giúp em với ạ
\(\widehat{ABD}=60^0\Rightarrow\Delta ABD\) đều \(\Rightarrow BD=a\) ; \(AO=\dfrac{a\sqrt{3}}{2}\) (trung tuyến tam giác đều)
\(2\overrightarrow{ID}+2\overrightarrow{DC}+\overrightarrow{ID}=\overrightarrow{0}\Rightarrow\overrightarrow{DI}=\dfrac{2}{3}\overrightarrow{DC}=\dfrac{2}{3}\overrightarrow{AB}\)
\(\overrightarrow{AO}.\overrightarrow{BI}=\overrightarrow{AO}.\left(\overrightarrow{BD}+\overrightarrow{DI}\right)=\overrightarrow{AO}.\overrightarrow{BD}+\overrightarrow{AO}.\overrightarrow{DI}\)
\(=\overrightarrow{AO}.\overrightarrow{DI}=\dfrac{2}{3}\overrightarrow{AO}.\overrightarrow{AB}=\dfrac{2}{3}AO.AB.cos30^0\)
\(=\dfrac{2}{3}.\dfrac{a\sqrt{3}}{2}.a.\dfrac{\sqrt{3}}{2}=\dfrac{a^2}{2}\)
