$ABCD$ is a trapezium, $MN=\dfrac{AB-CD}{2}$ where $M,N$ are the midpoints of $AB,CD$, respectively; show $\angle BAD+\angle ABC=90^\circ$ 
$ABCD$ is a trapezium in which $AB$ is parallel to $CD$. Let $M$ be the midpoint of $AB$ and $N$ the midpoint of $CD$. If $MN=\dfrac{AB-CD}{2}$, I should show that $\angle BAD+\angle ABC=90^\circ$.

Let $KP$ be the midsegment of $ABCD: K\in AD$ and $P\in BC$. We know $KP=\dfrac{AB+CD}{2}$. If $MN$ intersects $KP$ at $X$, $X$ is the midpoint of $MN$ and $KP$. How can I continue? Does this help?
Edit: I have another idea. Let $NN_2||AD$ and $NN_1||BC$. Can we show $\angle N_1NN_2=90^\circ$?

Let $NN_2$ $||$ $AD$ and $NN_1$ $||$ $BC$, then $AN_2ND$ and $NN_1BC$ are parallelograms. If we look at $\triangle N_2N_1N$, $N_2M=MN_1=NM \Leftrightarrow \triangle N_2N_1N$ is a right triangle. Therefore, $\angle BAD+\angle ABC=90^\circ$.
