Let N be midpoint of AC. Then BDMN cyclic. Suffices to prove AD.AC=2AD.AM=2AK.AB=AJ.AB=AO.AM <=> ODCM cyclic. From above AJ.AB=AD.AC => AJD=BCD => AOD-MAB=MCD-MCB=> AOD=MCD=>OMCD cyclic.

Wlog let $AB>AC$. Denote by $E$ the midpoint of side $AC$ and $S=BD\cap (ABC)$. We have, $\angle AEM=90=180-90=\angle MBS=180-\angle MBD$. So $BDEM$ is cyclic. Thus $\angle AEB=180-\angle BED=180-\angle BKD=\angle AKD$. So $\triangle AKD\sim \triangle AEB$ and therefore $\triangle AJD\sim \triangle ACB$. In particular $JDBC$ is cyclic. So, $\angle OMB=\angle AMB=180-\angle C=180-\angle OJB$. So $JBMO$ is cyclic and we are done.

Can anyone gives us a nice shape???

Let $AM$ intersect $(BDM)$ at E Let $F$ be midpoint of $AD$ Because $\angle AED=90$ we have $FA=FE=FD$ Let $\angle ABD=y=\angle DBC$ $\angle FED=90-\angle CAM=90-\angle CBM=90-(90-\frac{y}{2})=\frac{y}{2}=\angle KBD=180-\angle KED$ $\implies K, E, F$ collinear $KF\parallel DJ$ because $KA=KJ$ and $FA=FD$ $\implies EA=EO$ By power of a point $AK\cdot AB=AE\cdot AM$ $2\cdot AK\cdot AB=2\cdot AE\cdot AM$ $AJ\cdot AB=AO\cdot AM$ $\implies JBMO$ cyclic

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Let N be midpoint of AC. it is known that BMED is cyclic. ∠JBD = ∠KED = ∠JCA ---> JBCD is cyclic. ∠AJD = ∠BCA = ∠BMA ---> JBMO is cyclic.

Here is a very quick purely angle chasing solution without PoP or similar triangles. Clearly, $BD$ and $BM$ are internal and external bisectors of $\angle ABC$, so $\angle DBM = 90^{\circ}$. Hence $\angle DNM = 90^{\circ}$ and since $AM = MC$, we deduce that $N$ is the midpoint of $AC$. So $\angle JBD = \angle KND = \angle DCJ$ (the latter sine $KN$ is a midsegment in $ACJ$) and so $JBCD$ is cyclic. Therefore $\angle AJD = \angle BCD = \angle BMO$ and $JBMO$ is cyclic.