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. So, is the distance from P to the intersection of m and AB equal to itself? Yes, because of the reflexive property. Now, we can use LL to prove the two triangles are congruent. Then by CPCTC, we know AP = BP. A-ha!!

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