Olympiad Geometry Problem #97: Incircle, Tangency Points, Concurrence

Here is an enlightening problem with a simple configuration but which is still tricky to solve. It was posted on the Art of Problem Solving Forum last year, but I don't know the original source. Enjoy! Link below.

[quite_unknown_]

Construct perpendicular of EF through E on BC, call it Y. Construct Perpendicular from F on BC, call it Z. Then showing QRYZ is cyclic finishes because of Radical center is point of concurrency of 3 radical axes (FE, QR, YZ=BC).

[lorincmate]

I have shown this problem to one of my classmates and he found another easy solution. Project the whole figure while the incircle remains a circle and the intersection point of EF and BC became an ideal point (you can do this with stereographic projection), so BC will be parallel to EF). Thus you only have to solve the problem when ABC is an isosceles triangle with AB=AC.
And for that case, it is easy to see that RQ will be parallel to BC as well by symmetry.

[salahelaidi-gm]

did anyone notice thT BR QC and XD are concurrent

[김동욱-wtz]

Nice problem. I really wonder where this problem comes from.