In $\triangle ABC$, $AB=AC=28$ and $BC=20$.  Points $D,E,$ and $F$ are on sides $\overline{AB}$, $\overline{BC}$, and $\overline{AC}$, respectively, such that $\overline{DE}$ and $\overline{EF}$ are parallel to $\overline{AC}$ and $\overline{AB}$, respectively.  What is the perimeter of parallelogram $ADEF$?

[asy]
size(180);
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real r=5/7;
pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r);
pair bottom=(C.x+(D.x-A.x),C.y+(D.y-A.y));
pair E=extension(D,bottom,B,C);
pair top=(E.x+D.x,E.y+D.y);
pair F=extension(E,top,A,C);
draw(A--B--C--cycle^^D--E--F);
dot(A^^B^^C^^D^^E^^F);
label("$A$",A,NW);
label("$B$",B,SW);
label("$C$",C,SE);
label("$D$",D,W);
label("$E$",E,S);
label("$F$",F,dir(0));
[/asy]

$\textbf{(A) }48\qquad
\textbf{(B) }52\qquad
\textbf{(C) }56\qquad
\textbf{(D) }60\qquad
\textbf{(E) }72\qquad$
1. Given that $\triangle ABC$ is isosceles with $AB = AC = 28$ and $BC = 20$. Points $D, E,$ and $F$ are on sides $\overline{AB}$, $\overline{BC}$, and $\overline{AC}$, respectively, such that $\overline{DE}$ and $\overline{EF}$ are parallel to $\overline{AC}$ and $\overline{AB}$, respectively. This implies that $ADEF$ is a parallelogram.

2. Since $\overline{DE} \parallel \overline{AC}$ and $\overline{EF} \parallel \overline{AB}$, triangles $\triangle ADE$ and $\triangle AEF$ are similar to $\triangle ABC$ by AA similarity criterion.

3. Let $k$ be the ratio of similarity. Then $AD = k \cdot AB$ and $AF = k \cdot AC$. Since $AB = AC = 28$, we have:
   \[
   AD = k \cdot 28 \quad \text{and} \quad AF = k \cdot 28
   \]

4. Since $D$ and $F$ are points on $AB$ and $AC$ respectively, and $DE \parallel AC$ and $EF \parallel AB$, $DE$ and $EF$ are also scaled by the same factor $k$. Therefore:
   \[
   DE = k \cdot BC = k \cdot 20 \quad \text{and} \quad EF = k \cdot 28
   \]

5. The perimeter of parallelogram $ADEF$ is the sum of the lengths of its sides:
   \[
   \text{Perimeter} = AD + DE + EF + FA
   \]

6. Substituting the expressions for $AD$, $DE$, $EF$, and $FA$:
   \[
   \text{Perimeter} = k \cdot 28 + k \cdot 20 + k \cdot 28 + k \cdot 28
   \]

7. Simplifying the expression:
   \[
   \text{Perimeter} = k \cdot (28 + 20 + 28 + 28) = k \cdot 104
   \]

8. Since $k$ is the ratio of similarity and $D$ and $F$ are points on $AB$ and $AC$ respectively, $k$ must be such that $AD + DE = AB$ and $AF + FE = AC$. Given that $AB = AC = 28$, we have:
   \[
   AD + DE = 28 \quad \text{and} \quad AF + FE = 28
   \]

9. Therefore, $k = 1$ because $AD + DE = 28$ and $AF + FE = 28$ implies that $AD = DE = 14$ and $AF = FE = 14$.

10. Substituting $k = 1$ into the perimeter expression:
    \[
    \text{Perimeter} = 1 \cdot 104 = 104
    \]

However, this contradicts the initial problem statement. Re-evaluating the problem, we realize that the correct approach is to consider the lengths of $AD$ and $AF$ directly as half of $AB$ and $AC$ respectively, since $D$ and $F$ are midpoints.

11. Therefore, the correct perimeter calculation is:
    \[
    AD + DE + AF + FE = 14 + 14 + 14 + 14 = 56
    \]

The final answer is $\boxed{56}$