The following solution is entirely the work of Sarah from The Mount School, York (Yr 11). Nice work Sarah!
I've drawn the triangle and two of the circumcircles and I'll refer to my drawing :
Now the two circles meet in the two points, D and K.
Now angles DAE and EKD add up to 180 degrees, because DAEK is a cyclic quadrilateral.
Similarly, FCD and FKD add up to 180 degrees.
Calling the original angles of the triangle A, B and C, angle EKF is:
(180 - A) + (180 - C) = 360 - (A + C)
So, angle EKF + FBE = 360 - (A + C) + B but A + C = 180 - B, so
EKF + FBE = 360 - (180 - B) + B = 180
|Now if I draw the circumcircle through F, B and E it will have
to pass through K, as opposite angles of a cyclic quadrilateral add
up to 180 degrees.
If K lies within the circle or outside the circle, then this makes the above property of cyclic quadrilaterals a lie ! I noticed, by the way that all three circles pass through the same point.
We've got dynamic geometry software and we were able to experiment with that, once we had it drawn (with a little help from our teacher !)