Problem
GEO-B2-M07-P015 Finding the Third Cevian
#15
★★★★☆ Level 4 of 5
Point \(P\) lies inside triangle \(ABC\). Lines \(AP\), \(BP\), \(CP\) meet the sides at \(D,E,F\). It is known that \(BD:DC=5:4\) and \(CE:EA=3:5\). Find \(AF:FB\).
You can use Ceva or the areas around \(P\).
Since the cevians pass through one point, by Ceva \(\frac{5}{4}\cdot\frac{3}{5}\cdot\frac{AF}{FB}=1\). The first factors give \(\frac{3}{4}\), so \(\frac{AF}{FB}=\frac{4}{3}\). Answer: \(AF:FB=4:3\).
Shows the connection between areas and Ceva.