The respective intersections E, F, G and H of the angle bisectors of angles A, B, C and D of a cyclic quadrilateral ABCD, with the circumcircle, form a rectangle.
(Drag any of the red vertices/points).
1) Can you explain (prove) why EFGH is a rectangle?
2) Is the result also true if ABCD becomes a crossed quadrilateral? Unfortunately the sketch above does not allow you to transform ABCD into a crossed quadrilateral. So you'll need to construct your own dynamic geometry sketch to explore this case.
3) Can you apply or generalize the result to a cyclic hexagon? Click on the green 'Generalization to Cyclic Hexagon' button in the sketch above.
If stuck, read my paper An interesting cyclic quadrilateral result.
Michael de Villiers, created 16 May 2011; modified 25 Feb 2019.