Given a circle $(O)$ with chord $AB$. The perpendicular bisector of $AB$ intersects the circle $(O)$ at $M$ so that $\Delta MAB$ is acute. $C$ is a point on any position of the chord $AB$, except $A$ and $B$ themselves. $MC$ intersects the circle $(O)$ at $D$ (on the other side of the chord $AB$). Prove that MB is the tangent of the excircle of $\Delta BCD$.
Attempt:
Let $MM'$ be the diameter of the circle $(O)$.
$O'$ is a point on $M'B$ so that $O'B=O'C$.
A circle with the center $O'$ and the radius $O'B$ intersects $(O)$ at another point called $D$.
Let $BB'$ be the diameter of the circle $(O')$.
Having proven that $\Delta MHB$ is a right triangle with $\widehat{MHB}=90$ degrees and $\Delta MBM'$ is a right triangle with $\widehat{MBM'}=90$ degrees (Thales's theorem) and $\Delta BCB'$ is a right triangle with $BCB'=90$ degrees, this implies $\widehat{MBH}=\widehat{MM'B}$ and $MM'//CB'$.
This leads to $\widehat{CD'B}=\widehat{CB'B}$ (inscribed angle) and $\widehat{CB'B}=\widehat{MM'B}$ $(MM'//CB')$.
We also have $\widehat{MDB}$ or $\widehat{CDB}$ equals to $\widehat{MM'B}$ (inscribed angle), so $\widehat{CDB}=\widehat{CD'B}$
$\widehat{CDB}=\widehat{CD'B} \Rightarrow D$ and $D'$ are the same points, is this assumption correct?
I have some doubt about this, because there can be some other cases of $D$ and $D'$ are not same positions but $\widehat{CDB}=\widehat{CD'B}$ is still true.
$$\measuredangle ABM=\measuredangle BAM=\measuredangle BDM$$ and by the theorem about the angle between the tangent and the chord we are done!