Problem
GEO-B2-M10-P014 Finding the Third Ratio
#14
★★★★☆ Level 4 of 5
In triangle \(ABC\), points \(D\in BC\), \(E\in CA\). Lines \(AD\) and \(BE\) meet at \(P\), and line \(CP\) meets \(AB\) at \(F\). If \(BD:DC=2:1\), \(CE:EA=3:2\), find \(AF:FB\).
Since \(AD,BE,CF\) are concurrent, apply Ceva.
By Ceva's theorem, \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\). Hence \(2\cdot\frac{3}{2}\cdot\frac{AF}{FB}=1\), so \(\frac{AF}{FB}=\frac{1}{3}\). Therefore \(AF:FB=1:3\).
Useful as a reverse move: first understand that the third point is defined by concurrence.