A 1999 British Mathematics Olympiad Problem and its dual

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.