$ABCD$ is a trapezium, $MN=\dfrac{AB-CD}{2}$ where $M,N$ are the midpoints of $AB,CD$, respectively; show $\angle BAD+\angle ABC=90^\circ$

Question

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

Answer 1

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, $\angle BAD+\angle ABC=90^\circ$.

Answer 2

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.

Answer 3

Let $P$ and $Q$ be mid-points of $AC$ and $BD$ respectively.

Thus, since $PQ=MN$, we see that a parallelogram $PNQM$ is a rectangle, which ends a proof.