$\overrightarrow{AK}=\overrightarrow{AB}+\overrightarrow{BK}=\overrightarrow{AB}+\dfrac13\overrightarrow{BI}=\overrightarrow{AB}+\dfrac13\cdot\dfrac12(\overrightarrow{BA}+\overrightarrow{BC})=\overrightarrow{AB}+\dfrac16\overrightarrow{BA}+\dfrac16\overrightarrow{BC}=\dfrac{5}6\overrightarrow{AB}+\dfrac16\overrightarrow{BC}=\dfrac{5}6\overrightarrow{AB}+\dfrac16(\overrightarrow{AC}-\overrightarrow{AB})=\dfrac{2}3\overrightarrow{AB}+\dfrac16\overrightarrow{AC}$
$\overrightarrow{AH}=\overrightarrow{AB}+\overrightarrow{BH}=\overrightarrow{AB}+\dfrac15\overrightarrow{BC}$
Suy ra $\overrightarrow{AH}=\dfrac{6}5\overrightarrow{AK}$ nên $A,K,H$ thẳng hàng.
Nếu có thắc mắc bạn cứ hỏi tại đây nhé, tụi mình sẽ hỗ trợ.