[tex]\text{ Since the triangles are similar, we will use the Thales' theorem!}[/tex][tex]\begin{gathered} \text{ By thales' theorem, we have that } \\ \\ \frac{AB}{AC}=\frac{DE}{DF} \\ \frac{2}{4}=\frac{1.2}{DF} \\ \frac{DF}{4}=\frac{1.2}{2} \\ \frac{DF}{4}=0.6 \\ DF=0.6\cdot4 \\ DF=2.4 \end{gathered}[/tex][tex]\begin{gathered} \text{And again, by thales' theorem, we have} \\ \frac{AB}{BC}=\frac{DE}{EF} \\ \frac{2}{3}=\frac{1.2}{EF} \\ \frac{EF}{3}=\frac{1.2}{2} \\ \frac{EF}{3}=0.6 \\ EF=0.6\cdot3 \\ EF=1.8 \end{gathered}[/tex]
