The question: The diameter AB of a circle is extended past B, and at a point C on this extension CD perpendicular to AB is erected. If an arbitrary point M of this perpendicular is connected with A, and the other intersection point of AM with the circle is denoted A', then AM times AA' is a fixed quantity, i.e. it does not depend on the choice of M.
!http://www.hicksvillepublicschools.org/cms/lib2/NY01001760/Centricity/Domain/1129/KISELEV409.bmp
At first, I thought about dropping a perpendicular from A' to AC, labeling the point of intersection E. Then, trying to show that triangle A'AE is similar to triangle MAC. But after I did that, I noted, one that I didn't get anything that could be considered a fixed value (such as the radius or diameter of the circle) and, two that the proportions involving AM and AA' were not aligned so that I could get an expression for the product of AM and AA'.
Next, I considered drawing a tangent from M to the circle at the point F, then, using the secant-tangent theorem to get that (MF)^2 = AM times A'M. But, that didn't get me anything in terms of AA' and, once again, there were no constants involved (as the length of MF varies depending upon where M is chosen on CD).
So, what other auxiliary lines / circles must I construct in order to proceed? Or is there something else that I'm missing?
Hint:
The triangles $AA'B$ and $ACM$ are similar.