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Explanation ### 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 ΔAO'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. ΔNOB is therefore isosceles. OB = ON, which implies N = M.

Here's another way of looking at the configuration. The two circles are homothetic with center B. The lowest point A of the circle C' is mapped by that homothety onto the lowest point of the circle C, which is M. But any two points that correspond by a homothety lie on a line through the center of the homothety.

There are two more proofs that use the inversion transformation. ### Inversion - Introduction 