180 Days of Ideas for Discussion in Math Class.
(as of 9July2014, we're in overtime!)
Old Exam Questions #2 : My solution in a few days...
— solve my maths (@solvemymaths) May 24, 2015
The congruence of the triangles is trivial, then one of them turned 90 degrees relative to the other one, which means that GFE is a right angle. So the point F sits on the half circle for which GE is a diameter, the same is true for the point A. Then the radius of the circumcircle is half of the diameter 1/2(SQRT(9^2 + 6^2)). The whole problem is not exactly trivial, but pretty close to it.