Whenever I draw it neither point lies on the circle

This took longer than I'd like. Nice-ish problem. First suppose that $BEDI$ is cyclic. Then by radical axis on $(BEDI)$, $\Omega$ it follows that $IDFC$ is cyclic. Let $I_A$ be the $A$-excenter, note that $(BII_AC)$ is cyclic. Then note that $E - D - F$ are collinear since \[ \measuredangle EDI = \measuredangle EBI = \measuredangle I_ABI = \measuredangle I_ACI = \measuredangle FCI = FDI. \]As such, $\measuredangle GED = \measuredangle AEF = \measuredangle ACF = \measuredangle I_ACB = \measuredangle FCD = \measuredangle FID$ which finishes. Now consider the case where $BEGI$ is cyclic. Define $D' = (BGEI) \cap (ICF)$. Then by radaxis on $(BGEI)$, $\Omega$, and $(ICF)$, it follows that $D'$ is on $AI$. Then we can angle chase $E-D'-F$ again. Then by a similar angle chase we get $\measuredangle GID = \measuredangle FID = \measuredangle GED = \measuredangle FCB$.

Just note that both conditions are equivalent to $I$ being orthocentre of $\triangle AEF$ thru simple angle chase.