[tex]\it BD=0,(6)BC=\dfrac{\ 6^{(3}}{9}BC=\dfrac{2}{3}BC\\ \\ \\ Th. catetei \Rightarrow AB^2=BD\cdot BC \Rightarrow (3\sqrt6)^2=\dfrac{2}{3}BC\cdot BC \Rightarrow 54=\dfrac{2}{3}\cdot BC^2\Big|_{\cdot\dfrac{3}{2}} \Rightarrow \\ \\ \\ \Rightarrow BC^2=81=9^2 \Rightarrow BC=9\ cm[/tex]
[tex]\it Th.\ Pitagora \Rightarrow AC^2 =BC^2-AB^2=81-54=27=9\cdot3 \Rightarrow AC=3\sqrt3\ cm[/tex]
