Problem
GEO-B2-M07-P016 Position on a Cevian
#16
★★★★☆ Level 4 of 5
In triangle \(ABC\), points \(D\in BC\), \(E\in CA\), \(BD:DC=2:3\), \(CE:EA=3:2\). Lines \(AD\) and \(BE\) meet at \(P\). Find \(AP:PD\).
First find the three areas \([PAB]\), \([PBC]\), \([PCA]\).
Let \([PAB]=x\), \([PBC]=y\), \([PCA]=z\). From \(BD:DC=2:3\), \(x:z=2:3\). From \(CE:EA=3:2\), \(y:x=3:2\). Take \(x=2\); then \(z=3\), \(y=3\). The total area is \(8\) units, and \([PBC]=3\). Since \(P\in AD\), \(\frac{PD}{AD}=\frac{[PBC]}{[ABC]}=\frac{3}{8}\). Hence \(AP:PD=5:3\).
A good mass-points preview through areas.