Problem
GEO-B3-M02-P009 Pascal with One Point at Infinity
#9
★★☆☆☆ Level 2 of 5
B. New Original Problem. Points \(A,B,C,D,E,F\) lie on one circle, and \(AB\parallel DE\). Let \(Q=BC\cap EF\) and \(R=CD\cap FA\). Prove that \(QR\parallel AB\).
Inspired by Prasolov projective geometry method
C. Hint 1. Apply Pascal's theorem to the hexagon \(ABCDEF\).
D. Hint 2. The intersection of \(AB\) and \(DE\) is the point at infinity in the direction of \(AB\).
E. Full Solution.
By Pascal's theorem for the hexagon \(ABCDEF\), the points
\[ P=AB\cap DE,\quad Q=BC\cap EF,\quad R=CD\cap FA \]
are collinear. But \(AB\parallel DE\), so \(P\) is the point at infinity in the direction of \(AB\). Since the line \(QR\) passes through \(P\), it has the same direction. Hence \(QR\parallel AB\).
A useful task on the correct order of sides in Pascal and on using a point at infinity.