EF is parallel to BC (Midpoint theorem)
In ΔCED and ΔEDF,
∠FED =∠EDC (Alternate angles)
EF = EF (common side)
∠CED =∠EDF (Alternate angles)
Hence, ΔCED and ΔEDF are congruent by ASA condition.

Area of ΔCED = Area of ΔEDF.
Therefore, EFDC is a parallelogram
(Diagonals divide a parallelogram into two congruent triangles)
Similarly, FD is parallel to AC and ED is parallel to AB. So, it can be proved that AEFD and EFDB are also parallelograms.
Thus, Area of ΔCED = Area of ΔEDF = Area of ΔAEF = Area of ΔFDB
$\therefore \text{Area of}\left(DEF\right)=\frac{1}{4}\text{Area of}\left(\Delta ABC\right)$