Problem
GEO-B1-M08-P030 Concurrence From Ratios
#30
★★★★★ Level 5 of 5
In triangle \(ABC\), points \(D,E,F\) lie on sides \(BC,CA,AB\), respectively. It is known that \(BD:DC=2:3\), \(CE:EA=3:5\), \(AF:FB=5:2\). Prove that lines \(AD\), \(BE\), \(CF\) meet at one point.
Check the product of the three ratios.
We have \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=\frac{2}{3}\cdot\frac{3}{5}\cdot\frac{5}{2}=1\). By the converse form of the area version of Ceva's theorem, it follows that lines \(AD\), \(BE\), \(CF\) meet at one point.
Final challenge problem: strong, but a natural Book 1 conclusion using areas and ratios.