**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 that) the results are true? (*Hint*: try relating the two results to the angle bisectors of a specific triangle in each configuration).

**Further reading**: Go here.

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