See attachment for my solution

Let $[AB] = 2a, [BC] = 2b, [CD] = 2c, [DA] = 2d$ and WLOG, $a < c$. Let $(O_1,r_1), (O_2, r_2)$ be circumcircles of N-, M-isosceles $\triangle ABN, \triangle CDM$. Putting coordinate origin at midpoint $G$ of $[MN]$, centroid of trapezoid $ABCD$, and positive x-axis along ray $(AD$, line $BC$ and perpendicular to $BC$ through $G$ have equations: $y = \frac{c-a}{d} \cdot x + \tfrac{1}{2}(c+a)$ and $y = -\frac{d}{c-a} \cdot x$, respectively. x-coordinate of foot $K$ of perpendicular from $G$ to $BC$ is then $x_K = -\frac{d}{2} \cdot \frac{c^2 - a^2}{d^2 + (c-a)^2} = -\frac{d}{2} \cdot \frac{c^2 - a^2}{b^2}$. Circumcircle $(O_1,r_1) \equiv \odot(ABN)$ has equation $(x+r_1)^2 + \left(y+\tfrac{1}{2}(c-a)\right)^2 = r_1^2$. Substituting equation of $BC$ for $y$ yields quadratic equation for x-coordinates of $A, Q$: $x^2 + 2r_1 \cdot x + \left(\frac{c-a}{d} \cdot x + c\right)^2 = 0$ $\implies$ $x^2 + (...) \cdot x + \frac{c^2d^2}{b^2} = 0$. Since x-coordinate $x_A = -d$ of $A$ is known, x-coordinate of $Q$ is the other root, or $x_Q = -\frac{dc^2}{b^2}$. x-coordinate of $R$ is obtained similarly, or just by replacing $c \longleftrightarrow a$ and $-d \longleftrightarrow +d$ in formula for $x_Q$ $\implies$ $x_R = +\frac{da^2}{b^2}$. x-coordinate of midpoint of $[QR]$ is then $\tfrac{1}{2}(x_Q + x_R)= -\frac{d}{2} \cdot \frac{c^2 - a^2}{b^2} = x_K$ $\implies$ $K$ is midpoint of $[QR]$ $\implies$ $[GQ] = [GR]$.

There is a more geometrical approach once you note that midpoint of $\overline{MN}$ is in fact mid point of $\overline{UV}$ where $U$ is midpont of $\overline{AD}$ and $V$ is midpoint of $\overline{BC}$.

Okay, so I guess it's safe to discuss. [asy][asy] /* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(11cm); real lsf=1.0000; real lisf=2011.0; defaultpen(fontsize(10pt)); real xmin=-6.42; real xmax=6.00; real ymin=-1.70; real ymax=4.87; /* Initialize Objects */ pair A = (-3.5, 3.0); pair B = (0.0, 3.0); pair C = (1.5, -1.0); pair D = (-3.5, -1.0); pair M = midpoint(A--C); pair N = midpoint(B--D); pair Q = (2)*(foot(circumcenter(N,A,B),B,C))-B; pair R = (foot(circumcenter(M,D,C),B,C))*(2)-C; pair K = midpoint(M--N); pair L = midpoint(B--C); pair T = foot(B,C,D); pair P = midpoint(R--Q); /* Draw objects */ draw(A--B, rgb(0.0,0.8,0.2)); draw(B--C, rgb(0.0,0.8,0.2)); draw(C--D, rgb(0.0,0.8,0.2)); draw(D--A, rgb(0.0,0.8,0.2)); draw(A--C); draw(B--D); draw(circumcircle(A,B,N), rgb(0.0,0.2,0.6)); draw(circumcircle(C,D,M), rgb(0.0,0.2,0.6)); draw(Q--B); draw(L--midpoint(A--D)); draw(K--P); draw(B--T); /* Place dots on each point */ dot(A); dot(B); dot(C); dot(D); dot(M); dot(N); dot(Q); dot(R); dot(K); dot(L); dot(T); dot(P); /* Label points */ label("$A$", A, lsf * dir(135)); label("$B$", B, lsf * dir(60)); label("$C$", C, lsf * dir(35)); label("$D$", D, lsf * dir(135)); label("$M$", M, lsf * dir(45)); label("$N$", N, lsf * dir(135)); label("$Q$", Q, lsf * dir(60)); label("$R$", R, lsf * dir(60)); label("$K$", K, lsf * dir(120)); label("$L$", L, lsf * dir(60)); label("$T$", T, lsf * dir(45)); label("$P$", P, lsf * dir(60)); /* Clip the image */ clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $AB = 2x$, $CD=2y$, and assume without loss of generality that $x < y$. Let $L$ be the midpoint of $BC$ and denote $BC = 2\ell$. Let $P$ be the midpoint of $QR$. Let $T$ be the foot of $B$ on $DC$. Since $N$ is the midpoint of the hypotenuse of $\triangle ABD$, it follows that $AN = BN$. Since $MN \parallel AB$, we see that $MN$ is tangent to $(ABN)$. Similarly, it is tangent to $(BCM)$. Noting that $LM = %Error. "half" is a bad command. AB$ via $\triangle ABC$, we obtain \[ LR \cdot LC = LM^2 = \left( %Error. "half" is a bad command. AB \right)^2 = x^2 \implies LR = \frac{x^2}{\ell} \]Similarly, $LQ = \frac{y^2}{\ell}$. Then, \[ PL = \frac{LQ-LR}{2} = \frac{y^2-x^2}{2\ell} \text{ and } KL = \frac{ML+NL}{2} = x+y. \]But then, we find that \[ \frac{KL}{PL} = \frac{\frac{y^2-x^2}{2\ell}}{x+y} = \frac{y-x}{2\ell} = \frac{TC}{BC} \]Combined with $\angle KLP = \angle BCT$, we find that $\triangle KLP \sim \triangle BCT$. Therefore, $\angle KPL = \angle BTC = 90^\circ$. But $P$ is the midpoint of $QR$, so $KQ = KR$.

Such a good solution (sketch): Let $E, F$ be the feet of the perpendiculars from $N, M$ to $BC$ respectively. We will show that $QE=FR$ from which the result follows immediately. We have: \[\frac{EQ}{FR}=(\frac{EQ}{QN})(\frac{QN}{BN})(\frac{BN}{BD})(\frac{BD}{AD})(\frac{AD}{AC})(\frac{AC}{CM})(\frac{CM}{CR})(\frac{RM}{RF})\] \[=(\cos EQN)(\frac{\sin QBN}{\sin BQN})(\frac{1}{2})(\frac{BD}{AC})(2)(\frac{\sin CRM}{\sin MCR})(\frac{1}{\cos MRF})\] \[=(\cos ABD)(\frac{\sin CBD}{\sin ABD})(\frac{BD}{AC})(\frac{\sin ACD}{\sin BCA})(\frac{1}{\cos ACD})\] By Law of sines, we have $\sin CBD=\frac{(CD)(AD)}{(BC)(BD)}$ and $\sin BCA=\frac{(BA)(AD)}{(BC)(CA)}$. Then then plugging this in and simplifying gives: \[\frac{EQ}{FR}=\frac{(AB)(AC)^2(AD)^2(BC)(BD)^2(CD)}{(AB)(AC)^2(AD)^2(BC)(BD)^2(CD)}=1\] So $EQ=FR$ and the result immediately follows.

Line $MN$ is midparallel of $AB \parallel CD,$ cutting $\overline{BC},\overline{AD}$ at their midpoints $U,V,$ respectively. Circles $\odot(ABN),$ $\odot(CDM)$ cut $AD$ again at $G,H,$ respectively. $QG,RH$ meet $UV$ at $Y,Z,$ respectively. $\angle BQG$ and $\angle CRH$ are obviously right, i.e. $QRHG$ is a right trapezoid. $\angle ABN=\angle BAN=\angle ANV$ $\Longrightarrow$ $UV$ is tangent to $\odot(ABN)$ $\Longrightarrow$ $YN^2=YG \cdot YQ =YV \cdot YU.$ Hence, if $N'$ is the reflection of $N$ about $Y,$ it follows that $(U,V,N,N')=-1.$ Likewise, if $M'$ is the reflection of $M$ about $Z,$ we have $(U,V,M,M')=-1.$ Since $M,N$ are symmetric about the common midpoint $E$ of $\overline{MN},\overline{UV},$ then we deduce that $M',N'$ are symmetric about $E$ $\Longrightarrow$ $Y,Z$ are symmetric about $E$ $\Longrightarrow$ $E$ is on midparallel of $QG \parallel RH,$ in other words, $E$ is on perpendicular bisector of $\overline{QR},$ and the conclusion follows.

$Y,L,X$ are midpoints of $AB,MN,DC$ and $E,F,G,H,I$ are projections of $A,D,M,N,L$ on $BC$. since wwe need that $I$ is the midpoint of $RQ$ and $I$ is already the midpoint of $HG$ we only need $QH=GR$. since $\angle GRM=\angle XDM$ and $\angle MXD=\angle MGR=90$ $MGR\sim XDM$ so $RG=DX*MG/MX$. similarly $QH=AY*NH/YN$ since $YN=AD/2=XM$ we only need $XD*MG=AY*NH$ multiplying by $4$ both sides we only need $DC/AB=DF/AE$ swhich is true since $DCF\sim ABE$($\angle ABE=\angle DCF$ and $\angle AEB=\angle DFC=90$)

How is $PL=\frac{LQ-LR}{2}$ in v_Enhance's solution?

This is equivalent to proving that the perpendicular from the midpoint of $MN$ to $BC$ is the midpoint of $QR$. Let $X$ be the foot from $M$ and $Y$ the foot from $N$, then we are just proving $QX = RY$. We compute $QX = \frac{MX}{\tan MQX}$. Let $AM$ meet $BC$ at $Z$, then we see $\angle MQX = \angle MQZ = \angle BAZ = \angle BAM = \angle ABM = \angle ABD$. In addition, $MX = \frac{AC}{2} \sin DBC$. Thus we have $QX = \frac{BD \sin DBC}{2 \tan ABD}$. We can also write $BD = \frac{AD}{\sin ABD} $ We can obtain $RY$ by symmetry. Thus we want to show $\frac{AD \sin DBC \cos ABD}{2 \sin^2 ABD} = \frac{AD \sin ACD \cos ACD}{ 2 \sin^2 ACD}$, or just $\frac{\sin DBC \cos ABD}{\sin^2 ABD} = \frac{\sin ACD \cos ACD}{\sin^2 ACD}$. Now $\frac{CD}{\sin DBC} = \frac{BC}{\sin BDC}$, so we write $\sin DBC = \frac{CD \sin BDC}{BC}$, given $CD = AD \tan CAD$. Thus we finally want to show $\frac{\tan CAD \sin BDC \cos ABD}{BC \sin^2 ABD } = \frac{\tan BDA \sin BAC \cos ACD}{\sin^2 ACD} $. Note that $\angle BAC = \angle ACD, \angle BDC = \angle ABD$, and then we are done since the tans cancel out.