Simple Similarity

Draw the projection of $E'$ of $E$ in the segment $AD$ and the segment $EE'$ and intersect to $BD$ in $W$ . Then $\triangle ABG \sim \triangle GEW \sim \triangle GDF$ and $\triangle ABE \sim \triangle ECF$ then $\frac{AG}{GE}=\frac{6}{4}=\frac{AB}{EW}$ therefore $EW=\frac{2AB}{3}$ . If $CF=k$ then $DF=AB+k$ and $\frac{DF}{EW}=\frac{3(AB+k)}{2AB}$ . But as $\triangle GEW \sim \triangle GDF$ then $\frac{DF}{EW}=\frac{4+EF}{4}$ , therefore $\frac{3(AB+k)}{2AB}=\frac{4+EF}{4}$ then $6(AB+k)=AB(4+EF)\Longrightarrow 6AB+6k=4AB+EF\cdot AB \Longrightarrow 2AB=EF\cdot AB -6k$ . Given that $\frac{EF}{10}=\frac{k}{AB}\Longrightarrow EF \cdot AB=10k$ and substituting it in $2AB=EF\cdot AB -6k$ therefore $2AB=10k-6k=4k \therefore AB=2K$ . Then $\frac{AB}{k}=\frac{10}{EF}=2 \Longrightarrow EF=5$