In the Reader Investigations of the Amesa KZN Mathematics Journal, May 2004, Vol. 8, No. 1, p. 38, I posed the following problem as problem no. 4 (which had been discovered using dynamic geometry).
Problem
Given any equidiagonal quadrilateral ABCD with AC = BD, and similar isosceles triangles
constructed on the pairs of opposite sides as shown below, prove that EG ⟂ FH.
Investigate
Use the dynamic sketch below and investigate further by dragging.
1) Does the result still hold when the equidiagonal quadrilateral ABCD becomes concave or crossed?
2) Does the result still hold when one or both pairs of similar isosceles triangle lie 'inwards'?
Challenge
3) Can you explain why (prove that) the result is true for all the cases you observed above?
Hint: Click on the 'Show Hint' button. What do you notice?
Van Aubel property of an Equidiagonal Quadrilateral
Published Solution
Read the Solutions to Reader Investigations: May 2004 in AMESA KZN Mathematics Journal, December 2004, pp. 41-45, which gives two elegant proofs of this result (Problem 4) by Michael Fox from Leamington Spa, Warwickshire, UK.
Converse
Consider the following possible converse: Given any quadrilateral ABCD, and similar isosceles triangles constructed on the pairs of opposite sides as shown above, and EG ⟂ FH, then ABCD is equidiagonal.
4) Is this converse true or not? If true, give a proof. If not, give a counter-example.
Further Generalization
5) What happens if ABCD is not an equidiagonal quadrilateral? Can you generalize further to any quadrilateral?
6) Compare your conjecture/observation/exploration in 5) above with the similar rectangle generalization at: Van Aubel's Theorem and some Generalizations (click on the 'Link to Similar Rectangles' button).
7) The perpendicularity result can also be generalized differently to 5) as shown in Dào Thanh Oai's Perpendicular Lines Van Aubel Generalization.
More Recent Updates
When the above problem 1-3 was posted at the Facebook Group Romantics of Geometry on 8 Oct 2024, the following responses were received:
i) Kousit Sett from India gave a link to the following related results for a quadrilateral with equal and perpendicular diagonals and which was posted in the group on 7/06/2024. Go to: Marcio Perez Dedication
2) Marian Cucoanes from Focsani, Romania proved the equidiagonal Van Aubel result using complex coordinates which covers all cases. Go to: Marion Proof Equidiagonal Van Aubel.
3) Other interesting dual special cases for quadrilaterals with equal and perpendicular diagonals were also earlier posted in 2020 in the group by Stanley Rabinowitz. Go to: Rabinowitz Special Cases.
4) On 17/18 Oct 2024, Jean-Pol Coulon from Tournai, Belgium, highlighted a nice duality with the following interesting related problem and proof on 17/18 October 2024 in the Romantics of Geometry Facebook Group in relation to an earlier post by Ercole Suppa, RG 13475, about a quadrilateral that is both orthodiagonal and equidiagonal. Go to: Jean-Pol Coulon Posting.
Related Links
A Hierarchical Classification of Quadrilaterals
Definitions and some Properties of Quadrilaterals
Van Aubel's Theorem and some Generalizations
Dào Thanh Oai's Perpendicular Lines Van Aubel Generalization
A Van Aubel like property of an Orthodiagonal Quadrilateral
Some Corollaries to Van Aubel Generalizations
Klingens' theorem of two intersecting circles or two adjacent isosceles triangles
The Lux Problem
Quadrilateral Balancing Theorem: Another 'Van Aubel-like' theorem
More Properties of a Bisect-diagonal Quadrilateral
An associated result of the Van Aubel configuration and some generalizations
The Vertex Centroid of a Van Aubel Result involving Similar Quadrilaterals and its Further Generalization
Twin Circles for a Van Aubel configuration involving Similar Parallelograms
A Fundamental Theorem of Similarity
Finsler-Hadwiger Theorem plus Gamow-Bottema's Invariant Point
Pompe's Hexagon Theorem
Sum of Two Rotations Theorem
SA Mathematics Olympiad Problem 2016, Round 1, Question 20
SA Mathematics Olympiad 2022, Round 2, Q25
An extension of the IMO 2014 Problem 4
A 1999 British Mathematics Olympiad Problem and its dual
Dirk Laurie Tribute Problem
Extangential Quadrilateral
Triangulated Tangential Hexagon theorem
Theorem of Gusić & Mladinić
Conway's Circle Theorem as special case of Side Divider (Windscreen Wiper) Theorem
Japanese Circumscribed Quadrilateral Theorem
External Links
Equidiagonal quadrilateral (Wikipedia)
Dào Thanh Oai's (perpendicular lines) Generalization of Van Aubel's Theorem (Cut the Knot)
SA Mathematics Olympiad
Questions and worked solutions for past South African Mathematics Olympiad papers can be found at this link.
(Note, however, that prospective users will need to register and log in to be able to view past papers and solutions.)
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Created with WebSketchpad by Michael de Villiers, 7 Oct 2024; updated 9/10/12/26 Oct 2024.