# The Broken Chord TheoremWhat is this about? A Mathematical Droodle

### 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 https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

Explanation

The 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:

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 https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 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 AM = MF.

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: BC = CF.

(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.)