$ABCD$ is a square of center $O$. On the sides $DC$ and $AD$ the equilateral triangles DAF and DCE have been constructed. Decide if the area of the $EDF$ triangle is greater, less or equal to the area of the $DOC$ triangle.
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To determine whether the area of triangle \( \triangle EDF \) is greater, less, or equal to the area of triangle \( \triangle DOC \), we will calculate the areas of both triangles and compare them.

1. **Calculate the area of triangle \( \triangle DOC \):**
   - Since \(ABCD\) is a square with center \(O\), \(O\) is the midpoint of both diagonals \(AC\) and \(BD\).
   - The diagonals of a square bisect each other at right angles and are equal in length.
   - Let the side length of the square be \(a\). The length of the diagonal \(AC\) (or \(BD\)) is \(a\sqrt{2}\).
   - The area of triangle \( \triangle DOC \) is half the area of triangle \( \triangle AOC \), which is half the area of the square.
   - Therefore, the area of \( \triangle DOC \) is:
     \[
     S_{DOC} = \frac{1}{2} \times \frac{1}{2} \times a^2 = \frac{a^2}{4}
     \]

2. **Calculate the area of triangle \( \triangle EDF \):**
   - Triangles \( \triangle DAF \) and \( \triangle DCE \) are equilateral triangles constructed on sides \(AD\) and \(DC\) of the square \(ABCD\).
   - The side length of these equilateral triangles is equal to the side length of the square, \(a\).
   - The area of an equilateral triangle with side length \(a\) is given by:
     \[
     S_{\text{equilateral}} = \frac{\sqrt{3}}{4} a^2
     \]
   - To find the area of \( \triangle EDF \), we need to use the fact that \( \triangle EDF \) is formed by joining the vertices of the equilateral triangles.
   - The length of \(ED\) and \(FD\) is \(a\), and the angle \( \angle EDF \) is \(120^\circ\) (since it is the external angle of the equilateral triangles).
   - Using the formula for the area of a triangle with two sides and the included angle:
     \[
     S_{EDF} = \frac{1}{2} \times ED \times FD \times \sin(120^\circ) = \frac{1}{2} \times a \times a \times \sin(120^\circ)
     \]
   - Since \( \sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
     \[
     S_{EDF} = \frac{1}{2} \times a^2 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} a^2
     \]

3. **Compare the areas:**
   - The area of \( \triangle DOC \) is \( \frac{a^2}{4} \).
   - The area of \( \triangle EDF \) is \( \frac{\sqrt{3}}{4} a^2 \).

Since \( \sqrt{3} \approx 1.732 \), it is clear that \( \frac{\sqrt{3}}{4} a^2 \) is greater than \( \frac{a^2}{4} \).

Therefore, the area of \( \triangle EDF \) is greater than the area of \( \triangle DOC \).

The final answer is \( \boxed{ \text{The area of } \triangle EDF \text{ is greater than the area of } \triangle DOC } \).