A vector geometry method gets the answer easily.
For a diagram, imagine Melody's rectangle 'bashed' at A so that AB remains parallel to CD, so that we do have a parallelogram.
We have
\(\displaystyle \\ \underline{AF} = \underline{AB} + \underline{BF}= \underline{AB} + (1/2)\underline{BC} \dots(1)\\ \text{and}\\ \underline{EC}=\underline{EB}+\underline{BC} = (1/2)\underline{AB}+\underline{BC} \dots(2).\)
Let
\(\displaystyle \underline{AP}=\lambda\underline{AF}, \text{ and }\underline{EP}=\mu\underline{EC}, \\ 0< \lambda <1 \text{ and } 0< \mu <1. \)
From the triangle AEP,
\(\displaystyle \underline {AP} = \underline{AE}+\underline{EP},\)
and so substituting (1) and (2),
\(\displaystyle \lambda \{\underline{AB}+(1/2)\underline{BC}\}= (1/2)\underline{AB}+\mu\{(1/2)\underline{AB}+\underline{BC}\},\)
so
\(\displaystyle \underline{AB}\{\lambda - (1/2)-(\mu/2)\}=\underline{BC}\{\mu-(\lambda/2)\}.\)
AB and BC are not parallel, so for this to be true it's necessary that
\(\displaystyle \lambda -(1/2)-(\mu/2)=0\)
and
\(\displaystyle \mu - (\lambda/2)=0, \)
from which
\(\displaystyle \lambda = 2/3 \text{ and } \mu = 1/3.\)
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