Problem
GEO-B2-M10-P004 Checking Ceva
#4
★★☆☆☆ Level 2 of 5
In triangle \(ABC\), points \(D,E,F\) lie on \(BC,CA,AB\). Given \(BD:DC=2:3\), \(CE:EA=3:4\), \(AF:FB=2:1\). Prove that \(AD,BE,CF\) are concurrent.
For concurrence of three cevians, check Ceva's product.
\(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=\frac{2}{3}\cdot\frac{3}{4}\cdot 2=1\). By Ceva's theorem, lines \(AD,BE,CF\) are concurrent.
A warm-up on choosing between Ceva and Menelaus.