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September 3rd, 2010, 10:16 AM   #1
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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.
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September 3rd, 2010, 11:17 AM   #2
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Re: Geometry Proof

I think the proof is very simple, if you know some fundamental theorems of geometry.
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September 3rd, 2010, 01:05 PM   #3
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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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