smc

Note that because $\overline{DE}$ and $\overline{EF}$ are parallel to the sides of $△ABC$, the internal triangles $△BDE$ and $△EFC$ are similar to $△ABC$, and are therefore also isosceles triangles. It follows that $BD=DE$. Thus, $AD+DE=AD+DB=AB=28$. Since opposite sides of parallelograms are equal, the perimeter is $2\ast (AD+DE)=\boxed{56}$.