Problem
GEO-B2-M06-P009 A Transversal Through Two Sides
#9
★★★☆☆ Level 3 of 5
In triangle \(ABC\), point \(D\in AB\), \(AD:DB=2:3\), point \(E\in AC\), \(AE:EC=4:1\). Line \(DE\) meets the extension of \(BC\) at \(F\). Find \(BF:FC\).
Apply Menelaus to triangle \(ABC\) and line \(D E F\).
With one external point, use the positive form: \(\frac{AD}{DB}\cdot\frac{BF}{FC}\cdot\frac{CE}{EA}=1\). We have \(\frac{2}{3}\cdot\frac{BF}{FC}\cdot\frac{1}{4}=1\). Hence \(\frac{BF}{FC}=6\), so \(BF:FC=6:1\).
Menelaus in another cyclic notation.