Let $D$ be the circumcenter of $\triangle ABC$; note that $D$ lies on the perpendicular bisector of each side of $\triangle ABC$. Drop the perpendicular from $C$ onto $\overline{AB}$ at $E$.
Since $\overline{CE}$ and $\overline{DM}$ are perpendicular to $\overline{AB}$, they are parallel.
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In the diagram on the left:
Drop the perpendicular from $M$ onto $\overline{A_1A_2}$ at $F$.
Since $\overline{CA}$ and $\overline{FM}$ are perpendicular to $\overline{A_1A_2}$, they are parallel. Therefore, the right $\triangle MFD$ is similar to the right $\triangle CEA$. Thus, $ \frac{\overline{FM}}{\overline{DM}}=\frac{\overline{CE}}{\overline{CA}}\tag{1} $ Since $\overline{A_1A_2}=\overline{CA}$, we get that the area of $\triangle A_1A_2M$ is $ \tfrac12\overline{A_1A_2}\;\overline{FM}=\tfrac12\overline{CA}\;\overline{FM}=\tfrac12\overline{DM}\;\overline{CE}\tag{2} $ In the diagram on the right:
Drop the perpendicular from $M$ onto $\overline{B_1B_2}$ at $G$.
Since $\overline{CB}$ and $\overline{GM}$ are perpendicular to $\overline{B_1B_2}$, they are parallel. Therefore, right $\triangle MGD$ is similar to right $\triangle CEB$. Thus, $ \frac{\overline{GM}}{\overline{DM}}=\frac{\overline{CE}}{\overline{CB}}\tag{3} $ Since $\overline{B_1B_2}=\overline{CB}$, we get that the area of $\triangle B_1B_2M$ is $ \tfrac12\overline{B_1B_2}\;\overline{GM}=\tfrac12\overline{CB}\;\overline{GM}=\tfrac12\overline{DM}\;\overline{CE}\tag{4} $
Thus, the areas of $\triangle A_1A_2M$ and $\triangle B_1B_2M$ are both $\tfrac12\overline{DM}\;\overline{CE}$.