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?

Hint: The result follows almost directly from a useful Lemma proved on p. 190 of my book 'Some Adventures in Euclidean Geometry' which is available as downloadable PDF or in printed bookform from Some Adventures in Euclidean Geometry.

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 linked button 'Generalization to Cyclic Hexagon' in the sketch above.