Note that $AG:GL:LE = 2:1:1$ by well-known centroid ratios. This means that $L$ is the midpoint of both $BC$ and $GE$, so that $BGCE$ is always a parallelogram. The following angle chase solves the if part, since triange $BMG$ is isosceles: \[\angle BMC = \angle BMG = \angle BGM = 180^{\circ} - \angle BGC = 180^{\circ} - \angle BEC \]And the following angle chase solves the only if part: \[\angle BGM = 180^{\circ} - \angle BGC = 180^{\circ} - \angle BEC = \angle BMC = \angle BMG \]