Circle Inscribed in a Circular Segment

The applet suggests the following statement:

A chord ST is drawn in a circle C with center O. A circle C' is inscribed into the circular segment thus obtained that touches the chord ST at the point A and the circle C at the point B. Let M denote the midpoint of the arc defined by ST that does not include B. Then A, B, and M are collinear.

Let O' be the center of the inscribed circle. Then AO'B is isosceles. Extend AB beyond A and let it intersect the perpendicular OM to ST at point N. The two triangles AO'B and NOB are similar. Indeed they have a common angle at B and, since ON||O'A, their respective angles at O' and O are also equal. NOB is therefore isosceles. OB = ON, which implies N = M.