Complex bash with $\omega$ as the unit circle, so that $$|a|=|b|=|c|=1$$$$m = -b$$$$d = -\frac{ab}c$$$$e = \frac{ac(m+d) - md(a+c)}{ac-md} = \frac{b(a+c)}{b-c}$$$$o = \frac{mc(e\overline{e}-1)}{e-m-c+\overline{e}mc} = \frac{bc}{b-c}$$Let line $OM$ intersect $\omega$ again at $P$, so that $$p = \frac{m-o}{m\overline{o}-1} = -\frac{b^2}c$$$$j = \frac{bc(m+p) - mp(b+c)}{bc-mp} = \frac{b(b+c)}{b-c}$$Then $$\overline{\jmath} = -\frac{b+c}{b(b-c)} = \frac{j}{bm}$$so $J$ lies on the perpendicular bisector of $\overline{BM}$. $\blacksquare$

First of all, $\angle JCM = 180 - \angle MCB = 90 \implies J$ lies on $(ECM)$. Since $AM$ and $CD$ are both perpendicular to $AB$, $AMDC$ must be an isosceles trapezoid. It follow that $ED = EC$. We will now show that $MJB$ and $DEC$ are similar, from which the desired statement will follow. $\angle DEC = \angle MEC = \angle MJC = \angle MJB$, $\angle ECD = \angle ACD = \angle MDC = \angle MBC = \angle MBJ$. This solves the problem

Solution : Claim 1 : $J$ lies on $ (EMC) $ Proof : $ \angle MAB = \angle MCB = 90^\circ. \angle MCJ = 180^\circ - \angle MCB \Rightarrow \angle MCJ = 90^\circ \Rightarrow J $ lies on $ (EMC) $. Now consider the center of $ (MAB) $ as point $ Q $. Claim 2 : $ Q $ lies on $ (EMC) $ Proof : It's enough to prove that $ MECQ $ is cyclic quadrilateral. For this, we show that $ \angle BQC = \angle MEC $. We know that $ AM $ is parallel to $ CD \Rightarrow \angle ACD = \angle MDC = \angle EMA = \angle EAM $ and we know that $ \angle MDC = \angle MBC $. Note that $ QC = QB $(Circle radii) $ \Rightarrow \angle MBC = \angle QBC = \angle QCB $. So $ \angle BQC = 180^\circ - 2(\angle QBC) = \angle MEC \Rightarrow Q $ lies on $ (EMC) $. Now $ \angle JQM = \angle JCM = 90^\circ $ and $ MQ = BQ $ solves the problem.

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Claim: $J$ is the antipode of $M$ Re-define $J$ as the antipode of $M$ with respect to $(EMC)$. Then $J$, $O$, and $M$ are collinear and $J$, $B$, and $C$ are collinear as $JC\perp CM$ and $CM\perp CB$. Claim: $N$, the midpoint of $MB$, lies on $EMC$ This is just an angle chase $$\angle MEC=180^{\circ}-\angle ECD-\angle EDC=180^{\circ}-\angle MNC$$Claim: $JB=JM$ This follows from $JN\perp MB$.

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