Saudi Arabia Mathematical Competitions

We have $\widehat{SAC}=\widehat{SCA}=30^{\circ }$, so $SA=SC$. Since $OA=OC$, we get $SO\perp AC$, hence $LA=LC$. Moreover $\widehat{LAC}=60^{\circ }$, so triangle $ALC$ is equilateral.



Point $S$ is the centroid of $△LAC$, thus $\frac{LS}{SO}=2$. $DBCL$ is a parallelogram, hence $DQ=QC$. We get that $M$ is the centroid of $△DBC$, so $\frac{CM}{MO}=2$.

Finally, since $\frac{LS}{SO}=\frac{CM}{MO}$, we obtain that $SM\parallel CL$.