\(\widehat{B}+\widehat{C}=60^0\Rightarrow180^0-\widehat{A}=60^0\Rightarrow\widehat{A}=120^0\)
AD là phân giác \(\Rightarrow\widehat{BAD}=\widehat{CAD}=60^0\)
\(S_{ABD}=\dfrac{1}{2}AB.AD.\sin\widehat{BAD}=\dfrac{1}{2}AB.AD.\sin30^0=\dfrac{1}{4}AB.AD\)
\(S_{ABD}=\dfrac{1}{2}AC.AD.\sin\widehat{CAD}=\dfrac{1}{2}AC.AD.\sin30^0=\dfrac{1}{4}AC.AD\)
\(\Rightarrow S_{ABD}+S_{ACD}=\dfrac{1}{4}AD\left(AB+AC\right)\)
\(\Rightarrow\dfrac{1}{4}AD.\left(3+6\right)=S_{ABC}\)
\(\Rightarrow\dfrac{9}{4}AD=\dfrac{1}{2}AB.AC.\sin\widehat{BAC}\)
\(\Rightarrow\dfrac{9}{2}AD=3.6.\sin60^0\)
\(\Rightarrow\dfrac{9}{2}AD=3.6.\dfrac{\sqrt{3}}{2}\)
\(\Rightarrow AD=2\sqrt{3}\)
