G - centrul de greutate ⇒ GD = BG: 2 = 2,5 cm
BD = 5+2,5=7,5cm
AC=2· BD = 2· 7,5 = 15cm
[tex]\it \Delta BCA-dreptunghic,\ \hat B=90^o\ \stackrel{TP}{\Longrightarrow}\ BC^2=AC^2-AB^2=15^2-9^2=\\ \\ =(15-9)(15+9)=6\cdot24=6\cdot6\cdot4=6^2\cdot2^2=12^2 \Rightarrow BC=12cm[/tex]
[tex]\it GF||AC \Rightarrow GF||DC\ \stackrel{T.Thales}{\Longrightarrow} \dfrac{BF}{FC}=\dfrac{BG}{GD} \Rightarrow \dfrac{BF}{FC}=\dfrac{5}{2,5} \Rightarrow \\ \\ \\ \Rightarrow \dfrac{BF}{FC+BF}=\dfrac{5}{2,5+5} \Rightarrow \dfrac{BF}{BC}=\dfrac{^{2)}5}{\ 7,5} \Rightarrow \dfrac{BF}{12}=\dfrac{10}{15} \Rightarrow BF=\dfrac{12\cdot10}{15}=8[/tex]
