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 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: 6,766 Thanks: 697 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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# prove pointwise characterization of perpendicular bisector

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