tiếp: Xét tam giác ABD có ID/IA = BD/AB= (ac/b+c)/c=a/b+c

=> ID=(a/b+c)IA

=> \(\overrightarrow{ID}=-\frac{a}{b+c}\overrightarrow{IA}\)

Thế vào (1) ta đc: 

\(-\frac{a}{b+c}\overrightarrow{IA}=\frac{b}{b+c}\overrightarrow{IB}+\frac{c}{b+c}\overrightarrow{IC}\)

\(\Leftrightarrow\frac{1}{b+c}\left(a\overrightarrow{IA}+b\overrightarrow{IB}+c\overrightarrow{IC}\right)=0\)

<=> \(a\overrightarrow{IA}+b\overrightarrow{IB}+c\overrightarrow{IC}=0\): đpcm

giả sử AI kéo dài cắt BC tại D.

ta có: \(\frac{BD}{CD}=\frac{c}{b}\Rightarrow BD=\frac{c}{b}CD\Leftrightarrow\overrightarrow{DB}=-\frac{c}{b}\overrightarrow{DC}\Leftrightarrow\overrightarrow{DI}+\overrightarrow{IB}=-\frac{c}{b}\left(\overrightarrow{DI}+\overrightarrow{IC}\right)\Leftrightarrow\left(1+\frac{c}{b}\right)\overrightarrow{DI}=-\overrightarrow{IB}-\frac{c}{b}\overrightarrow{IC}\Leftrightarrow\overrightarrow{ID}=\frac{b}{b+c}\overrightarrow{IB}+\frac{c}{b+c}\overrightarrow{IC}\)