Problem
GEO-B3-M02-P013 The Sixth Point via Pascal
B. New Original Problem. Five points \(A,B,C,D,E\) lie on one conic. A line \(e\) through \(E\), not tangent to the conic, is drawn. Let \(K=AB\cap DE\), \(L=e\cap BC\), \(M=KL\cap CD\), and \(F=AM\cap e\). Prove that \(F\) lies on the same conic.
C. Hint 1. Imagine that \(F\) already lies on the conic and apply Pascal.
D. Hint 2. Show that the constructed point is the second intersection of \(e\) with the conic.
E. Full Solution.
Let \(F_0\) be the second intersection of the line \(e\) with the conic. Apply Pascal to the hexagon \(A,B,C,D,E,F_0\). The points
\[ AB\cap DE=K,\quad BC\cap EF_0=L,\quad CD\cap F_0A \]
are collinear. Since \(E,F_0,L\) lie on \(e\), the point \(BC\cap EF_0\) is indeed \(L\). Hence the third point \(CD\cap F_0A\) lies on the line \(KL\), so it coincides with \(M=KL\cap CD\).
Therefore the line \(AM\) passes through \(F_0\). But the constructed point \(F\) is \(AM\cap e\), and \(F_0\in e\). Thus \(F=F_0\), and \(F\) lies on the conic.
This is a good constructive version of Pascal: the student sees not only a collinearity proof, but also a straightedge construction of a conic point.