September 3rd, 2010, 10:16 AM  #1 
Senior Member Joined: Apr 2009 Posts: 106 Thanks: 0  Geometry Proof
The perpendicular bisector of a segment AB is a line l such that l intersects AB at its midpoint and l is perpendicular to AB. Prove the following theorem: Pointwise Characterization of Perpendicular Bisector. A point P lies on the perpendicular bisector of AB if and only if PA = PB. 
September 3rd, 2010, 11:17 AM  #2 
Newbie Joined: Aug 2010 Posts: 25 Thanks: 0  Re: Geometry Proof
I think the proof is very simple, if you know some fundamental theorems of geometry.

September 3rd, 2010, 01:05 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,559 Thanks: 604  Re: Geometry Proof
A point on the perpendicular bisector connected to the end points gives a figure consisting of two right triangles with the legs equal. Therefore the hypotenuses are equal. For a point (P) off the bisector, compare to the point (Q) on the bisector at the same distance from the line. Assume P is on the A side of the bisector, the PA < QA and PB > QB. Meanwhile QA = QB. All these statements follow from the pythagorean theorem. 

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