Problem
GEO-B3-M06-P020 A Transversal in a Cyclic Configuration
In cyclic quadrilateral \(ABCD\), let \(E=AB\cap CD\), \(F=AD\cap BC\). Prove that for any line through \(E\) meeting \(AD\) and \(BC\) at \(X,Y\), the collinearity of \(X,Y,E\) can be written by trigonometric Menelaus in triangle \(AFB\).
C. Hint 1. View \(X\in AF\), \(Y\in BF\), \(E\in AB\).
D. Hint 2. Angles with the transversal can be replaced by equal cyclic angles of \(ABCD\).
In triangle \(AFB\), the points \(X,Y,E\) lie on \(AF,BF,AB\), respectively. Their collinearity is equivalent to Menelaus' condition.
The trigonometric Menelaus form expresses this through the angles made by line \(EXY\) with sides \(AF,BF,AB\). Since \(ABCD\) is cyclic, angles between \(CD\), \(AD\), \(BC\), \(AB\) can be replaced by equal inscribed angles. Thus the entire collinearity check reduces to a product of sines of circle angles.
This teaches choosing the right triangle for Menelaus, not just the given quadrilateral.