my solution: let $N$ midpoint of $BC$ note that $\angle DBM=\angle MND=90\longrightarrow BKDN$ is cyclic so $AD\cdot AN=AK\cdot AB\longrightarrow AD\cdot AC=AB\cdot AJ$ so $BDCJ$ is cyclic quadrilateral and we have $\angle AJD=\angle ACB=\angle BNA$ so $JBMO$ is cyclic. DONE

Attachments:

Dear Mathlinkers, 1. KN//CJ 2. according to a converse of the Reim theorem, B, D, C, J are concyclic 3. according to the pivot theorem, we are done... Sincerely Jean-Louis

Nice problem, and just realised that my solution somewhat matches to the above solutions, but since I worked a bit hard on it, I am gonna post Since, $M$ is the mid-point of arc $CBA$, therefore, $\angle AMN=\angle CMN=\frac{1}{2}\angle AMC=\frac{B}{2}$ and, $\angle BMA=\angle BCA=C$ $\implies \angle DBM=\angle DNM=90^{\circ} \implies BDNM \text{ is cyclic}$, Let $MN \cap JC =X$ Since, $AK=KJ \text{ and } AN=CN \implies KN||JC$, Hence, $$\angle KNM =180^{\circ}-\angle ABM=\angle JXM \implies BJXM \text{ is cyclic } \implies \angle ACJ=\frac{B}{2} \implies BJDC \text{ is cyclic }$$Therefore, $\angle BJO=180^{\circ}-C=180^{\circ}-\angle BMO \implies JBMO \text{ is also cyclic }$

AZE JBMO TST wrote: There is a triangle $ABC$ that $AB$ is not equal to $AC$.$BD$ is interior bisector of $\angle{ABC}$($D\in AC$) $M$ is midpoint of $CBA$ arc.Circumcircle of $\triangle{BDM}$ cuts $AB$ at $K$ and $J,$ is symmetry of $A$ according $K$.If $DJ\cap AM=(O)$, Prove that $J,B,M,O$ are cyclic. Solved with A-Thought-Of-God First, let $N$ denote the midpoint of minor $\overarc{AC}$ and $E$, the midpoint of $AC$. Claim. $BMED$ is cyclic. Proof. By Fact 5 we know that $B, D, N$ are collinear, and since $\{M,N\}$ are reflections over circumcenter of $ABC$, so $B, D, N$ are collinear as well. Thus, $$\angle NBM = \angle DBM = 90^{\circ} = \angle MED. \quad \square$$ Claim. $BJDC$ is cyclic. Proof. Now, by PoP and our above claim, we have: \begin{align*} &AK \cdot AB=AD \cdot AE\\ \implies &(2AK) \cdot AB=AD \cdot (2AE) \\ \implies &AJ \cdot AB=AD \cdot AC. \quad \square \end{align*}Now, the angle chase!! We note that $$\measuredangle BJO = \measuredangle BJD = \measuredangle DCB = \measuredangle ACB = \measuredangle AMB = \measuredangle OMB \implies \text{J, B, M, O are cyclic}. \quad\blacksquare$$

Trivial by complex bash (will edit tmr) Let $(I)$ be the unit circle s.t $a=x^2$,$b=y^2$ and $c=z^2$ $D=BF\cap AC =>d=-\frac{y^2(x^2+z^2)-xyz(y-z)}{y^2+xz}$ Since $2I=0=m+f$ =>$m=-f=-(-xz)=xz$ $K$ is a points such that $A,B,K$ are collinear and $BMDK$ is harmonic.From here get the equation for $k$ then $j=2k-a$.Then $DJ\cup AM=O\neq I$ and proving $JBMO$ is cyclic.Just a bit patience required...$\boxed{\lambda }$