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.
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.
ReplyDelete