[tex]\displaystyle\\\text{AM este mediana ipotenuzei.} \\AM=5cm\\\\De\!finitie:\text{Mediana ipotenuzei are lungimea egala cu jumatate}\\ \text{din lungimea ipotenuzei.}\\\\AM=\frac{BC}{2}\\\\\implies~BC=2\times AM=2\times5=\boxed{10~cm}\\\\AD~\text{este inaltimea perpendiculara pe ipotenuza.}\\CD=6,4~cm\implies~BD=BC-CD=10-6,4=\boxed{3,6~cm}\\\\\text{Aplicam teorema inaltimii:}\\\\AD=\sqrt{BD\times CD}=\sqrt{3,6\times 6,4}=\sqrt{23,04}=\boxed{\bf4,8~cm}[/tex]
