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 $J$.

I would first consider a inversion with center $G$ and radius 1. Denoting an inverted point $P$ with $P'$, we get that $A',B',C',D',E'$ and $F'$ are still concyclic with lines $A'C'$, $B'D'$ and $E'F'$ concurring at $G$. More importantly, we have $I'=A'B' \cap E'F'$ and $J'=C'D' \cap E'F'$.
Motivated by the projective proof of the Butterfly theorem, we get
$$(E',G;I',F')=(E',D';C',F')=(E',J';G,F')$$
$$\Rightarrow \frac{E'I'}{ \frac{GI'}{GF'}}= \frac{E'G}{ \frac{J'G}{J'F'}}.$$
On the other hand,$$IE=IG-EG=\frac{1}{I'G}-\frac{1}{E'G}= \frac{E'I'}{I'G \cdot E'G}$$ and
$$JF=JG-FG=\frac{1}{J'G}-\frac{1}{F'G}= \frac{F'J'}{J'G \cdot F'G}$$ and now using the last result we are done.
There probably exist a nicer proof without inversion and projective geometry, but this is the first solution that came to my mind.
