Problem
GEO-B2-M07-P021 Two Cevians and Both Divisions
#21
★★★★★ Level 5 of 5
In triangle \(ABC\), points \(D\in BC\), \(E\in CA\), \(BD:DC=2:5\), \(CE:EA=3:4\). Lines \(AD\) and \(BE\) meet at \(P\). Find \(AP:PD\) and \(BP:PE\).
First recover the three areas around \(P\).
Let \([PAB]=x\), \([PBC]=y\), \([PCA]=z\). From \(BD:DC=2:5\), \(x:z=2:5\). From \(CE:EA=3:4\), \(y:x=3:4\). Take \(x=8\); then \(z=20\), \(y=6\), total \(34\). On cevian \(AD\): \(\frac{PD}{AD}=\frac{[PBC]}{[ABC]}=\frac{6}{34}=\frac{3}{17}\), so \(AP:PD=14:3\). On cevian \(BE\): \(\frac{PE}{BE}=\frac{[PCA]}{[ABC]}=\frac{20}{34}=\frac{10}{17}\), so \(BP:PE=7:10\).
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