Ratio and Mensuration of Figures

Question

Is there any quick solution to the problem? Thanks.

Answer 1

Using Menelaus' theorem twice, we have $$\frac{CB}{BD}\times\frac{DP}{PA}\times\frac{AE}{EC}=1\Rightarrow \frac{AE}{EC}=\frac{6}{5}$$ and $$\frac{CD}{DB}\times\frac{BP}{PE}\times\frac{EA}{AC}=1\Rightarrow BP:PE=11:4$$

Added : Another way without using Menelaus' theorem.

Draw a line $DQ$ where $Q$ is on $CA$ such that $DQ$ is parallel to $BE$. Since we have $$\frac 32=\frac{BD}{DC}=\frac{EQ}{CQ}\ \ \text{and}\ \ \frac 21=\frac{AP}{PD}=\frac{AE}{QE},$$ we have $\frac{AE}{EC}=\frac 65$. Also, draw a line $ER$ where $R$ is on $BC$ such that $ER$ is parallel to $AD$. Since we have $$\frac{6}{5}=\frac{AE}{EC}=\frac{DR}{CR}\ \ \text{and}\ \ \frac{BP}{PE}=\frac{BD}{DR}\ \ \text{and}\ \ \frac 32=\frac{BD}{DC},$$ we have $\frac{BP}{PE}=\frac{BD}{DR}=\frac{11}{4}$.