The Broken Chord Theorem
by Paper Folding

The applet below is an illustration of an elegant theorem credited to Archimedes.

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.

To get an insight into the likely origins of the problem, we'll start with an isosceles triangle APF in which AP = PF. The altitude from the apex naturally cuts the base AF at the midpoint M. Now choose any point C on MF. Draw line PC and reflect F in PC to obtain B. Triangles PCB and PCF are equal and so are their corresponding elements. Importantly, the angles at B and F are equal. But the latter is one of the base angles in an isosceles triangle. Thus it is equal to the other base angle. By transitivity, we see that angles CAP and CBP are equal. If so, the quadrilateral ABCP is cyclic.

This shows that the configuration in the Broken Chord theorem is naturally arrived at by paper folding. For a proof one only needs to unfold the paper. Or so it seems ...

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?

We may start with extending AC to F, as in another proof. But first note that the problem presents several pairs of inscribed angles subtended by the same arcs. For this proof, we are interested in three pairs (all different from the one used elsewhere):

α = ∠PAC = ∠PBC,
β = ∠ABP = ∠ACP,
γ = ∠BAC = ∠BPC.

The three angles are not independent. Since ΔAPB is isosceles we observe that

(2)	α + γ = β.

We now proceed with the construction. Let F lie on the extension of AC so that CF = BC. ΔBCF is isosceles. Its apex angle is exterior to ΔABC, from which we conclude that

∠BCF = α + β + γ = 2β.

If T is the point (not shown in the applet) where PC crosses BF, then ∠FCT is vertical to ∠ACP = β. This implies that PC is the bisector of ∠BCF and that F is indeed the reflection of B in PC. And the unfolding is completed.

Nathan Bowler came up with a shorter and a more direct argument. First observe that

∠BAP + ∠BCP = π.

But

∠BAP = ∠ABP = β = ∠ACP.

So it follows that PC is the external angle bisector at C of ΔBCF. Which, in turn, implies that the reflection F of B falls on the extension of AC. A shorter way to unfolding!

As an extra insight, observe that we might have reflected in PC vertex A instead of B. This would produce another isosceles triangle BPG equal to APF. A proof may now be built based the fact that the triangles are obtained from each other by a rotation around P.

References

S. E. Louridas, M. Th. Rassias, Problem-Solving and Selected Topics in Euclidean Geometry, Springer, 2013 (69-71)