\(Ta\text{ }có\text{ }:3\overrightarrow{IA}+2\overrightarrow{IC}-2\overrightarrow{ID}=0\\ \Rightarrow3\overrightarrow{IA}+2\left(\overrightarrow{IC}-\overrightarrow{ID}\right)=0\\ \Rightarrow3\overrightarrow{IA}=-2\left(\overrightarrow{IC}-\overrightarrow{ID}\right)\\ \Rightarrow3\overrightarrow{IA}=-2\overrightarrow{DC}=2\overrightarrow{BA}\\ \Rightarrow\overrightarrow{IA}=\frac{2}{3}\overrightarrow{BA}\\ \Rightarrow I;B;A\text{ thẳng hàng},I\text{ nằm giữa }A;B\left(\frac{2}{3}>0;IA< BA\right)\)
\(\text{Lại có }:\overrightarrow{JA}-2\overrightarrow{JB}+2\overrightarrow{JC}=0\\ \Rightarrow\overrightarrow{JA}=2\left(\overrightarrow{JB}-\overrightarrow{JC}\right)\\ \Rightarrow\overrightarrow{JA}=2\overrightarrow{CB}=2\overrightarrow{DA}\\ \Rightarrow J;D;A\text{ thẳng hàng},D\text{ nằm giữa }J;A\left(2>0;JA>DA\right)\)
\(\text{Lại có }:O\text{ là trung điểm }AC;BD\left(\text{Tính chất hình bình hành}\right)\\ \Rightarrow\overrightarrow{JO}=\overrightarrow{JA}+\overrightarrow{AO}=-2\overrightarrow{AD}+\frac{1}{2}\left(\overrightarrow{AD}+\overrightarrow{AB}\right)\\ =-2\overrightarrow{AD}+\frac{1}{2}\overrightarrow{AD}+\frac{1}{2}\overrightarrow{AB}=-\frac{3}{2}\overrightarrow{AD}+\frac{1}{2}\overrightarrow{AB}\)
\(\text{Mặt khác }:\overrightarrow{JI}=\overrightarrow{JA}+\overrightarrow{AI}=-2\overrightarrow{AD}+\frac{2}{3}\overrightarrow{AB}=\frac{4}{3}\left(-\frac{3}{2}\overrightarrow{AD}+\frac{1}{2}\overrightarrow{AB}\right)\\ \Rightarrow\overrightarrow{JI}=\frac{4}{3}\overrightarrow{JO}\\ \Rightarrow J;I;O\text{ thẳng hàng}\)
