Problem
GEO-B1-M08-P027 Find the Third Ratio
#27
★★★★★ Level 5 of 5
In triangle \(ABC\), lines \(AD\), \(BE\), \(CF\) meet at one point. Points \(D,E,F\) lie on \(BC,CA,AB\), respectively. It is known that \(BD:DC=2:3\), \(CE:EA=3:4\). Find \(AF:FB\).
Use the product of the three ratios.
By the area form of Ceva, \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\). Thus \(\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{AF}{FB}=1\), so \(\frac{1}{2}\cdot\frac{AF}{FB}=1\). Therefore \(AF:FB=2:1\).
A difficult ratio problem: the student must recognise area as the tool.