If a pont lies on the perpendicular bisector of a segment, then the point is equidistant from the endpoints of the segment.

What is this theorem saying? We have a segment AB. If we construct the perpendicular bisector, m, then any point, P on m is equidistant from A and B. In other words, d(AP)=d(BP).

How can we prove this? Hint: what about the theorems from day 6? What do we know?We know the distance from A to the perpendicular bisector is the same as the distance from B to the perpendicular because of the definition.