MathDB
Geometric location: |PA| * |PC| = |PB| * |PD|

Source: AIMO 2008, TST 4, P3, Suggested by Christian Reiher

January 4, 2009
symmetrygeometryperpendicular bisectorgeometry unsolved

Problem Statement

Let ABCD ABCD be an isosceles trapezium. Determine the geometric location of all points P P such that |PA| \cdot |PC| \equal{} |PB| \cdot |PD|.
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