Bradley's Theorem, its Generalization & an Analogue Theorem

"Generalising is at the heart of mathematics." - Sue Johnston Wilder & John Mason in Developing Thinking in Geometry, London: Paul Chapman Publishing, 2005, p. 93.

"Generalization is probably the easiest and most obvious way to enlarging mathematical knowledge." - W.W. Sawyer in A Mathematician's Delight, Penguin Books, 1943.

"When we generalize a result, we make it more useful. It may strike you as strange that generalization nearly always makes the result simpler too. The more powerful result is easier to learn than the less powerful one. - W.W. Sawyer in Prelude to Mathematics, Pelican Books, 1955, p. 25.

The following dynamic investigation is suitable for high school learners of geometry at any level. It was stimulated by Result 30 in a provocative paper by Christopher Bradley, namely Cyclic Quadrilaterals, in the November 2004 issue of The Mathematical Gazette.

Conjecture
Given a tangential quadrilateral ABCD with its sides AB, BC, CD and DA, respectively touching its incircle at K, L, M and N.
1) Click on the 'Show Incentres' button to show the incentres of triangles AKN, BLK, CML and DNM. What do you notice?
2) Can you formulate a conjecture? Check your conjecture by dragging.
3) Does your conjecture hold if ABCD is dragged into the shape of a concave or crossed quadrilateral?
4) Click on the 'Show Angles' button to display the measurements of angles NKL, KLM, and PQR. What relationship do you notice between these angles?

Bradley's Theorem, its Generalization & an Analogue Theorem

Bradley's Theorem
In the investigation above you should've discovered the following theorem by Bradley (2004):
If ABCD is a tangential quadrilateral with its sides AB, BC, CD and DA, respectively touching its incircle at K, L, M and N, then the respective incentres P, Q, R and S of triangles AKN, BLK, CML and DNM lie on its incircle (and obviously form a cyclic quadrilateral). In addition, ∠PQR = (∠NKL + ∠KLM)/2, etc.
Challenge
5) Can you explain why (prove that) this theorem is true?
6) Can you explain/prove the theorem in more than one way?

Analogue to Bradley's Theorem
7) Can you formulate a possible analogue to Bradley's theorem for a cyclic quadrilateral ABCD and the analogous construction of four circumcentres?
8) Check your answer by clicking on 'Link to Bradley analogue' button to navigate to a new sketch which shows such an analogous result: here the circumcentres of triangles AOB, BOC, COD and DOA of cyclic quadrilateral ABCD form a tangential quadrilateral. In addition, ∠SPQ = ∠DAB + ∠ABC - (∠RSP + ∠PQR)/2, etc.
Challenge
9) Can you explain why (prove that) this analogue result is true?
Hint: Note as shown in the dynamic sketch that the angle bisectors (the dashed rays) of PQRS are concurrent at O. Try to use this observation to construct a proof.
10) Can you explain/prove it in more than one way?
Note: This analogue to Bradley's theorem can be seen as a variation of a more general Angle Divider Theorem for a Cyclic Quadrilateral (De Villiers, 1994, 1996, 2023).

Further Generalization
11) Can you generalize Bradley's theorem further to other tangential polygons?
12) Click on the 'Link to Tangential pentagon' button to navigate to a new sketch, which shows a generalization to a tangential pentagon.
13) Can you also prove the result in 12)? Can you generalize further (including to triangles)?
14) Can you generalize the above analogue of Bradley's theorem further to other cyclic polygons?
15) Click on the 'Link to Cyclic pentagon' button to navigate to a new sketch, which shows a generalization to a cyclic pentagon.
16) Can you also prove the result in 15)? Can you generalize further (including to triangles)?

Published Paper
Read my paper Bradley's Theorem, an Analogue & their Generalization about the above results that has been published in the June 2024 issue of the Learning & Teaching Mathematics journal of AMESA.

References
Bradley, C.J. (2004). Cyclic Quadrilaterals. The Mathematical Gazette, Vo. 88, No. 513, pp. 417–431.
De Villiers, M. (1994, 2007). Some Adventures in Euclidean Geometry. Morrisville, NC: Lulu Press, pp. 58-61; p. 67; 195-198; 202.
De Villiers, M. (1996). An Interesting Duality in Geometry. AMESA Proceedings, Peninsula Technikon, 1-5 July 1996, pp. 345-350.
De Villiers, M. (2023). Conway’s Circle Theorem as a Special Case of a More General Side Divider Theorem. Learning & Teaching Mathematics, No. 34, 2023, pp. 37-42.

Related Links
Angle Divider Theorem for a Cyclic Quadrilateral
Concurrent Angle Bisectors of a Quadrilateral
Conway’s Circle Theorem as special case of Side Divider Theorem
Tangential Quadrilateral Theorem of Gusić & Mladinić
Enrichment for the Gifted: Generalizing Some Geometrical Theorems & Objects
Semi-regular Angle-gons and Side-gons: Generalizations of rectangles and rhombi
Alternate sides cyclic-2n-gons and Alternate angles circum-2n-gons: Generalizations of isosceles trapezia and kites
Golden Quadrilaterals: Generalizing the concept of a golden rectangle

In Memoriam: Christopher J. Bradley (1938-2013)
Christopher Bradley was Deputy Leader of the British Mathematical Olympiad team and Secretary of the British Mathematical Olympiad Committee, and was known for his elegant Olympiad problems. He was involved with the UK Maths Trust and the Mathematical Association, and published many articles in The Mathematical Gazette, as well as introductory texts on topics and problem solving for the UK Maths Trust. Here are two sample problems from some of his publications: Prove that one member of a Pythagorean triple is always divisible by 5, and that the area of any right-angled triangle with integer sides is divisble by 6.

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Created by Michael de Villiers with WebSketchpad, 18 Feb 2024; updated 9 June 2024.