Question about isosceles trapezium ABCD, AB ∥ DC and AD = BC. Question about isosceles trapezium ABCD, AB ∥ DC and AD = BC. AC and BD intersect at M. ∠AMB = 60. P, Q, R are the midpoints of MA, MD, BC respectively. Prove that ∆PQR is equilateral.
I tried solving using midpoint theorem, but I am getting stuck. Please guide.

 A: Connect mid point of AB (E) to that of DC(H).Draw circucircle of triangle PQR, it intersect MB and MC at N and T respectively. We have:
$$\angle AMB=\frac{\overset{\large\frown}{PN}+\overset{\large\frown}{QT}}2=60^o$$
$$\Rightarrow \overset{\large\frown}{PN}+\overset{\large\frown}{QT}=120^o$$
Since trapezoid is symmetric about EH, so we have:
$$\overset{\large\frown}{NT}=\overset{\large\frown}{PQ}=120^o$$
Which gives:
$$\angle PRQ=60^o$$
Now draw diameter RE, it bisects angle PRQ because E is point where DA is tangent to the circumcircle. therefore:
$\overset{\large\frown}{EP}=60^o$
$\Rightarrow \overset{\large\frown}{PR}=120^o $
Which gives:
$$\angle PQR=60^o$$
Hence triangle PQR is equilateral.
A: Here is a solution using complex numbers geometry.
We start from the fact that triangle $MAB$ and, as a consequence $MDC$ are equilateral triangles. We will consider WLOG that their resp. sidelengths are $1$ and $s$.
Let us take $M$ as the origin and the parallel to lines $AB$ and $DC$ passing through $M$ as the $x$ axis (the $y$ axis being the line passing through midpoints of $AB$ and $DC$, resp.)
Let us use complex numbers representation associating the lowercase letters to the uppercase letters attributed to points), with $$w:=b=e^{i \pi/3}.$$
To points $A,D,C,P,Q,R$ resp. we associate:
$$a=w^2, d=sw^4, c=sw^5, p=\tfrac12w^2, q=\tfrac12sw^4, r=\tfrac12(w+sw^5)$$
Triangle $PQR$ is equilateral iff the $\pi/3$ rotation centered in $P$ maps $Q$ to $R$, i.e.,
$$\mathscr{R}_{\pi/3}(\vec{PQ})=\vec{PR} \ \iff \ w(q-p)=r-p$$
This relationship is equivalent to:
$$w(\tfrac12sw^4-\tfrac12w^2)=\tfrac12(w+sw^5)-\tfrac12w^2 \ \iff \  w=1+w^2   $$
which is true.
