Solution:

$E$ can be inside, or outside the circle (Figure 3) but the proof below holds in both cases; notice that $E$ cannot be on the circle as $A, B, C, D$ are distinct. Let lines $AC$ and $NE$ meet at point $P$. Then $EN = DN = BN$ (median in a right triangle), so $\angle PEC = \angle NED = \angle NDE = \angle BDC = \angle BAC = \angle EAP$. Now $AB \perp CD$ so $EN \perp AC$. But $OM \perp AC$ so $OM \parallel EN$. Similarly $ON \parallel EM$ so $NEMO$ is a parallelogram (possibly degenerated). As $MN \perp OE$, this parallelogram is a rhombus. Then the chords $AC$ and $BD$, being equidistant from $O$, are equal. Hence their minor arcs are equal, which means that either $AD \parallel BC$ or $AB \parallel CD$; the latter contradicts the fact that $AB$ and $CD$ meet at $E$.