Proof that an arbitrary point of a circle keeps a given ratio First of all, let me enunciate my problem. Here it is:

Let $ABC$ be a triangle ($AB\subset r$, where $r$ is a line) such that $CM$ and $CN$ are respectively the internal and external angle bisector of angle $\angle ACB$ ($M, N \in r$). Let $P$ be a point belonging to a circumference $\Gamma$ with diameter $MN$. Prove that $\displaystyle\frac{PA}{PB}$ doesn't change its value for any specific choice of $P$.

So, here's a picture of a specific case for this problem:

Here's some things I know about this composition:

*

*$C$ belongs to the circle, for $\angle MCN$ is a right angle;


*$M$ and $N$ are trivial for this property as a consequence of the angle bisector theorem.
This'd be the same as proving that the Circle of Apollonius is the locus of all the points whose distances from $A$ and $B$ have the ratio $\displaystyle\frac{CA}{CB} = k$. But I couldn't do it.
 A: HINT.- Given three points $A=(a,0), \space M=(b,0),\space B=(c,0)$ with $a\lt b\lt c$ the locus of points $P=(x,y)$ such that $PM$ is an internal angle bisector of triangle $\triangle{PAB}$ is easily given by
$$\frac{\sqrt{(x-a)^2+y^2}}{\sqrt{(x-b)^2+y^2}}=\frac{b-a}{c-b}$$ This locus is a circumference in which any point (say in the first quadrant) can play the rol of vertex $C$ in the problem so we have the particular case of triangle $\triangle{CAB}$ proposed by the owner. What remains is just to apply the same property used to determine the aforementioned locus.In other words, for all $P$ the segment $\overline{PM}$ is a bisector angle of triangle $\triangle{PAB}$.
A: It's easy to prove this without using any coordinate geometry. Consider, instead, the following problem. Given a line segment $\overline{AB}$ and a point $M$ on it, find the locus of a point $C$ such that $CM$ is the angle bisector of $\angle ACB$.
For a particular $C$, let $N$ be the point of intersection of the external angle bisector of $\angle ACB$ and $AB$. Then
$$ \frac{AM}{BM} = \frac{AC}{BC} =\frac{AN}{BN} $$
Since $N$ doesn't lie within $\overline{AB}$, $\frac{AM}{BM} = \frac{AN}{BN}$ uniquely defines a point $N$ on the line $AB$, that externally divides $A,B$ in the given ratio. Also, $MC \perp NC$ as they are angle bisectors. Hence, the circle with diameter $MN$ is in fact the locus of a point $P$ whose internal angle bisector is $PM$.
Thus for any point $P$ on the circle, $\frac{PA}{PB}$ is a constant.
