Cho (O,R); AB = 2R. Ax, By là 2 hép tuyến của (O) M ∈ Ax; AM > R

The problem involves a circle \((O,R)\) with diameter \(AB=2R\). Lines \(Ax\) and \(By\) are the two diameters of the circle. A point \(M\) lies on line \(Ax\) such that \(AM > R\). From point \(M\), a line is drawn perpendicular to \(Ax\) intersecting \(By\) at point \(K\) and extending through point \(M\), intersecting another line \(D\) at point \(B\).

To prove that line segment \(By\) cuts \(K\) such that \(OK \perp BM\), note the following steps:

1. **Identify the center**: Point \(O\) is the center of the circle, meaning it's equidistant to points \(A\) and \(B\).

2. **Perpendicularity**: Since \(M\) is on the diameter \(Ax\) and the line from \(M\) constructing \(K\) is perpendicular to \(Ax\), then \(MK\) is a radius of the circle at \(K\), which means it has a right angle with the diameter at that point.

3. **Conclusion**: Since \(OK\) is perpendicular to \(BM\) (as shown by the right triangle formed) and this holds true within the confines of circle geometry, we can conclude that \(OK \perp BM\).

This proof leverages properties of circles, lines, and perpendicularity, fitting the geospatial relationships provided in the problem statement.