Problem
GEO-B2-M06-P012 Intersection of Two Lines
#12
★★★☆☆ Level 3 of 5
In triangle \(ABC\), points \(D\in AB\), \(E\in AC\), \(AD:DB=1:2\), \(AE:EC=2:3\). Lines \(CD\) and \(BE\) meet at \(P\), and \(AP\) meets \(BC\) at \(F\). Find \(BF:FC\).
For cevians \(CD\), \(BE\), \(AF\), use \(\frac{AD}{DB}\cdot\frac{BF}{FC}\cdot\frac{CE}{EA}=1\).
By Ceva, \(\frac{AD}{DB}\cdot\frac{BF}{FC}\cdot\frac{CE}{EA}=1\). Here \(\frac{AD}{DB}=\frac{1}{2}\), and \(\frac{CE}{EA}=\frac{3}{2}\). Thus \(\frac{1}{2}\cdot\frac{BF}{FC}\cdot\frac{3}{2}=1\). Hence \(\frac{BF}{FC}=\frac{4}{3}\). Answer: \(BF:FC=4:3\).
The same Ceva theorem, with points placed differently.