Let $ABC$ be an equilateral triangle inscribed in circle $O$. $M$ is a point on arc $BC$. 
Lines $\overline{AM}$, $\overline{BM}$, and $\overline{CM}$ are drawn. Then $AM$ is: 

$\textbf{(A)}\ \text{equal to }{BM + CM}\qquad  \textbf{(B)}\ \text{less than }{BM + CM}\qquad \\ \textbf{(C)}\ \text{greater than }{BM+CM}\qquad \\ \textbf{(D)}\ \text{equal, less than, or greater than }{BM + CM}\text{, depending upon the position of } {M}\qquad  \\ \textbf{(E)}\ \text{none of these}$
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 $\triangle ABC$ is equilateral, we cancel out $AB$, $AC$, and $BC$ to get that \[BM + CM = AM \implies \boxed{\textbf{(A)}}.\]