**Theorem 1**: If the perpendicular bisectors of the alternate sides *AB*, *CD* & *EF* of a (convex) hexagon *ABCDEF* circumscribed around a circle are concurrent at the centre of the circle, then the lines *PY*, *XR* & *QZ* connecting opposite tangential points are concurrent.

**Theorem 2 (dual)**: If the angle bisectors of the alternate angles *B*, *D* & *F* of a (convex) hexagon *ABCDEF* inscribed in a circle are concurrent at the centre of the circle, then the lines *AD*, *BE* & *CF* connecting opposite vertices are concurrent.

**Corollary 1**: If the perpendicular bisectors of the alternate sides *AB*, *CD* & *EF* of a (convex) hexagon *ABCDEF* circumscribed around a circle are concurrent at the centre of the circle, then it has 3 pairs of equal adjacent angles (e.g. angle *A* = angle *B*, angle *C* = angle *D* , and angle *E* = angle *F*).

**Corollary 2**: If the angle bisectors of the alternate angles *B*, *D* & *F* of a (convex) hexagon *ABCDEF* inscribed in a circle are concurrent at the centre of the circle, then it has 3 pairs of equal adjacent sides (e.g. *EF* = *FA*, *AB* = *BC*, and *CD* = *DF*).

A 1999 British Mathematics Olympiad Problem and its dual

**Challenge**: Can you explain why (prove) the results are true? If stuck, have a look at my 2002 *Math Gazette* paper at *A dual to a BMO problem*.

**Note**: Theorem 2 follows directly from Corollary 2 with the application of this lovely theorem by Cartensen (2000-2001): The main diagonals of a cyclic hexagon are concurrent, if and only if, the two products of alternate sides are equal; i.e. *NB⋅LC⋅MA = BL⋅CM⋅AN*. This theorem also appears in Gardiner & Bradley (2002, p. 96; 99) and also on the Math Stack Exchange (2013).

*References*

Cartensen, J. (2000-2001). About hexagons. *Mathematical Spectrum*, 33(2), pp. 37–40.

Gardiner, A.D. & Bradley, C.J. (2005). *Plane Euclidean Geometry: Theory and Problems*. Leeds, University of Leeds: The United Kingdom Mathematics Trust.

*Math Stack Exchange*. (2013). Accessed on 31 July 2021 at:
diagonals of cyclic hexagon.

**Further Result**: Clicking on the '*Show Diagonals*' button, reveals another interesting corollary of the 1st result, namely, that the main diagonals are concurrent at the same point of concurrency as the lines *PY*, *XR* & *QZ*. In general, according to Brianchon's theorem the diagonals of any hexagon circumscribed around a conic are concurrent. So in this case, we have that the lines *PY*, *XR* & *QZ* are concurrent at the Brianchon point.

**Further Result Challenge**: Can you explain why (prove) the BMO-Brianchon result above is also true? If stuck, have a look at the neat 2005 proof of my friend & colleague, Michael Fox, from the United Kingdom at *BMO-Brianchon Concurrency problem*. An accompanying dynamic sketch that
illustrates the proof is at *dynamic proof illustration*.

**Further Generalizations**

1) The first result can be generalized to a hexagon circumscribed around a conic with the midpoints of the alternate sides *AB*, *CD* & *EF* of a (convex) hexagon *ABCDEF* touching the conic. For a dynamic sketch illustrating this generalization go to *BMO Conic General problem.*

2) Corollary 2 of the second result can be generalized to a cyclic hexagon with concurrent diagonals, which then has its two products of alternate sides equal, as shown at *BMO 'Dual' General problem*.

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Created by Michael de Villiers, 30 October 2010; updated to *WebSketchpad* 6 May 2021.