Converse of the Perpendicular Bisector Theorem
If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.
The converse of the perpendicular bisector theorem states that if a point is equidistant from the endpoints of a segment, then it lies on the perpendicular bisector of the segment.
In other words, if we have a point P that is equidistant from two points A and B, then we can say that P lies on the perpendicular bisector of segment AB. This means that the line passing through P and perpendicular to AB will cut AB into two equal parts.
To prove the converse of the perpendicular bisector theorem, we need to show that if a point P lies on the perpendicular bisector of segment AB, then it is equidistant from A and B.
Proof:
Given: Point P lies on the perpendicular bisector of segment AB
To Prove: AP = BP
Let’s draw a diagram:
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A ———P———B
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M
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We know that PM is perpendicular to AB (since P lies on the perpendicular bisector of AB). So, we can say that triangles PAM and PBM are right triangles.
Therefore, we have:
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AP^2 = PM^2 + AM^2 (by Pythagoras’ theorem)
BP^2 = PM^2 + BM^2 (also by Pythagoras’ theorem)
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Since we want to show that AP = BP, we need to use these equations to eliminate PM.
Adding the two equations above, we get:
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AP^2 + BP^2 = PM^2 + AM^2 + PM^2 + BM^2
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Simplifying, we have:
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AP^2 + BP^2 = 2PM^2 + AM^2 + BM^2
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But we know that PM is the perpendicular bisector of segment AB, so AM = BM (since M is the midpoint of AB). Therefore, we can substitute AM = BM into the equation above:
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AP^2 + BP^2 = 2PM^2 + 2AM^2
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But we also know that PM is perpendicular to AB, so we can use the Pythagorean theorem again to substitute 2PM^2 = AB^2:
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AP^2 + BP^2 = AB^2 + 2AM^2
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Finally, we can use the fact that M is the midpoint of AB to substitute AB = 2AM:
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AP^2 + BP^2 = (2AM)^2 + 2AM^2
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Simplifying, we get:
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AP^2 + BP^2 = 4AM^2 + 2AM^2
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AP^2 + BP^2 = 6AM^2
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But we know that AM is half the length of AB (since M is the midpoint of AB). Therefore:
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AM^2 = (AB^2)/4
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Substituting this into the equation above, we get:
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AP^2 + BP^2 = (6/4)*AB^2
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AP^2 + BP^2 = (3/2)*AB^2
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But we also know that P lies on the perpendicular bisector of AB, so it cuts AB into two equal parts. Therefore, we have:
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AP = BP = AB/2
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Substituting this into the equation above, we get:
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AP^2 + BP^2 = 2(AP^2)
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Simplifying, we get:
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AP^2 = BP^2
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Taking the square root of both sides, we get:
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AP = BP
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Which is what we wanted to show. Therefore, the converse of the perpendicular bisector theorem is true: if a point is equidistant from the endpoints of a segment, then it lies on the perpendicular bisector of the segment.
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