Find the area ratio of the $ABNM$ and $ABCD$ trapezoids

For reference:

In the trapezoid $ABCD$ ($AB \parallel CD$), $MN$ is the median. $NP$ is traced parallel to $AD$ ($P \in CD$). The area of ​​$MNCD$ is $16\ \mathrm{m^2}$ and that of the triangle $NCP$ is $4\ \mathrm{m^2}$. Calculate the area ratio of the trapezoids $ABNM$ and $ABCD$. (Answer:$\frac{1}{3}$)

My progress: I found the following relationships.

$h$ = height of trapezoid:

$S_{CPN}=4=\frac{CP\cdot h}{4}\implies CP\cdot h =16$

$S_{ABCD} = MN\cdot h = \frac{AB+CD}{2}\cdot H$

$S_{AMBN} = \frac{AB+MN}{2}\cdot\frac{h}{2}$

$S_{PDMF} = S_{AEMF}=FM\cdot\frac{h}{2}$

$S_{PCFN} = \frac{FN\cdot h}{4}$

$\frac{S_{DCMN}}{S_{MNAB}}=\frac{MN^2-CD^2}{AB^2-MN^2}(\text{ property})$

$\frac{S_{AMNB}}{S_{ABCD}}=\frac{\frac{(AB+MN)\cdot h}{4}}{MN\cdot h}= \frac{AB+MN}{4MN} = \frac{3AB+CD}{4(AB+CD)}$

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Solution 1:

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