Problem
GEO-B2-M06-P002 Checking Concurrence
#2
★★☆☆☆ Level 2 of 5
In triangle \(ABC\), points \(D,E,F\) lie on \(BC,CA,AB\). Given \(BD:DC=4:5\), \(CE:EA=5:6\), \(AF:FB=3:2\). Prove that \(AD\), \(BE\), \(CF\) meet at one point.
Check the product of the three ratios.
Compute: \(\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{3}{2}=1\). Therefore, by Ceva's theorem, lines \(AD\), \(BE\), \(CF\) are concurrent.
A good problem for recognising Ceva.