How to find a point$M$ so that $(MC)$ is the bisector of the angle $\widehat{AMB}$ Let $A,B $ be  two distinct points in plane and $C \in [AB]$
Is there a way to find and draw a point $M$ in plane so  that $MC$ bisects $\widehat{AMB}$ ?
 A: If $C$ is a mid-point of $AB$ so any point on a perpendicular bisector of $AB$ without a point $C$ is valid.
Let $\Delta MAB$ be a triangle. 
Thus, $$\frac{MA}{MB}=\frac{AC}{CB}.$$
Also, if $AD$ is a bisector of a supplementary angle to $\angle AMB$ and $D$ is placed on the line $AB$, so
$$\frac{MA}{MB}=\frac{AD}{DB}.$$
Easy to find this point $D$.
Now, since $\measuredangle DMC=90^{\circ},$ we see that $M$ is placed on the circle with diameter $DC$.
A: Let's treat the case when $C$ is the midpoint. In this case, the geometric locus of $M$ is in the bisector of $AB$.
Assume $C$ is not the midpoint of $AB$:
Let $D$ be the (unique) point in the line $\overline{AB}$ such that the cross ratio $(A,C,B,D)$ equals $1$, i.e. such that $\frac{(AB)(CD)}{(CB)(AD)}=1$. Then, if $M$ is a point with $\angle AMC=\angle CMB$, then $\angle CMD=90$. This means that the geometric locus of the points $M$ such that $\angle AMC=\angle CMB$, is in the circle with diameter $CD$. 
So, to construct this locus (the circle with diameter $CD$), we should construct $D$ first.  We can even construct this using only a ruler. Given three colinear points $A,C,B$, pick any random point $P$ outside the line $\overline{AB}$. Choose any point $Q$ in $[PC]$, let $R=\overline{BQ}\cap \overline{PA}$ and $S=\overline{AQ}\cap \overline{PB}$. Then $D=\overline{RS}\cap \overline{AB}$ is the point such that $(A,C,B,D)=1$.
