Problem
GEO-B2-M07-P014 Intersection of Two Cevians
#14
★★★☆☆ Level 3 of 5
In triangle \(ABC\), points \(D\in BC\), \(E\in CA\), \(BD:DC=3:5\), \(CE:EA=2:3\). Lines \(AD\) and \(BE\) meet at \(P\). Find \([PAB]:[PBC]:[PCA]\).
Use \( [PAB]:[PCA]=BD:DC\) and \( [PBC]:[PAB]=CE:EA\).
Let \([PAB]=x\), \([PBC]=y\), \([PCA]=z\). Then \(x:z=3:5\), and \(y:x=2:3\). Take \(x=9\). Then \(z=15\), \(y=6\). Answer: \([PAB]:[PBC]:[PCA]=9:6:15=3:2:5\).
Another problem on recovering the three areas.