let $G$ be the antipode of $B$ so it suffices to show that $D,E,G$ are collinear which is clear since simple angle chase leads to $\angle GDC=\angle EDC$.

A bit lengthier than the above solution. We let $F = (OCD) \cap \overline{BC} \neq C$. Now, we have the following key claim. Claim : Lines $BC$ and $OE$ are parallel. Also, lines $OF$ and $DE$ are parallel. Proof : Simply note that, \[2\measuredangle OEA = 2\measuredangle ODC = \measuredangle DOC = 2\measuredangle DAC = 2(90+\measuredangle BCA)= 2\measuredangle BCA\]from which it is clear that $\measuredangle OEA = \measuredangle BCA$ and thus, $BC \parallel OE$ as desired. Also, \[\measuredangle OFC = \measuredangle ODC = \measuredangle DCO = \measuredangle DEO\]from which using the previous result it follows that $OF \parallel DE$, proving the claim. Observe that from this it follows that $OECR$ and $OEDR$ are isosceles trapezoids. Now, we also note the following. Claim :$\triangle BFD$ is isosceles. Proof : Note that, \[\measuredangle CFD = \measuredangle COD = 2\measuredangle CBD = 2\measuredangle FBD\]from which the claim is clear. Now, let $M' = \overline{OF} \cap \overline{BE}$. Since $\triangle OM'E \sim \triangle FM'B$ due to the previously proved parallel lines, we have that \[\frac{BM'}{M'E}=\frac{BR}{OE}=\frac{BR}{RD}=1\]from which it is clear that $M'$ must be the midpoint of $BE$. Thus, $M'=M$ and we conclude that $DE$ is parallel to $OM$ as desired.