Let $P$ be the point symmetric to point $C$ with respect to point $O$ (Fig. 1). Since triangle $ABC$ is acute,
points $A$, $P$, $B$, and $C$ lie on a circle with center $O$ in this exact order.
om57_1r_img_1.jpg
The areas of triangles $COE$ and $POE$ are equal, as these triangles have a common height dropped from vertex $E$, and segments $CO$ and $OP$ are of equal length. Similarly, the areas of triangles $COF$ and $POF$ are equal. Therefore, we obtain $[EOFC] = \frac{1}{2} \cdot [EPFC]$, where the symbol $[\mathcal{F}]$ denotes the area of figure $\mathcal{F}$.
To complete the solution, we need to prove that $[EPFC] = [ABC]$.
Segment $CP$ is the diameter of the circumcircle of triangle $ABC$, so lines $AP$ and $AC$ are perpendicular. Line $DE$ is
perpendicular to line $AC$, so it is parallel to line $AP$. Therefore, the areas of triangles $DEA$ and $DEP$ are equal. Similarly,
the areas of triangles $DFB$ and $DFP$ are equal. Thus,

which completes the solution of the problem.