Problem
GEO-B3-M02-P017 Brianchon for a Tangential Hexagon
B. New Original Problem. Hexagon \(ABCDEF\) is circumscribed about one circle: each of its sides is tangent to the circle. Prove that the lines \(AD\), \(BE\), and \(CF\) are concurrent.
C. Hint 1. This is the dual theorem to Pascal.
D. Hint 2. Pass from tangent sides to contact points and polars.
E. Full Solution.
Consider polarity with respect to the given circle. Each tangent side of the hexagon corresponds to its point of tangency, and each vertex of the hexagon corresponds to the line joining two neighbouring contact points in the dual configuration.
Pascal's theorem for the six contact points on the circle, in dual form, says that the three lines joining opposite vertices of the tangential hexagon are concurrent.
These three lines are exactly \(AD\), \(BE\), and \(CF\). Hence they pass through one point.
This is a preview: the full pole-polar technique belongs to the next module, but here it is important to see the duality of Pascal.