The Broken Chord Theorem
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A Mathematical Droodle
Explanation
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Copyright © 1996-2012 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
Extend AC to F so that CF = BC. ΔBCF isosceles. Let α be its base angle: α = ∠CFB = ∠BFC. It then follows that the exterior angle ACB = 2α. In the given circle inscribed angles ACB and APB are equal as subtended by the same arc. Therefore,
| (2) |
∠APB = 2α = 2∠CFB = 2∠AFB.
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In the circumcircle of ΔABF, P, being the midpoint of the arc ACB, lies on the perpendicular to chord AB and also ∠APB = 2∠AFB. 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
- R. Honsberger, Episodes in Nineteenth and Twentieth Century Euclidean Geometry, MAA, 1995, pp. 1-2
The Broken Chord Theorem
- The Broken Chord Theorem: Proof Close to Archimedes'
- The Broken Chord Theorem: proof by Gregg Patruno
- The Broken Chord Theorem by Paper Folding
- The Broken Chord Theorem: proof by Stuart Anderson
- The Broken Chord Theorem: proof by Bui Quang Tuan
- The Broken Chord Theorem: proof by Mariano Perez de la Cruz
- Pythagoras' from the Star of David
- Pythagoras' from Broken Chords
- Extremal Problem in a Circular Segment
|Activities|
|Contact|
|Front page|
|Contents|
|Geometry|
|Store|
Copyright © 1996-2012 Alexander Bogomolny
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