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.
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Michael de Villiers, created 16 May 2011; modified 25 Feb 2019.