In a $$\triangle ABC,D,E$$ and $$F$$ are the mid-points of the sides AB, BC and CA respectively. Then the ratio of the area of a $$\triangle DEF$$ and the area of a $$\triangle ABC$$ is:

We are given :

Now In triangle ABC 
using the mid point theorem 
we get DE || BC and DE = $$\frac{1}{2}\left(BC\right)$$
Now ADE is similar to ABC 
So Area of ADE : Area of ABC = (DE/BC)^2 = 1:4
Similarly Area of BDF :Area of ABC = 1:4
and Area of CEF : Area of ABC = 1:4
Now we can say if area of ABC = 4A
then area of ADE = area of BFD = area of CEF = a
Therefore area of DEF = 4a-3a =a
so area of DEF : area of ABC = 1:4