Question 569330
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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
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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