Olympiad Geometry Problem #18: Altitudes, Circumcircle, Equal Segments

This is the first problem I have done from an IMO Shortlist. It was the easiest geometry problem on the 2010 IMO Shortlist, and I hope to do more such problems. See the link below, hope you all enjoy!

[mathlegendno]

My motivation for constructing G was to reflect Q over AB, due to <AQB=<ACB and both meeting at AB, so they have the same size circumcircle, followed by the reflection forcing the 2 circles together. Then I noticed that it seemed to lie on EF and taking a leap of faith, I did a phantom point sort of thing where it ended to exactly your construction of G, namely EF meeting the circumcircle, followed by angle chasing leading to congruency.

[mathlegendno]

An alternate way to show the perpendicular lines is <AEF+<OAE=90 which is trivial

[alextkach]

Thank you, maestro. I really enjoyed it

[helo]

can you please explain how to get AF is angle bisector of QFE?