A 'Trap'ezium

A cyclic trapezium is also an isosceles trapezium, and by symmetry the center of the circle must be on the perpendicular bisector of the bases of the trapezium. Since $OM = 2$ , place the trapezium on a coordinate plane so that the center of the circumscribed circle is at the origin and $M$ is at $(0, 2)$ .

Since $\angle AMD = 60°$ , $\angle DMC = 180° - 60° = 120°$ , and since $\triangle DMC$ is an isosceles triangle, $\angle MDC = \angle MCD = \frac{180° - 120°}{2} = 30°$ . This means the slope of $BD$ is $\tan 30° = \frac{\sqrt{3}}{3}$ , and since its $y$ -intercept is $2$ the equation of its line is $y = \frac{\sqrt{3}}{3}x + 2$ . Likewise, the slope of $AC$ is $\tan 150° = -\frac{\sqrt{3}}{3}$ , and since its $y$ -intercept is $2$ the equation of its line is $y = -\frac{\sqrt{3}}{3}x + 2$ .

Let $a$ be the $x$ -coordinate of $B$ . Since $B$ is on $y = \frac{\sqrt{3}}{3}x + 2$ , its coordinates are $(a, \frac{\sqrt{3}}{3}x + 2)$ . Using the distance between $B$ and the origin gives the radius $r$ of the the circle, so $r^2 = a^2 + (\frac{\sqrt{3}}{3}x + 2)^2$ , which makes the equation of the circle $x^2 + y^2 = a^2 + (\frac{\sqrt{3}}{3}x + 2)^2$ .

$C$ is on both the circle $x^2 + y^2 = a^2 + (\frac{\sqrt{3}}{3}x + 2)^2$ and on the line $y = -\frac{\sqrt{3}}{3}x + 2$ , and combining these equations to eliminate $y$ gives an $x$ -coordinate of $x = a + \sqrt{3}$ .

By symmetry, if $a$ is the x-coordinate of $B$ , then $AB = 2a$ , and if $a + \sqrt{3}$ is the $x$ -coordinate of $C$ , then $CD = 2a + 2\sqrt{3}$ . Therefore, $|AB - CD| = |2a - (2a + 2\sqrt{3})| = 2\sqrt{3}$ , so $m = 2$ , $n = 3$ , and $m + n = \boxed{5}$ .

As the trapezium is cyclic, it is also isosceles.

Let $X$ and $Y$ be midpoints of diagonals $AC$ and $BD$ respectively.

$XY$ intersects $OM$ at $L$

Join $OX$ and $OY$ .

$OX$ and $OY$ are perpendicular to $AC$ and $BD$ ( Why? )

Angle $XMY=120°$ , So angle $XMO=$ Angle $YMO=60°$ (By symmetry)

In $∆OXM$ , $sinM=sin60°=\frac{OX}{2}=\frac{\sqrt{3}}{2}$

$OX=\sqrt{3}$ , so, $XM=1$

$OM$ is perpendicular to $XY$ (By symmetry)

So, we get $XL=YL=\frac{\sqrt{3}}{2}$ , so, $XY=\sqrt{3}$

We know that $|AB-CD|=2×XY$

So, answer us $2\sqrt{3}$

$n+m=2+3=\boxed{5}$