jmc

Draw line $PB$, and let $[PEB]=2b$, $[PDB]=a$, and $[CAP]=c$, so $[CPD]=3a$ and $[APE]=3b$. Because $△CAE$ and $△CEB$ share an altitude,$$c+3b=\frac{3}{2}(3a+a+2b)$$$$c+3b=6a+3b$$$$c=6a$$Because $△ACD$ and $△ABD$ share an altitude,$$6a+3a=3(a+2b+3b)$$$$9a=3a+15b$$$$6a=15b$$$$a=\frac{5}{2}b$$Thus, $[CAP]=15b$, and since $[APE]=3b$, $r=\frac{CP}{PE}=\boxed{5}$.