Ta có:
\(\overrightarrow {IB} = 2\overrightarrow {IC} \Leftrightarrow \overrightarrow {IB} = 2\left( {\overrightarrow {IB} + \overrightarrow {BC} } \right) \Leftrightarrow - \overrightarrow {IB} = 2\overrightarrow {BC} \Leftrightarrow \overrightarrow {BI} = 2\overrightarrow {BC} \)
\(\overrightarrow {JC} = - \dfrac{1}{2}\overrightarrow {JA} \Leftrightarrow \overrightarrow {JB} + \overrightarrow {BC} = - \dfrac{1}{2}\left( {\overrightarrow {JB} + \overrightarrow {BA} } \right)\) \( \Leftrightarrow \dfrac{3}{2}\overrightarrow {JB} = - \dfrac{1}{2}\overrightarrow {BA} - \overrightarrow {BC} \Leftrightarrow \overrightarrow {JB} = - \dfrac{1}{3}\overrightarrow {BA} - \dfrac{2}{3}\overrightarrow {BC} \)
\( \Rightarrow \overrightarrow {BJ} = \dfrac{1}{3}\overrightarrow {BA} + \dfrac{2}{3}\overrightarrow {BC} \)
\( \Rightarrow \overrightarrow {IJ} = \overrightarrow {BJ} - \overrightarrow {BI} = \dfrac{1}{3}\overrightarrow {BA} + \dfrac{2}{3}\overrightarrow {BC} - 2\overrightarrow {BC} = \dfrac{1}{3}\overrightarrow {BA} - \dfrac{4}{3}\overrightarrow {BC} \)
\(\overrightarrow {KA} = - \overrightarrow {KB} \Leftrightarrow \overrightarrow {KB} + \overrightarrow {BA} = - \overrightarrow {KB} \Leftrightarrow {\rm{2}}\overrightarrow {KB} = - \overrightarrow {BA} \) \( \Rightarrow 2\overrightarrow {BK} = \overrightarrow {BA} \Rightarrow \overrightarrow {BK} = \dfrac{1}{2}\overrightarrow {BA} \)
\( \Rightarrow \overrightarrow {JK} = \overrightarrow {BK} - \overrightarrow {BJ} = \dfrac{1}{2}\overrightarrow {BA} - \dfrac{1}{3}\overrightarrow {BA} - \dfrac{2}{3}\overrightarrow {BC} = \dfrac{1}{6}\overrightarrow {BA} - \dfrac{2}{3}\overrightarrow {BC} = \dfrac{1}{2}\left( {\dfrac{1}{3}\overrightarrow {BA} - \dfrac{4}{3}\overrightarrow {BC} } \right) = \dfrac{1}{2}\overrightarrow {IJ} \)
Vậy \(I,J,K\) thẳng hàng
