The perpendicular bisectors of the sides of a quadrilateral circumscribed around a circle (a tangential quad) form another quadrilateral circumscribed around a circle (another tangential quad). Alternatively, but equivalently formulated, the circumcentres of triangles ABC, BCD, CDA and DAB of a quadrilateral ABCD circumscribed around a circle form another quadrilateral circumscribed around a circle.
Perpendicular-Bisectors (or Circumcentres) of Circumscribed Quadrilateral Theorem
Challenge: Can you prove the result?
Background
1) This interesting result was experimentally discovered by the author about 1991/92 using dynamic geometry, and so far have not yet seen/found earlier references to this result. Unable at the time to prove the result myself, I wrote to Jordan Tabov from the Bulgarian Academy of Sciences, who produced a neat trigonometric proof of the result, and which appears on p. 192-193 of my Some Adventures in Euclidean Geometry book, first published in 1994. An abridged version of Tabov's proof is available: here.
2) A little later after my discovery, round about 1993/94, I also wrote to the well-known geometer HSM Coxeter (1907-2003) at the University of Toronto about this result & he confirmed that he'd not seen it before. A colleague of his, John Wilker, then produced a proof similar to that of Tabov (link given above).
3) A few years later, round about 2004/2005, it was apparently independently rediscovered by Marcello Tarquini, and then proved by Darij Grinberg. A copy of Grinberg's paper involving his synthetic proof is available: here.
4) In 2006, Alexei Myakishev in his excellent article On Two Remarkable Lines Related to a Quadrilateral, Forum Geom., 6, 289–295 proved in general, that the quadrilateral formed by the circumcentres of the subdividing triangles of any quadrilateral is affine equivalent to the original. Similarly, but independently, Maria Mammana & Biagio Micale in their 2008 paper Quadrilaterals of triangle centres in the Mathematical Gazette, Vol. 92, No. 525 (November), pp. 466-475 (see Theorem 4), also proved this affine relationship. So for the above result, this implies that the two tangential quadrilaterals are affine equivalent.
5) More recently, Stanislaw Hauke found the problem listed as unsolved Problem 1351 at Antonio Gutierrez's site: GoGeometry, and in March 2018, he produced the following neat geometric proof: here.
More Properties
1) Another interesting property of the configuration shown above is that the angle bisectors of EFGH are respectively parallel to the angle bisectors of CBAD. This is easy to prove & left to the reader.
2) In an article published by Branko Grünbaum (1929-2018) in 1993, he used Mathematica to show, that in general, if the perpendicular bisectors of any quadrilateral Q form another quadrilateral Q₁, and the perpendicular bisectors of Q₁ are also constructed to form quadrilateral Q₂, then Q₂ is similar to Q. This theorem implies that for the above dynamic configuration that the perpendicular bisectors of EFGH form another circumscribed (tangential) quadrilateral similar to the original.
Reference: Grünbaum, B. (1993). Quadrangles, Pentagons and Computers. Geombinatorics, 3, pp. 4-9.
3) A generalization of Grünbaum's theorem mentioned in 2) above, to any (non-cyclic) quadrilateral, and that Q₂ is homothetic to Q, was proved purely geometrically by Maria Mammana & Biagio Micale in their 2008 paper Quadrilaterals of triangle centres in the Mathematical Gazette, Vol. 92, No. 525 (November), pp. 466-475 (see Theorem 5). Morever, as pointed out to me in a personal communication from Vladimir Dubrovsky, Kolmogorov School of Moscow State University, The Russian Federation, it's not hard to see that the ratio of the homothety equals
-(cot A + cot C)(cot B + cot D)/4.
Explore More: Read & explore more about the properties of circumscribed (tangential) quadrilaterals at this interactive webpage.
********************************Michael de Villiers, created 19 Dec 2009; updated to WebSketchpad, 22 March 2021; 14; 18 July 2022.