Given the equilateral triangle $ABC\ :\ A(-a,0),B(a,0),C(0,\sqrt{3}\ a)$. 1. Point $M$ is the incenter, $M$ is the circumcenter: $AM=MB$ and $AM=MC$. 2. Point $M$ lies on the circumcircle, $\angle BAM=30^{\circ}$ : $AM=AB$ and $AM=AC$. 3. Point $M$ is a vertex of the paralellogram $ABMC\ :\ AB=BM=MC=CA$.