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 $KN$?