Problem
GEO-B2-M07-P022 Recovering the Intersection Point
#22
★★★★★ Level 5 of 5
In triangle \(ABC\), points \(D,E,F\) are chosen on the sides so that \(BD:DC=3:4\), \(CE:EA=2:3\), \(AF:FB=2:1\). Prove that cevians \(AD\), \(BE\), \(CF\) are concurrent, and find \([PAB]:[PBC]:[PCA]\), where \(P\) is the intersection point.
First check Ceva, then recover the areas.
The product \(\frac{3}{4}\cdot\frac{2}{3}\cdot 2=1\), so by Ceva the cevians are concurrent. Let \(x=[PAB]\), \(y=[PBC]\), \(z=[PCA]\). From \(BD:DC=3:4\), \(x:z=3:4\). From \(CE:EA=2:3\), \(y:x=2:3\). Take \(x=9\); then \(z=12\), \(y=6\). Answer: \([PAB]:[PBC]:[PCA]=9:6:12=3:2:4\).
The problem combines Ceva and area reconstruction.