Problem: Let $ABCD$ be a convex quadrilateral
with $\angle DAB = \angle BCD = 90\dg$ and $\angle ABC > \angle CDA$.
Let $Q$ and $R$ be points on segments $BC$ and $CD$,
respectively, such that line $QR$ intersects lines $AB$ and $AD$
at points $P$ and $S$, respectively.
It is given that $PQ=RS$.
Let the midpoint of $BD$ be $M$ and the midpoint of $QR$ be $N$.
Prove that the points $M$, $N$, $A$ and $C$ lie on a circle.

Solution: The condition is equivalent to $N$ being the midpoint
of both $\ol{PS}$ and $\ol{QR}$ simultaneously.
(Thus triangles $BAD$ and $BCD$ play morally dual roles.)

\begin{center}
\begin{asy}
import graph; size(12cm);
pen zzttqq = rgb(0.6,0.2,0.); pen cqcqcq = rgb(0.75294,0.75294,0.75294);
draw((-1.83576,5.20298)--(-4.,1.5)--(-0.44779,-2.69232)--(4.5,1.5)--cycle, linewidth(0.6) + zzttqq);

draw(circle((0.25,1.5), 4.25), linewidth(0.6));
draw((-5.40339,-0.90118)--(6.45435,0.35776), linewidth(0.6));
draw((-1.83576,5.20298)--(-4.,1.5), linewidth(0.6) + zzttqq);
draw((-4.,1.5)--(-0.44779,-2.69232), linewidth(0.6) + zzttqq);
draw((-0.44779,-2.69232)--(4.5,1.5), linewidth(0.6) + zzttqq);
draw((4.5,1.5)--(-1.83576,5.20298), linewidth(0.6) + zzttqq);
draw((-4.,1.5)--(-5.40339,-0.90118), linewidth(0.6));
draw((4.5,1.5)--(6.45435,0.35776), linewidth(0.6));
dot("$D$", (4.5,1.5), dir(40));
dot("$B$", (-4.,1.5), dir(180));
dot("$M$", (0.25,1.5), dir(90));
dot("$A$", (-1.83576,5.20298), dir(130));
dot("$C$", (-0.44779,-2.69232), dir(270));
dot("$Q$", (-2.24919,-0.56630), dir(80));
dot("$R$", (2.67884,-0.04308), dir(135));
dot("$P$", (-5.40339,-0.90118), dir(270));
dot("$S$", (6.45435,0.35776), dir(270));
dot("$N$", (0.21482,-0.30469), dir(80));
\end{asy}
\end{center}

The rest is angle chasing.
We have
\begin{align*}
  \dang ANC &= \dang ANP + \dang QNC \\
  &= 2\dang ASP + 2\dang QRC \\
  &= 2 \dang DSR + 2 \dang DRS = 2 \dang RDS \\
  &= 2 \dang ADC = \dang AMC.
\end{align*}