Problem based on area of triangle In the figure, E,C and F are the mid points of AB, BD and ED respectively. Prove that: $8\triangle CEF=\triangle ABD$

From the given,
$ED$ is the median of $\triangle ABD$
So, $\triangle AED=\triangle BED$
Also, by mid point Theorem 
$EC||AD$ and $CF||AB$.
Now, what should I do next?

Well $\triangle$AED=$\frac{1}{2}\triangle ABD$.
So the problem reduces to showing that
$$\triangle CEF = \frac{1}{4} \triangle BED$$
Since F is the midpoint of ED, we have by similarity that:
$$|EB| = 2|FB|$$
in other words: 
$$\triangle CFD = \frac{1}{4} \triangle BED$$
Likewise, also by similarity,
$$\triangle BEC = \frac{1}{4} \triangle ABD = \frac{1}{2} \triangle BED$$
Which implies immediately that $\triangle CEF = \frac{1}{4} \triangle BED$, and hence we are done.
Note that the factors of 1/4 come from the scaling of both the base and the height by 1/2 due to similarity, hence the area is scaled by $(1/2)^2 = 1/4$ due to similarity.
The scaling factor of 1/2 for the bases and heights follows from the fact that the relevant line segments are medians and the midpoint theorem and the subsequent similarity.
