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In a trapezoid $ABCD(AB||CD)$, $\angle ADC = 120^\circ$ and $AB=AD=2CD$. $M$ is the midpoint of $AD$. Find the length of the midsegment $MN$ if $ h_{MB}$ in $\triangle MBC$ is $3$ cm.

$\angle ADC + \angle BAD = 180 ^\circ$, thus $\angle BAD = 60 ^\circ$. Let $DD_1$ is the height of $ABCD$ through $D$. $\triangle AD_1M$ is equilateral, so $AM=AD_1=D_1M$. Now I am trying to show $BD_1=AD_1$ but don't see how to do it.

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Hint:

It could be reasoned that $\triangle BMC$ is an equilateral triangle. So that $MN = h = 3cm$

Use the fact that $\triangle CDM$ and $\triangle BD'M$ are both isosceles triangles and, then, $\angle ABM = \angle MCD = 30$.

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  • $\begingroup$ This will come after I show $AD_1 = BD_1$, but I do not know how to do it. I wrote it in the post. $\endgroup$ Commented Sep 1, 2019 at 17:11
  • $\begingroup$ The 60-degree right ADD' gives you $AD_1=AD/2=DC$ $\endgroup$
    – Quanto
    Commented Sep 1, 2019 at 17:14
  • $\begingroup$ Got it. I'd forgotten $AB=AD$. $\endgroup$ Commented Sep 1, 2019 at 18:09
  • $\begingroup$ I've just looked at the answers and it is saying $8$ $cm$. $\endgroup$ Commented Sep 1, 2019 at 18:53
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    $\begingroup$ it can be easily seen that triangle BCM is equilateral and its heights MN and $CC_1 $ are equal. that is MN=3 Cm. $\endgroup$
    – sirous
    Commented Sep 1, 2019 at 19:54

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