Problem
GEO-B2-M07-P007 Three Ratios from Three Areas
#7
★★★☆☆ Level 3 of 5
Point \(P\) lies inside triangle \(ABC\). Given \([PBC]=9\), \([PCA]=6\), \([PAB]=12\). Lines \(AP\), \(BP\), \(CP\) meet sides \(BC\), \(CA\), \(AB\) at \(D,E,F\). Find \(BD:DC\), \(CE:EA\), \(AF:FB\).
Use \(S_A=[PBC]\), \(S_B=[PCA]\), \(S_C=[PAB]\).
We have \(\frac{BD}{DC}=\frac{[PAB]}{[PCA]}=\frac{12}{6}=2:1\). Next, \(\frac{CE}{EA}=\frac{[PBC]}{[PAB]}=\frac{9}{12}=3:4\). Finally, \(\frac{AF}{FB}=\frac{[PCA]}{[PBC]}=\frac{6}{9}=2:3\).
Checks the correct order of the three formulas.