Problem
GEO-B2-M10-P005 A Ratio from Areas
#5
★★★☆☆ Level 3 of 5
Point \(P\) lies inside triangle \(ABC\). Line \(AP\) meets \(BC\) at \(D\). Given \([PAB]:[PBC]:[PCA]=2:3:4\), find \(BD:DC\).
Compare areas \([PAB]\) and \([PCA]\); they give the ratio \(BD:DC\).
Triangles \(PAB\) and \(PCA\) are conveniently compared using the common base \(AP\). Their altitudes to line \(AP\), drawn from \(B\) and \(C\), are in the ratio \(BD:DC\), because \(B,D,C\) are collinear. Hence \(BD:DC=[PAB]:[PCA]=2:4=1:2\).
If needed, one can prove the equality using altitudes to line \(AP\).