# The Broken Chord Theorem

What is this about?

A Mathematical Droodle

What if applet does not run? |

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander BogomolnyThe applet purports to remind 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:

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

(1) | AM = MC + BC. |

What if applet does not run? |

### Proof by Stuart Anderson

Extend PM to meet the circle at Q. Assume Q is different from B and let BQ and AC extended intersect in F. In ΔAQF, QM is the altitude from vertex Q. It is also the angle bisector at Q because P is the midpoint of chord APB, so that the angles at Q subtend equal arcs:

∠AQP = ∠FQP (= ∠BQP). |

We see that ΔAQF is isosceles and

Now, since quadrilateral AQBC is cyclic its opposite angles add to π. On the other hand, its angles at B and C are supplementary to those in ΔAQF. It follows that ΔBCF is similar to ΔAQF and, hence, is also isosceles:

(At the outset we assumed that Q is different from B. If they coincide then instead of BQ we consider the tangent to the circle at B.)

### 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|

Copyright © 1996-2018 Alexander Bogomolny64973782 |