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.
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, $\angle ABM = \angle MCD = 30$.