[tex]\it BM=MC \Rightarrow M\ -\ mijlocul\ lui\ BC \Rightarrow BM=MC=\dfrac{BC}{2}\\ \\ \\ AM=BM=MC \Rightarrow AM=\dfrac{BC}{2} \Rightarrow \Delta ABC-dreptunghic,\ \widehat A=90^o\\ \\ \\ tgC=\dfrac{AB}{AC} \Rightarrow tg30^o=\dfrac{8}{AC} \Rightarrow \dfrac{\sqrt3}{3}=\dfrac{8}{AC} \Rightarrow\\ \\ \\ \Rightarrow AC=\dfrac{^{\sqrt3)}3\cdot8}{\ \ \sqrt3}=\dfrac{\not3\cdot8\sqrt3}{\not3}=8\sqrt3[/tex]
