What Might This Be About?
8 October 2014, Created with GeoGebra
Points $D,$ $E,$ $F,$ $G,$ $Q$ are on the circumcircle $(ABC)$ such that $\angle BAF=\angle FAC,$ $\angle BAD=\angle QAC,$ $\angle ABG=\angle GBC,$ and $AD\parallel BE.$
Then $\angle EBG=\angle GBQ.$
The proof is by simple angle (or, perhaps, more appropriately, arc) chasing.
Since $AD\parallel BE,$ $BD=AE.$ Now, in terms of arcs, it is given that $BD=CQ$ and $AG=CG.$ From here, $EG=GQ$ and, therefore $\angle EBG=\angle GBQ.$
The lemma could be reformulated as follows:
Let $Q\in (ABC)$ and $AD$ is the isogonal conjugate of $AQ$ in $\angle BAC.$ Let $BE\parallel AD.$ Then $BE$ is the isogonal conjugate of $BQ$ in $\angle ABC.$
Replacing $B$ with $C,$ drawing a parallel to $AD$ through $C$ and its isogonal conjugate with respect to $\angle ACB,$ we see that the latter too passes through $Q.$ For this reason, $Q$ could be said to be the isogonal conjugate of the point at infinity corresponding to the direction of $AD.$