Olympiad geometry: Prove equal segments 
$A$, $B$, $C$, $D$, $E$ and $F$ are six concyclic points. $AC$, $BD$ and $EF$ are concurrent at $G$. Line $EF$ intersects $\odot(ABG)$ and $\odot(CDG)$ at $I$ and $J$ respectively. Show that $IE=FJ$.


My idea was the trigonometry version of Ptolemy theorem. Let $\angle AGE=\angle CGF=\beta$ and $\angle BGE=\angle DGF=\alpha$. I got
$$\begin{aligned}GI=\frac{AG\sin\alpha+BG\sin\beta}{\sin(\alpha+\beta)};\\GJ=\frac{CG\sin\alpha+DG\sin\beta}{\sin(\alpha+\beta)}.\end{aligned}$$
$EI=FJ$ is equivalent to $GI-GJ=GE-GF$. So we need only prove that $$GE-GF=\frac{AG\sin\alpha+BG\sin\beta-CG\sin\alpha-DG\sin\beta}{\sin(\alpha+\beta)}.$$This is not any much easier, although it becomes independent of $I$ and $J$.

Hint:
Join J and I to center of big circle O,they intersct this circle at M and N respectively. Extend  JO and IO to meet the circle at K and L respectively. You have to show that triangle OIJ is isosceles. In this case we have:
$JF\times JE=JM\times JK=IN\times IL=IE\times IF$
Since:
$IF=IE+Ef$
and:
$JE=JF+EF$
We conclude that:
$IE=JF$
Note that this true if DC is diameter of circle DJC and AB is the diameter of circle ABI, this may helps.
