Proving . . .

Circle k(S; r)is touching point A of line AB. Circle l(T; s)is touching point B of line AB and intersects circle k in the edge points C, D of its diameter. Prove that the intersection M of lines CD and AB is the centre of line AB.

Hi!
Reading the problem over and over and can´t quite grash it. C;D are antipods of large circle? And inside that circle are two other circles?

See this (I know it is not accurate but it helps)

http://content.imagesocket.com/thumbs/sssssss723.JPG

If you know it, written form will be sufficient for me.

Here is a sketch of the geometry.

The large blue circle has the same radius as circle k.
Using the radius of the smaller circle,
next draw right-angled triangles to show |TF|=|TG|.

The show |BM|=|BA|

I had to shrink my sketch and it has become a little skewed.
Those lines are meant to be perpendicular.

This only works if CD is a centreline.

Is this somewhat understandable?

Simplest is the following...

Triangle TBE is isosceles,
from this you can show |BM|=|MA|

Again, CD needs to be a centreline of the smaller circle.

A general proof, when CD is not a diameter of the smaller circle
can be derived from the attached sketch.

Here is a geometric proof.