$\begin{array}{l}
a)\overrightarrow {CI} = \overrightarrow {DI} - \overrightarrow {DC} = - \overrightarrow {ID} - \dfrac{1}{2}\overrightarrow {BC} \\
= \dfrac{2}{3}\overrightarrow {DA} - \dfrac{1}{2}\overrightarrow {BC} = \dfrac{2}{3}\left( {\overrightarrow {DC} + \overrightarrow {CA} } \right) - \dfrac{1}{2}\overrightarrow {BC} \\
= \dfrac{2}{3}\left( {\dfrac{1}{2}\overrightarrow {BC} + \overrightarrow {CA} } \right) - \dfrac{1}{2}\overrightarrow {BC} \\
= \dfrac{1}{3}\overrightarrow {BC} + \dfrac{2}{3}\overrightarrow {CA} - \dfrac{1}{2}\overrightarrow {BC} = - \dfrac{1}{6}\overrightarrow {BC} - \dfrac{2}{3}\overrightarrow {AC} \\
b)\overrightarrow {KC} = \overrightarrow {KB} + \overrightarrow {BC} = - \dfrac{4}{5}\overrightarrow {BA} + \overrightarrow {BC} \\
= - \dfrac{4}{5}\left( {\overrightarrow {BC} + \overrightarrow {CA} } \right) + \overrightarrow {BC} = - \dfrac{4}{5}\overrightarrow {BC} - \dfrac{4}{5}\overrightarrow {CA} + \overrightarrow {BC} \\
= \dfrac{1}{5}\overrightarrow {BC} + \dfrac{4}{5}\overrightarrow {AC} = - \dfrac{6}{5}\left( { - \dfrac{1}{6}\overrightarrow {BC} - \dfrac{2}{3}\overrightarrow {AC} } \right) = - \dfrac{6}{5}\overrightarrow {CI}
\end{array}$
Vậy ba điểm C, I, K thẳng hàng.
