\(\overrightarrow{AI}=\dfrac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\)
\(\overrightarrow{BH}=x\overrightarrow{BC}\rightarrow\overrightarrow{BA}+\overrightarrow{AH}=x\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\rightarrow\overrightarrow{AH}=x\overrightarrow{AC}+\left(1-x\right)\overrightarrow{AB}\)
Để A,I,H thẳng hàng thì
\(\overrightarrow{AI}=k.\overrightarrow{AH}\)
\(\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}=k\left(1-x\right)\overrightarrow{AB}+kx\overrightarrow{AC}\)
Hay \(\left\{{}\begin{matrix}\dfrac{1}{3}=k\left(1-x\right)\\\dfrac{1}{3}=kx\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}k=\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
vậy x=1/2 thì thoả mãn
