Problem
GEO-B2-M06-P010 Finding a Point on a Side
#10
★★★☆☆ Level 3 of 5
In triangle \(ABC\), point \(D\in BC\), \(BD:DC=3:4\), point \(F\) lies on the extension of \(AB\) beyond \(B\), \(AF:FB=7:2\). Line \(DF\) meets \(CA\) at \(E\). Find \(CE:EA\).
Points \(D,E,F\) are collinear.
By Menelaus, \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\). Thus \(\frac{3}{4}\cdot\frac{CE}{EA}\cdot\frac{7}{2}=1\). We get \(\frac{21}{8}\cdot\frac{CE}{EA}=1\), hence \(CE:EA=8:21\).
Students must track which point is external.