Problem
GEO-B1-M06-P028 Find the Third Ratio
#28
★★★★★ Level 5 of 5
In triangle \(ABC\), lines \(AD\), \(BE\), \(CF\) meet at one point, where \(D\in BC\), \(E\in CA\), \(F\in AB\). It is known that \(BD:DC=2:3\), \(CE:EA=3:4\). Find \(AF:FB\).
Use the product \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\).
By the area form of Ceva's theorem, \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\). Substitute: \(\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{AF}{FB}=1\). We get \(\frac{1}{2}\cdot\frac{AF}{FB}=1\), hence \(AF:FB=2:1\).
Can be given after the previous problem as an application of areas to concurrence.