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.