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The Broken Chord Theorem
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| 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
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 ...
| Buy this 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):
a =  PAC = PBC,b = ABP = ACP,g = BAC = BPC.
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The three angles are not independent. Since DAPB is isosceles we observe that
| (2) | a + g = b. |
We now proceed with the construction. Let F lie on the extension of AC so that CF = BC. DBCF is isosceles. Its apex angle is exterior to DABC, from which we conclude that
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BCF = a + b + g = 2b.
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If T is the point (not shown in the applet) where PC crosses BF, then FCT is vertical to ACP = b.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
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BAP + BCP = p.
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But
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BAP = ABP = b = ACP.
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So it follows that PC is the external angle bisector at C of DBCF. 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.
The Broken Chord Theorem
- A Proof Close to Archimedes'
- A Proof by Gregg Patruno
- A Proof by Paper Folding
- A Proof by Stuart Anderson
- Angle Trisection by Paper Folding
- Angles in Triangle Add to 180o
- Broken Chord Theorem by Paper Folding
- Dividing a Segment into Equal Parts by Paper Folding
- Egyptian Triangle By Paper Folding
- My Logo
- Paper Folding And Cutting Sangaku
- Parabola by Paper Folding
- Radius of a Circle by Paper Folding
- An Interesting Example of Angle Trisection by Paperfolding
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