Problem
GEO-B2-M06-P007 Two Cevians Determine the Third
#7
★★★☆☆ Level 3 of 5
In triangle \(ABC\), points \(D\in BC\), \(E\in CA\), \(BD:DC=3:2\), \(CE:EA=5:6\). Lines \(AD\) and \(BE\) meet at \(P\), and \(CP\) meets \(AB\) at \(F\). Find \(AF:FB\).
Now \(AD\), \(BE\), \(CF\) are concurrent.
By Ceva, \(\frac{3}{2}\cdot\frac{5}{6}\cdot\frac{AF}{FB}=1\). The first factors equal \(\frac{5}{4}\), hence \(\frac{AF}{FB}=\frac{4}{5}\). Answer: \(4:5\).
A standard problem on the intersection of two cevians.