## Pompe's Hexagon Theorem

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:

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

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

#### .sketch_canvas { border: medium solid lightgray; display: inline-block; } 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 ΔFAB counter-clockwise around centre F through angle γ and ΔDCB clockwise around centre D through angle β. Then determine the angles of ΔBDF in terms of α, β and γ.

Note: Pompe's 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.

Some applications: Pompe's 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) easily solves this interesting Dirk Laurie Tribute Problem involving a cyclic hexagon with a set of alternate sides equal to the radius of the circle.

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 = 360o, 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.

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 with Hans Humenberger in the free online journal At Right Angles in the March 2021 issue at Ghosts of a Problem Past.

Created 30 March 2020 by Michael de Villiers, modified 6 April 2020; 17 April 2021.