smc

Since quadrilateral $ABMC$ is inscribed in circle $O$, thus it is a cyclic quadrilateral. By Ptolemy's Theorem, $$AC\cdot MB+MC\cdot AB=BC\cdot AM.$$ Because $△ABC$ is equilateral, we cancel out $AB$, $AC$, and $BC$ to get that $$BM+CM=AM⟹\boxed{\text{equal to }BM+CM}.$$