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vecto AN+vecto BP+vecto CM
=vecto AB+vecto BN+vecto BC+vecto CP+vecto CA+vecto AM
=vecto AB+1/3vecto BC+vecto BC+1/3vecto CA+vecto CA+1/3vecto AB
=4/3 vecto AB+4/3vecto BC+4/3vecto CA
=vecto 0
\(\overrightarrow{AI}=\dfrac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\)
\(\overrightarrow{BH}=x\overrightarrow{BC}\rightarrow\overrightarrow{BA}+\overrightarrow{AH}=x\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\rightarrow\overrightarrow{AH}=x\overrightarrow{AC}+\left(1-x\right)\overrightarrow{AB}\)
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
\(BC^2=AB^2+AC^2-2AB\cdot AC\cdot\cos A\)
\(=AB^2+AC^2-2\cdot AB\cdot AC\cdot\cos60\\ =AB^2+AC^2-2\cdot AB\cdot AC\cdot\dfrac{1}{2}\\ =AB^2+AC^2-AB\cdot AC\)
\(\overrightarrow{AP}+\overrightarrow{AM}\)
\(=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\)
\(=\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\)
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)
\(=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)