Assuming the measure of the radius is strictly greater than the measure of AB divided by 2, construct a circle of the radius centered at A and then construct another circle with the same radius centered at B. The two circles will intersect at two points. Construct a line segment passing through the two points of intersection. The fact that each of the points is equidistant from A and B guarantees that the constructed segment will not only be perpendicular to AB, but will also bisect AB.
John
My calculator said it, I believe it, that settles it