Problem
GEO-B3-M02-P011 A Second Degenerate Pascal
#11
★★★☆☆ Level 3 of 5
B. New Original Problem. Points \(A,B,C,D\) lie on one circle. The tangents at \(B\) and \(D\) meet at \(X\). Let \(Y=AB\cap CD\) and \(Z=BC\cap AD\). Prove that \(X,Y,Z\) are collinear.
Inspired by Prasolov projective geometry method
C. Hint 1. The order of repeated vertices is different from the previous problem.
D. Hint 2. Try the hexagon \(A,B,B,C,D,D\).
E. Full Solution.
Consider the degenerate hexagon \(A,B,B,C,D,D\) on the circle. Its side \(BB\) is the tangent at \(B\), and its side \(DD\) is the tangent at \(D\).
The pairs of opposite sides give:
\[ AB\cap CD=Y,\quad BB\cap DD=X,\quad BC\cap DA=Z. \]
By Pascal's theorem, \(X,Y,Z\) are collinear.
The student must choose the order of vertices; that makes the task meaningfully less mechanical.