# Find the midsegment of a rectangular trapezoid

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.

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, $$\angle ABM = \angle MCD = 30$$.
• 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. Commented Sep 1, 2019 at 17:11
• The 60-degree right ADD' gives you $AD_1=AD/2=DC$ Commented Sep 1, 2019 at 17:14
• Got it. I'd forgotten $AB=AD$. Commented Sep 1, 2019 at 18:09
• I've just looked at the answers and it is saying $8$ $cm$. Commented Sep 1, 2019 at 18:53
• it can be easily seen that triangle BCM is equilateral and its heights MN and $CC_1$ are equal. that is MN=3 Cm. Commented Sep 1, 2019 at 19:54