$\measuredangle XCE=\measuredangle XBE$ and $\measuredangle XDE=\measuredangle XAE$, so $\triangle AXC \sim \triangle DXB$ but as $M,N$ are corresponding midpoints in the triangles, we have $\measuredangle DNX= \measuredangle AMX$ as desired.

Let the circumcircles of $ADE$ and $BCE$ meets again at $Y \neq E$, and $AB \cap CD=P, AD\cap BC=Q$.$O$ be the circumcenter of quadrilateral $ABCD$. It is well-known that $E$ is the orthocenter of $\triangle OPQ$ and $X,Y$ are perpendicular foot from $P,Q$ to $OQ,OP$ ,respectively. Hence $OEXY$ are cyclic and $OE$ is the circle’s diameter. Also , $\measuredangle OMA = \measuredangle OND =\frac{\pi}{2}$ ,so $OEXYMN$ are cyclic, we are done.

it isnt given that $ABCD$ has to be cyclic, so it has no circumcenter.