Join the points D and E as shown in the diagram below:




Now DEAF is a paralleogram and DF is a diagonal. Since lined passing through midpoint of two sides of a triangle is paralle to the third side. So DE is paralle to AC and DF is parallel to BC and so on.

So Area of triangle DAF (shaded triangle on the left)=area of the triangle DEF. A diagonal of a parallelogram divides it into two triangles of equal area.

Similarly in parallelogram DECF the area of triangle DEF= ECF. Same logic as above. Hence the two shaded areas are equal and also equal to the unshaded triangle DEF.

Now the final paralleogram BDFE we have triangle BDE and triangle FDE. They are also equal in area as DE is diagonal of paralleogram BDEF.

Hence all 4 areas of the above triangle are equal. Thus the shaded area is equ\l to the unshaded area.

Hence option C is the best fit!