Let $T = EH \cap FI$. Let $K$ and $L$ be the feet of the perpendiculars from $A$ to $FI$ and from $B$ to $EH$, respectively. Then $K$ and $L$ are on the circumcircle $\Gamma$ of $ABIH$. Note that $K$ is also on the circumcircle of $ACFG$ and $L$ is also on the circumcircle of $BDEC$. Let $CK$ extended intersect $\Gamma$ in $M$, and let $CL$ extended intersect $\Gamma$ in $N$. Then the lines $HM$ and $IN$ intersect in $S$. $[\angle EBC = \angle ELC = \angle HLN = \angle HIN]$. But $\angle HIS = \angle EBC$, so $S$ is on $IN$. Similarly, $S$ is on $HM$.

Consider the self-crossing hexagon $IKMHLN$ inscribed in the circle $\Gamma$. By Pascal's Theorem, the points $IK \cap HL = T$, $KM \cap LN = C$, and $MH \cap NI = S$ are collinear. This shows that $T$ is a point that belongs to all three lines $FI$, $EH$ and $CS$, i.e., these lines are concurrent.