Problem
GEO-B2-M06-P013 External Point on the Base
#13
★★★☆☆ Level 3 of 5
In triangle \(ABC\), point \(D\in AB\), \(AD:DB=3:2\), point \(E\in AC\), \(AE:EC=5:1\). Line \(DE\) meets the extension of \(BC\) at \(F\). Find \(BF:FC\).
Write Menelaus for \(D,E,F\).
We have \(\frac{AD}{DB}\cdot\frac{BF}{FC}\cdot\frac{CE}{EA}=1\). Thus \(\frac{3}{2}\cdot\frac{BF}{FC}\cdot\frac{1}{5}=1\), so \(\frac{BF}{FC}=\frac{10}{3}\). Answer: \(BF:FC=10:3\).
The order \(CE:EA\), not \(AE:EC\), is important.