Problem
GEO-B2-M06-P019 Two Unknown Points on One Side
#19
★★★★☆ Level 4 of 5
In triangle \(ABC\), points \(D\in BC\), \(E\in CA\), \(BD:DC=2:3\), \(CE:EA=3:5\). Lines \(AD\) and \(BE\) meet at \(P\), \(CP\) meets \(AB\) at \(F\), and \(DE\) meets the extension of \(AB\) at \(X\). Find \(AF:FB\) and \(AX:XB\).
Use Ceva for \(F\) and Menelaus for \(X\).
By Ceva, \(\frac{2}{3}\cdot\frac{3}{5}\cdot\frac{AF}{FB}=1\), so \(\frac{2}{5}\cdot\frac{AF}{FB}=1\), and \(AF:FB=5:2\). By Menelaus for \(D,E,X\): \(\frac{2}{3}\cdot\frac{3}{5}\cdot\frac{AX}{XB}=1\), hence \(AX:XB=5:2\).
A numerical version of the previous result.