We have trapezoid $ABCD$ $(AB||CD)$. $\angle BAD + \angle ABC = 90^\circ$. $M,N,P$ and $Q$ are the midpoints of $AB,CD,AC$ and $BD$, respectively. $AD$ intersects $BC$ in $K$. Are $M,N,K$ collinear?
I tried to make the graph in GeoGebra, but I didn't succeed. I am sorry. $\angle BAD+\angle ABC=90^\circ$ is the same as $\angle CDK+\angle DCK=90^\circ$, thus $\angle AKB=90^\circ$. How to show that $M$ lies on $KN$?
$P,Q$ and the angles $ABC,BAD$ are irrelevant (except insofar as they guaraantee that the lines $AD,BC$ intercept).
Since $AB$ is parallel to $DC$ the triangles $KAB,KDC$ are similar. $N$ is the midpoint of $DC$ and $M$ is the midpoint of $AB$, so $K,N,M$ are collinear. [An expansion centre $K$, by a factor $KA/KD$ takes $N$ to $M$.]