if possible can you draw it to see clearly the exercise thank you very much.

Let $OM$ and $ON$ cut line segment $BC$ at $E$ and $F$ respectively.And let $P$ and $Q$ be projections of $O$ to $AB$ and $CD$ respectively.$AB=CD$ and $ABCD$-cyclic implies that $ABCD$ is isosceles trapezoid.$BL=CL \implies AL=DL \implies \angle BAD=\angle CDA = \angle DCB = \angle ABC$. $OP \perp AB$ and $OQ \perp CD$ implies that $AP=PB$ and $QD=QC;AL=DL$ implies that $PL=QL \implies \angle LPQ = \angle LQP= \angle ABC$ which implies that $BC || PQ ;OL \perp PQ \implies \angle MOP= \angle NOL$ and $OL \perp BC \implies \angle DCB = \angle NOE \implies CENO$-cyclic and similarly we get that $BFMO$ is cyclic and from $OB=OC$ we get that $\angle OBC= \angle OCB \implies \angle OCN =\angle OBM=\angle OEN=\angle OFM$ which implies that $MEFN$ is cyclic $\implies \angle OFE=\angle OMN$ and $\angle MOP=\angle NOL$ implies that $\angle OFE=\angle OMP \implies \angle OMP=\angle OMN \implies \angle OMX= \angle OMB$ if $MN$ cuts circle at $X$ and $Y$. Let angle bisector of $\angle BOX$ cut $XM$ at $K$ and let $WLOG$ $K$ lies between $X$ and $M \implies \angle OMX+\angle KOM =\angle OKX=\angle OKB$ $\implies \angle OKB >\angle OMX=\angle OMB$ which is contradiction thus we get that $M=K \implies MB=MX \implies AXBY$ is isosceles trapezoid which implies that $XY=AB$. _____________________________________________________ by Elshad Mustafayev[Sumgayit]

It's enough to prove that $O$ is center of the excircle of $LMN$ and this is obvious because $2MON=ALC$ and $DLO=OLA$