[tex]a)Aplicam \ teorema \ bisectoarei \ in \ \Delta ABC:\frac{AD}{DC}=\frac{BA}{BC}=\frac{8}{10}=\frac{4}{5}[/tex]
[tex]b)AM \ mediana=\textgreater BM=MC=\frac{BC}{2}=5cm[/tex]
[tex]AM \ mediana=\textgreater AM=\frac{BC}{2}=5cm[/tex]
[tex]AM=AN+NM=\textgreater NM=AM-AN[/tex]
[tex]Aplicam \ teorema \ bisectoarei \ in \ \Delta BAM:\frac{AN}{NM}=\frac{BA}{BM}[/tex]
[tex]\frac{AN}{AM-AN}=\frac{BA}{BM}[/tex]
[tex]BA(AM-AN)=AN*BM[/tex]
[tex]BA*AM-BA*AN=AN*BM[/tex]
[tex]BA*AM=AN*BM+BA*AN[/tex]
[tex]BA*AM=AN(BM+BA)[/tex]
[tex]AN=\frac{BA*AM}{BM+BA}=\frac{40}{13}cm[/tex]
[tex]\frac{AN}{AM}=\frac{\frac{40}{13}}{5}=\frac{40}{65}=\frac{8}{13}[/tex]
