$ABCD$ 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 $\angle N_1NN_2=90^\circ$?


Extend $AD$ and $BC$ to meet at $P$.  We are given that $m=x-y$.  Since $CD\parallel AB$, $\triangle PDN\sim\triangle PAM$ and thus
$$\frac yn=\frac x{n+m}=\frac x{n+x-y}\\nx=y(n+x-y)=yn+xy-y^2\\nx-xy=yn-y^2\\x(n-y)=y(n-y)\\x=y\quad\text{or}\quad n-y=0$$
Since generally $x\neq y$, we conclude that $n-y=0$.  Thus $y=n$ and $m+n=(x-y)+y=x$.  
Therefore, $2PM=AB$, so $M$ is the circumcenter of $\triangle PAB$.  This is only possible in a right triangle, so $\angle P=90^\circ$ and thus $\angle A+\angle B=90^\circ$ as desired.
