$\begin{array}{l}
\overrightarrow {IJ} = \overrightarrow {IC} + \overrightarrow {CJ} = \overrightarrow {CB} + \frac{1}{3}\overrightarrow {CA} = \overrightarrow {AB} - \overrightarrow {AC} - \frac{1}{3}\overrightarrow {AC} = \overrightarrow {AB} - \frac{4}{3}\overrightarrow {AC} \\
\overrightarrow {IK} = \overrightarrow {IB} + \overrightarrow {BK} = 2\overrightarrow {CB} + \frac{1}{2}\overrightarrow {BA} = 2\left( {\overrightarrow {AB} - \overrightarrow {AC} } \right) - \frac{1}{2}\overrightarrow {AB} \\
\,\,\,\,\,\,\,\, = \frac{3}{2}\overrightarrow {AB} - 2\overrightarrow {AC} = \frac{3}{2}\left( {\overrightarrow {AB} - \frac{4}{3}\overrightarrow {AC} } \right) = \frac{3}{2}\overrightarrow {IJ}
\end{array}$
\( \Rightarrow 3\) điểm \(I,\,\,J,\,\,K\) thẳng hàng.
