Problem
GEO-B2-M06-P021 Internal and External Points
In triangle \(ABC\), points \(D\in BC\), \(E\in CA\) are internal. Lines \(AD\) and \(BE\) meet at \(P\), and \(CP\) meets \(AB\) at \(F\). Line \(DE\) meets the extension of \(AB\) at \(X\). Prove that \(F\) lies on segment \(AB\), \(X\) lies outside segment \(AB\), and \(\frac{AF}{FB}=\frac{AX}{XB}\).
Use Ceva and Menelaus, and explain the positions geometrically.
Since \(D\) and \(E\) are internal, lines \(AD\) and \(BE\) meet inside the triangle, so line \(CP\) meets side \(AB\) at an internal point \(F\). A line through two internal points \(D\) and \(E\) meets the third side \(AB\) on its extension, so \(X\) is external. By Ceva, \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\), and by Menelaus for \(D,E,X\): \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AX}{XB}=1\). Therefore \(\frac{AF}{FB}=\frac{AX}{XB}\).
A strong problem on the meaning of internal and external division.