Denote midpoints of $AC,BC,CD,EF$ by $K,L,M,N$ respectively. Composition of spiral similarities $\overrightarrow{AB}\mapsto \overrightarrow{BC}\mapsto \overrightarrow{CA}$ also sends $\overrightarrow{DB}\mapsto \overrightarrow{EC}\mapsto \overrightarrow{FA}$, so observe that vectors $\overrightarrow{DB},\overrightarrow{EC},\overrightarrow{FA}$ form a triangle (which is similar to $\triangle ABC$). Hence $\overrightarrow{KN}=\frac{1}{2}(\overrightarrow{AE}+\overrightarrow{CF})=\frac{1}{2}(\overrightarrow{AF}+\overrightarrow{CE})=\frac{1}{2}\overrightarrow{DB}=\overrightarrow{ML}$, i.e. $KMLN$ is a parallelogram. Remaining part of problem is trivial since triangles $ADB,KML$ are homothetic with center $C$.