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The Broken Chord Theorem: What is this about?
A Mathematical Droodle


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Explanation

Copyright © 1996-2009 Alexander Bogomolny

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The applet purports to remind of a theorem going under the name of The Broken Chord Theorem. The theorem is credited to Archimedes himself, although it does not appear in his Book of Lemmas:

  On the circumcircle of triangle ABC, point P is the midpoint of the arc ACB. PM is perpendicular to the longest of AC or BC. Prove that M divides the broken line ABC in half.

Assume for the definiteness' sake that AC > BC, so that M lies on AC and the theorem states that

(1) AM = MC + BC.


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


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Extend AC to F so that CF = BC. DBCF isosceles. Let a be its base angle: a = CFB = BFC. It then follows that the exterior angle ACB = 2a. In the given circle inscribed angles ACB and APB are equal as subtended by the same arc. Therefore,

(2) APB = 2a = 2CFB = 2AFB.

In the circumcircle of DABF, P, being the midpoint of the arc ACB, lies on the perpendicular to chord AB and also APB = 2AFB. Which implies that APB is the central angle subtending chord AB. P therefore is the center of that circle in which AF is a chord. The perpendicular from P to AF (read AC) crosses AF at its midpoint and (1) follows.

References

  1. R. Honsberger, Episodes in Nineteenth and Twentieth Century Euclidean Geometry, MAA, 1995, pp. 1-2

The Broken Chord Theorem

  1. A Proof Close to Archimedes'
  2. A Proof by Gregg Patruno
  3. A Proof by Paper Folding
  4. A Proof by Stuart Anderson

Copyright © 1996-2009 Alexander Bogomolny

34222730Page copy protected against web site content infringement by Copyscape


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