Problem
GEO-B2-M06-P008 A Cevian and a Median
#8
★★★☆☆ Level 3 of 5
In triangle \(ABC\), point \(D\in BC\), \(BD:DC=2:1\). Point \(M\) is the midpoint of \(AB\). Lines \(AD\), \(BE\), \(CM\) are concurrent, where \(E\in CA\). Find \(CE:EA\).
For point \(M\), \(AM:MB=1:1\).
By Ceva, \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AM}{MB}=1\). Thus \(2\cdot\frac{CE}{EA}\cdot 1=1\), so \(\frac{CE}{EA}=\frac{1}{2}\). Answer: \(CE:EA=1:2\).
The problem combines Ceva with a midpoint.