Midpoint of $O_1O_2$ coincides with the center of a circumcircle

Question

Two circles $O_1$ and $O_2$ intersect at $A$ and $B$. Through $A$ a line cuts $O_ 1$ at $C$ and $O_2$ at $D$. Denote $M$ is the midpoint of $CD$.

Prove that the center of circumcircle of triangle $ABM$ bisects $O_1O_2$.

Answer 1

Preliminary : Let $O$ be defined as the midpoint of $O_1O_2$.

Let us reformulate the question into the more natural one :

Prove that $M$ belongs to the circle $(C')$ with center $O$ and radius $OA$.

This issue can be treated by 2 methods 1) and 2) :

1) a 3D understanding method : Consider the 2 figures :

The first one looks very similar (at least for a part of the line segments) to the initial figure. The second one is a 3D "expansion " of the first figure where these line segments become generating lines of a [one sheeted hyperboloid] (https://mathworld.wolfram.com/One-SheetedHyperboloid.html) intersected by a horizontal plane at half distance between the bottom and the top circles.

2) Using powers of points with respect to circles. The set of points such that the sum of powers of $M$ with respect to $(C_1)$ and $(C_2)$ is equal to $k$ is a circle (see here). It is not difficult to see that circle $(C')$ is the circle of this family associated with constant $k=0$.