**NOTE**: Please WAIT while the applet below loads.

The following interesting and very useful hexagon theorem is attributed to my colleague and friend of several years, Waldemar Pompe, from the University of Warsaw, Poland.

**Pompe's Hexagon Theorem**

Given a hexagon *ABCDEF* with *AB* = *BC*, *CD* = *DE* and *EF* = *FA*, and angles α + β + γ = 360^{o}, then the respective angles of Δ*BDF* are α/2, β/2 and γ/2.

**Reference**: Pompe, W. (2016). *Wokół obrotów: przewodnik po geometrii elementarnej*. Wydawnictwo Szkolne Omega, Kraków, Poland.

Pompe's Hexagon Theorem

**Challenge**

1) While checking the angle measurements, drag points *B* and *D* until the hexagon becomes convex. What do you notice?

1) Can you explain *why* (prove that) this hexagon theorem is true?

2) If you get stuck, press the '**Proof Hint**' button in the *dynamic sketch* at the top, which illustrates the rotation of Δ*FAB* counter-clockwise around centre *F* through angle γ and Δ*DCB* clockwise around centre *D* through angle β. Then determine the angles of Δ*BDF* in terms of α, β and γ.

**Note**: Pompe's hexagon theorem can also be viewed as a configuration showing that the sum of the rotation of *A* through γ around *F* to map onto *E*, and the subsequent rotation of *E* through β around *D* to map onto *C*, is equivalent to a rotation of *A* through α around *B* to map onto *C* - see for example, the Sum of Two Rotations Theorem.

**Some applications**

Pompe's hexagon theorem certainly deserves to be much better known as it applies directly in so many geometric results. For example, it:

1) provides an immediate proof for the *Lux Problem* as well as the equivalent *Klingens' Theorem*.

2) immediately proves *Napoleon’s theorem* (the centres of equilateral triangles on the sides of any triangle form another equilateral triangle), as well as some of its generalizations.

3) directly shows that in *Van Aubel’s quadrilateral theorem*, the angle formed by the centres of two squares on adjacent sides, say *AB* and *BC*, and the midpoint of the diagonal *AC*, is a right angle

4) solves this interesting *Dirk Laurie Tribute Problem* involving a cyclic hexagon with a set of alternate sides equal to the radius of the circle.

5) together with another result, can be used to prove *Jha and Savaran’s generalisation of Napoleon’s theorem* to a hexagon.

**Egamberganov's Theorem (2017)**

This theorem appears in A generalization of Napoleon's Theorem, and is a generalization of Pompe's Hexagon Theorem. It can be formulated in general as follows in relation to the sketch above:

Given a hexagon *ABCDEF* with *AB** *CD* * *EF* = *BC* * *DE* * *FA*, and angles α + β + γ = 360^{o}, then ∠*DBF* = ∠*EAF* + ∠*DCE*, ∠*FDB* = ∠*EDF* + ∠*BDC*, and ∠*BFD* = ∠*DFE* + ∠*AFB*.
(A dynamic geometry sketch of this theorem will be done in due course & posted in the next update).

In the sketch above click on the '**Show Angles for Egamberganov's Theorem**' to display the respective angles above.

**Proof Hints or Comparisons**

a) If still stuck for a proof of Pompe's Hexagon theorem, or to compare your own solution(s), a proof is given in the Appendix of my joint paper *Ghosts of a Problem Past* with Hans Humenberger in the free online journal *At Right Angles* in the March 2021 issue.

b) Alternatively, look at the Sum of Two Rotations Theorem.

**Related Links**

Lux Problem

Klingens' Theorem

Sum of Two Rotations Theorem

Van Aubel's Theorem and some Generalizations

Fermat-Torricelli Point Generalization

Napoleon's Theorem: Generalizations & Converses

Jha and Savaran’s generalisation of Napoleon’s theorem

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Release: 2019Q4, Semantic Version: 4.6.2, Build Number: 1039, Build Stamp: dn.kcptech.com/20191209222013

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Created 30 March 2020 by Michael de Villiers, modified 6 April 2020; updated 17 April 2021; 9 June 2022; 20 Nov 2023.