[tex]\it \Delta ABC-echilateral,\ BB_1-bisectoare \Rightarrow\ BB_1-median\breve a\\ \\ AA_1\cap BB_1=\{M\} \Rightarrow M\ -\ centrul\ de\ greutate\ al\ triunghiului\\ \\ \mathcal{A}_{A_1BM}=\dfrac{1}{3}\cdotr\mathcal{A}_{A_1AB}=\dfrac{1}{3}\cdot\dfrac{1}{2}\cdot\mathcal{A}_{ABC}=\dfrac{1}{6}\cdot216=36\ cm^2[/tex]
