<Qua-2025-759>, hint

Triangle AOD is equilateral, <ASD=60. Hence AOSD is cyclic.

<AOL=90, DA=DO=DL, SO=SD. Line BML cuts DC in N, then  2NC =CD=BA. Thus

AM/MC=2. MC=AC/3. Hence OM/MC=1/2. Triangle SOC, <OCS=30, CSO=60

Hence SO=SC/2=SD. Thus DS/SC=1/2=OM/MC.

Therefore MS//BD//CL