Problem
GEO-B3-M03-P017 Brianchon for a Tangential Hexagon
B. New Original Problem. Hexagon \(ABCDEF\) is circumscribed about a circle \(\omega\). Prove that the lines \(AD\), \(BE\), and \(CF\) are concurrent.
C. Hint 1. This is the dual statement to Pascal.
D. Hint 2. Pass from the six tangent sides to the six contact points and their polars.
E. Full Solution.
Let the sides of the hexagon touch \(\omega\) at \(A_1,B_1,C_1,D_1,E_1,F_1\), respectively. Pascal's theorem for the six contact points on the circle states the collinearity of three intersections of opposite sides of the corresponding inscribed hexagon.
Passing to poles and polars turns the collinearity of three points into the concurrence of the corresponding three lines. In the original tangential hexagon, these lines are exactly \(AD\), \(BE\), and \(CF\). Hence they meet at one point.
This is a full preview of Pascal duality; a diagram is recommended.