$\frac{CD}{BD}=3$ and $\frac{AE}{EB}=\frac{3}{2}$, Find $\frac{CP}{PE}$.

Question

We have $\frac{CD}{BD}=3$ and $\frac{AE}{EB}=\frac{3}{2}$, Find $\frac{CP}{PE}$. I tried a line parallel to $AB$ to see if it could be solved by Thales's theorem but it didn't seem to help much.

Answer 1

Let $F\in DB$ such that $EF||AD$ and let $DB=5x$.

Thus, $CD=15x$ and $$\frac{DF}{FB}=\frac{AE}{EB}=\frac{3}{2},$$ which gives $DF=3x$ and we obtain: $$\frac{CP}{PE}=\frac{CD}{DF}=\frac{15x}{3x}=5.$$