Problem
GEO-B2-M06-P006 Ceva, Not Menelaus
#6
★★☆☆☆ Level 2 of 5
Points \(D,E,F\) lie respectively on sides \(BC,CA,AB\) of triangle \(ABC\), and \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\). Prove that lines \(AD\), \(BE\), \(CF\) are concurrent.
The condition matches Ceva's theorem exactly.
By Ceva's theorem, the product being \(1\) for points on the sides of a triangle is equivalent to concurrence of the cevians. Therefore \(AD\), \(BE\), \(CF\) meet at one point.
This problem reinforces the distinction between the two theorems.