Cho tam giác ABC gọi I, J,K lần lượt là được xác định bởi IB=2IC , JC=-1:2JA, KA=-KB chứng minh I, J, K thẳng hàng

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