• You see cyclics. You see lengths. You use POWER OF POINT.
• Beware of where you have to go. Your goal is to reach $GE.BE = FE.CE$ at any cost.

$AC$ and $BD$ intersect at E. So we use the power of point. 

We have,

$BE.DE = CE.AE$

$\Rightarrow \frac{DE}{CE} = \frac{AE}{BE} $

$ \Rightarrow \frac{DG + GE}{CE}=\frac{AF + FE}{BE}$

$\Rightarrow \frac {CE + GE}{CE}=\frac{BE + FE}{BE}$

$\Rightarrow 1 + \frac{GE}{CE}= 1 + \frac{FE}{BE}$

$\Rightarrow \frac{GE}{CE}= \frac{FE}{BE}$

$\Rightarrow GE.BE = FE.CE$

So, by power of point we have that $BFGC$ has to be cyclic.