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?
Let (area of) $\triangle ECF = x$.

Then $\triangle FCD = \triangle ECF = x$ (same base, same height - I'll just call this SBSH for brevity).

So $\triangle ECD = 2x$ and $\triangle EBC = \triangle ECD = 2x$ (SBSH)

Hence $\triangle EBD = 4x$ and $\triangle AED = \triangle EBD = 4x$ (SBSH)

Therefore $\triangle ABD = 8x = 8\triangle CEF \ \ (QED)$
ED is the median of △(ABD) and therefore △(AED)=△(BED)=1/2 x △(ABD). 
EC is the median of △(BED) and therefore △(EBC)=△(ECD)=1/2 x △(BED). 
CF is the median of △(ECD) and therefore △(CEF)=△(CDF)=1/2 x △(ECD). 

From the three equations above △(CEF)=1/8 x △(ABD)
or 8 x △(CEF)=△(ABD)