[tex]\it Not\breve am\ BD=x,\ \ CD=4x,\ \ BC=5x\\ \\ T.\ catetei \Rightarrow AB^2=BD\cdot BC \Rightarrow (5\sqrt2)^2=x\cdot5x \Rightarrow 50=5x^2 \Rightarrow \\ \\ \Rightarrow x^2=10 \Rightarrow x=\sqrt{10} \Rightarrow BC=5\sqrt{10}\ cm[/tex]
[tex]\it \Delta ABC-dreptunghic\ \stackrel{T.P.}{\Longrightarrow} AC^2=BC^2-AB^2=(5\sqrt10)^2-(5\sqrt2)^2=\\ \\ \\ =250-50=200 \Rightarrow AC=\sqrt{200}=\sqrt{100\cdot2}=10\sqrt2\ cm[/tex]
