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 09/08/2021 Bởi Nevaeh 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 Bình luận
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