Let $ABCD$ be a cyclic quadrilateral and let $AB$ and $CD$ meet at $E$. Let $M= (EBC)cap (EAD)$. Prove that $OMperp EM$

You are very close!

Just note that M is the spiral center of the spiral similarity $S$ sending $AB$ to $DC$ . And hence the spiral similarity $S$ also take the midpoint of $AB$ to midpoint of $DC$.

So $S:M_1 rightarrow M_2 $

So $S:BM_1 rightarrow CM_2$ .

So $M$ is the spiral center of the spiral symmetry which takes $BM_1$ to $CM_2$.

But notice that $BM_1cap CM_2=E implies M =(EBC) cap (EM_1M_2)$

So $M in (EM_1M_2)$ and by your observation, we get $Min (EM_1OM_2)$ , and hence we have $OMperp EM$.

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Here M is called the miquel point and if we define $F=BCcap DA$ , then we have $Min EF$ if $ABCD$ is cyclic .