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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 $\measuredangle ABC=20^\circ$. enter image description here

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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  • $\begingroup$ Where is point $N$? Also, can you use trig? $\endgroup$ Commented Feb 12, 2022 at 22:08

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Extend $AD$ and $BC$, they intersect at $E$. Then $\triangle ABE$ is a right triangle and $EM$ is the median to hypotenuse. $EM=\frac{AB}{2}$. Let $X$ be the intersection of $EM$ and $PQ$. Again $EX$ is the median in $\triangle PEQ, EX=\frac{PQ}{2}=2$.

Now, $KX=\frac{KM}{2}=1 \implies EK=2-1=1$. But $EK$ is a median in $\triangle EDC$ so $DC=2EK=2$. Also, $AB+DC=2PQ=8$ so $AB=6$.

Let $h$ be the height of the trapezoid. We have that $AB - DC = 4=h \tan 70^\circ + h \tan 20^\circ, h=\frac{4}{\tan 70^\circ + \tan 20^\circ}=2\sin 40^\circ$.

Finally, $AD=\frac{h}{\sin 70^\circ}$ and $BC=\frac{h}{\sin 20^\circ}$

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  • $\begingroup$ Thank you! I really can't. Can you give me one more hint? $\endgroup$ Commented Feb 13, 2022 at 12:03
  • $\begingroup$ @Hipo. I know it's been a while, I finally noticed your comment and posted a solution. Better late than never :) $\endgroup$ Commented Jan 24 at 22:24

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