**Theorem 1**

Given a hexagon *ABCDEF* with *AB* = *BC*, *CD* = *DE*, *EF* = *FA*, and ∠*A* = ∠*C* = ∠*E*, then *AD*, *BE*, and *CF* are concurrent at *P*.

**Theorem 2**

The point *P* defined as in Theorem 1, the powerpoint *T* of △*BDF*, and the circumcenter *Q* of △*ACE* are collinear.

**Corollary of Theorem 2**: The line *PTQ* is perpendicular to the line of perspective related to the two perspective triangles *ACE* and *DFB*.

In the sketch below, click on the *Show Line of Perspective* button to view & interact with the sketch dynamically illustrating this corollary.

**Theorem 3**

The points *A*, *B*, *C*, *P*, *T* and the orthocentre *H*_{1} of △*ACE* lie on a rectangular hyperbola. Likewise, the points *A*, *B*, *C*, *P*, *Q* and the orthocentre *H*_{2} of △*DFB* lie on a rectangular hyperbola.

In the sketch below, click on the *Show Hyperbola* buttons to view & interact with the sketch dynamically illustrating this theorem.

Concurrency, collinearity and other properties of a particular hexagon

**Theorem 4** (special case)

If ∠*A* = ∠*C* = ∠*E* = 120^{o}, then the line *PTQ* is parallel to the Euler line of △*BDF*.

In the sketch above, click on the *Link to Hung special case* button to view & interact with the sketch dynamically illustrating this theorem.

**Challenge**

Can you prove the theorems above?

**Reference**

A joint paper Concurrency, collinearity and other properties of a particular hexagon by myself and Tran Quang Hung, Vietnam, with proofs of the above results, has been published in the *Mathematics Competitions Journal*, Vol 35, No 1, 2022, pp. 82-91 of the World Federation of National Mathematics Competitions (WFNMC). *All rights reserved*.

**Analogous Results for a Haag Hexagon**

Though different from the above hexagon, a Haag Hexagon, which can be defined as a hexagon *ABCDEF* with *AB = AF, CB = CD, ED = EF*, and *∠A = ∠C = ∠E* = 120^{o}, has several analogous properties to the hexagon above. Some of these properties can be proved in exactly the same way as described in my joint paper with Tran Quang Hung. Go HERE to view a dynamic sketch of a Haag hexagon and to compare its properties with the above hexagon.

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**Additional properties**

Revisiting the hexagon *ABCDEF* with *AB* = *BC*, *CD* = *DE*, *EF* = *FA*, and ∠*A* = ∠*C* = ∠*E* shortly after the publication of our paper in the *Mathematics Competitions* Journal, more of its properties were found as given below.

**Theorem 5**

Given a hexagon *ABCDEF* with *AB* = *BC*, *CD* = *DE*, *EF* = *FA*, and ∠*A* = ∠*C* = ∠*E*, then *ABCDEF* has an incircle.

In the sketch above, click on the *Link to Hexagon Incircle* button to view & interact with a sketch dynamically illustrating this theorem.

An equivalent formulation of Theorem 5 is the following: Given a hexagon *ABCDEF* with *AB* = *BC*, *CD* = *DE*, *EF* = *FA*, and ∠*A* = ∠*C* = ∠*E*, then the angle bisectors of ∠*A*, ∠*C*, and ∠*E* are concurrent at the circumcentre, *Q*, of △*ACE*.

**Important Note**: The concurrency of the main diagonals of the hexagon in Theorem 1 above follows directly from Theorem 5 with the application of Brianchon's theorem, and therefore provides a much easier proof than the one given by us in our paper in the link at the top (which uses a theorem by Anghel (2016)).

**Acknowledgement**: It has also come to our attention that the concurrency Theorem 1 is apparently attributed to A. Zaslavsky, and a diagram (without proof) of it is given in Arseniy Akopyan's (2011) diagrammatic book Geometry in Figures. It also appeared earlier as a problem in the Third Sharygin Olympiad in Geometry, 2007, Final Round, Grade 9, Problem 3. Though in Russian, it's easy to see that the given solution on p. 6 to Problem 3 of the Sharygin Solutions, is via Theorem 5 above (see p. 6, Fig. 9.3).

The second formulation of Theorem 5 now gives us the following neat converse: Given a hexagon

Given a hexagon

In the sketch above, click on the

**Challenge**

Can you prove the additional Theorem 5 and its converse, as well as Theorem 6?

**Further Generalization**

Can you generalize Theorem 5 and its converse to an octagon, decagon, etc.?

Go to the dynamic sketches at Further Hexagon Properties and click on the *Link to Octagon by reflection*, *Link to Octagon Check* and *Link to Decagon* buttons to explore further generalizations.

**Challenge**: Can you prove your generalizations?

**Note**

A paper discussing & proving these additional properties, and generalizations, has been submitted for publication. All rights reserved.

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Created by Michael de Villiers, 3 November 2021 with *WebSketchpad*; updated 11 January 2022, 19 & 20 June 2022; 2, 23, 26 Sept 2022.