I need help with this geometric problem. Given triangle ABC. CM is the bisector of $\angle ACB, M\in AB$ and $CN, N\in AB$ is the bisector of the suplementary angle of $\angle ACB$. B is between M and N. A circle with diameter MN is drawn. For any point P on the circle prove that:
$$\frac{PA}{PB}= \frac{AC}{BC}.$$
There is a two-focus definition of a circle as all points $X$ such that $\frac{XF_1}{XF_2}=k$ for a constant $k$, which can be applied here. First, since $CM$ is the internal bisector of $\angle ACB,$ we have $\frac{CA}{CB}=\frac{MA}{MB}.$
So if we define $k=\frac{MA}{MB}$ there is a circle $\frac{XA}{XB}=k$ using the two foci $A,B$ with $k$ as above, and this circle contains the point $C$ by choice.
Under your assumption the external bisector at $C$ happens to cross the (extended) side $AB$ of the given triangle at a point $N$ where $B$ is between $M$ and $N$.
We also have a right angle $MCN$ (using that the two angles are supplementary), making $MN$ the diameter of some circle going through point $C$, and in your assumptions this is the circle through $M,N$ you're considering. So that's the same as the above circle with general formula $$\frac{XA}{XB}=k,$$ which means we get $\frac{PA}{PB}=\frac{CA}{CB}$ for any point $P$ on the circle, by taking $P=X$ in that formula, and $k=\frac{MA}{MB}=\frac{CA}{CB}.$
Edit: To make this idea work we need a separate proof that the point $N$ also satisfies the circle equation $\frac{XA}{XB}=k.$ When this is brought down to length ratios it becomes (in one form) $\frac{CA}{NA}=\frac{CB}{NB},$ which one can show fairly easily using the law of sines. The need for this extra step is that we know we have a circle through $M,N$ containing the point $A,$ but in order to conclude this is the same as the circle $\frac{XA}{XB}=k$ we need both ends of the line segment to lie on that circle.