=\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}\)
\(\overrightarrow{EI}=\frac{3}{4}\overrightarrow{IF}=\frac{3}{4}\left(\overrightarrow{IE}+\overrightarrow{EF}\right)\Rightarrow\overrightarrow{EI}=\frac{3}{7}\overrightarrow{EF}\)
\(\overrightarrow{AI}=\overrightarrow{AE}+\overrightarrow{EI}=\overrightarrow{AE}+\frac{3}{7}\overrightarrow{EF}=\overrightarrow{AE}+\frac{3}{7}\left(\overrightarrow{EA}+\overrightarrow{AF}\right)=\frac{4}{7}\overrightarrow{AE}+\frac{3}{7}\overrightarrow{EF}\)
\(\overrightarrow{AI}=\frac{4}{7}.\frac{1}{2}\overrightarrow{AB}+\frac{3}{7}.\frac{2}{3}\overrightarrow{AC}=\frac{2}{7}\overrightarrow{AB}+\frac{2}{7}\overrightarrow{AC}=\frac{4}{7}\left(\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}\right)=\frac{4}{7}\overrightarrow{AM}\)
\(\Rightarrow A;M;I\) thẳng hàng
\(\overrightarrow{AK}=\frac{1}{2}\overrightarrow{AE}+\frac{1}{2}\overrightarrow{AF}=\frac{1}{4}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\)
Gọi P là điểm trên BC sao cho \(\overrightarrow{BP}=k.\overrightarrow{BC}\)
\(\overrightarrow{AP}=\overrightarrow{AB}+\overrightarrow{BP}=\overrightarrow{AB}+k.\overrightarrow{BC}=\overrightarrow{AB}+\overrightarrow{k}.\overrightarrow{BA}+k.\overrightarrow{AC}\)
\(=\left(1-k\right)\overrightarrow{AB}+k\overrightarrow{AC}=3k\left(\frac{1-k}{3k}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\right)\)
A;K;P thẳng hàng khi và chỉ khi: \(\frac{1-k}{3k}=\frac{1}{4}\Rightarrow k=\frac{4}{7}\)
Vậy điểm P thỏa mãn \(\overrightarrow{BP}=\frac{4}{7}\overrightarrow{BC}\) thì A;K;P thẳng hàng
a: \(\overrightarrow{BK}=\overrightarrow{BA}+\overrightarrow{AK}\)
\(=\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{AC}\)
\(=\overrightarrow{BA}-\dfrac{1}{3}\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{BC}\)
\(=\dfrac{2}{3}\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{BC}\)
