A trapezoid $ABCD$ $(AB\parallel CD)$ is given with midsegment $PQ=4$. $K$ is the midpoint of $CD$ and $M$ is the midpoint of $AB$ and $KM=2$. Find the perimeter $P_{ABCD}$ of the trapezoid $ABCD$ if $\measuredangle BAD=70^\circ$ and $\measuredangle ABC=20^\circ$.
As we can see on the diagram, the points $K,O,M$ lie on the same line. I tried to use the similarity $$\triangle AOB\sim\triangle COD \Rightarrow \dfrac{AB}{CD}=\dfrac{OM}{OK}$$ If we denote $KO=x$, then $OM=2-x$ and $\dfrac{AB}{CD}=\dfrac{2-x}{x}$ I don't see if we can use that. Also $MNKM$ is a parallelogram.
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$\begingroup$Hint: extend $AD$ and $BC$, they intersect at $E$. Then $\triangle ABE$ is a right triangle and $M$ is the center of circle that inscribes $\triangle ABE$. Thus, $\angle AMK=40^{\circ}$. Also, $AB+DC=2PQ=8$. Can you take it from here?
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