[tex]Aplicam ~teorema~bisectoarei~in ~\Delta ABD:\\\\
\frac{AB}{AD} = \frac{BE}{ED} = \frac{1}{2} \\\\
BE+ \frac{ED}{ED} = \frac{3}{2}= \frac{BD}{ED}\\\\
\Delta FED \sim \Delta CBD\\\\
\frac{ED}{BD} = \frac{EF}{BC}= \frac{2}{3} \\\\
EF= \frac{2}{3} *12\\\\
\bold{EF=8}[/tex]
