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Theorem of the Broken Chord


Date: 12/14/2001 at 11:21:07
From: James Wise
Subject: Theorem of the Broken Chord

I am trying to prove the theorem of the broken chord, which asserts 
that if AB and BC make up a broken chord in a circle, where BC > AB, 
and if M is the midpoint of arc ABC, the foot F of the perpendicular 
from M on BC is the midpoint of the broken chord.

I know that, given that M is the midpoint of arc ABC, I have to show 
that point F is the midpoint of the broken chord ABC. I have 
approached this problem by looking at triangles, sectors, and 
arclengths, and am having problems coming to any conclusion that 
includes the entire broken chord. I think I have proved a special case 
of the theorem where the BC portion of the broken chord is equivalent 
to the diameter of the circle; however, I have not made any headway in 
any other case.


Date: 12/14/2001 at 17:13:38
From: Doctor Rick
Subject: Re: Theorem of the Broken Chord

Hi, James, thanks for writing to Ask Dr. Math.

Here is an answer I wrote to another correspondent. I hope it helps 
you. She wrote:

>Dr. Math,
>
>I just read the theorem of the broken chord, attributed to 
>Apollonius. Take a circle with two chords, AB and BC, where BC>AB,
>which constitute a "broken chord." Locate the midpoint M of arc ABC,
>and drop a perpendicular to chord BC, which intersects BC at the 
>point F. F is the midpoint of the broken chord ABC; in other words,
>AB+BF = FC. How can this be?
>
>I've tried to use the inscribed angle theorem, as well as 
>perpendicular bisectors of chords, and tried to find 
>similar/congruent triangles. Also, I tried extending CM and AB to 
>intersect at point O outside the circle. Then I could use the fact 
>that AB*BO = CM*MO. However, I still got stuck. What circle theorems 
>am I missing that could help me? 

You've got all the theorems you need - especially the inscribed angle 
theorem. I know of two different proofs of the theorem (which I have 
seen attributed to Archimedes, not Apollonius). Each begins by 
constructing an additional point that will help you recast the 
conclusion in terms of the equality of two straight line segments.

                              M
                      B   **********
                   *******   /|     *******
                ***  /  -- /  |            ***
              **    /     *   |               **
           ***     /     F  --|                 ***
          *       /           +-                   *
         *       /            | --                  *
       **       /             |   --                 **
      *        /              |     --                 *
     *        /               |       --                *
    *        /                |         --               *
    *       /                 |           --             *
   *       /                  |             --            *
   *      /                   |               --          *
   *     /                  --O--               --        *
   *    /              -----     ------           --      *
   *   /         ------                -----        --    *
   *  /     -----                           -----     --  *
    */ -----                                     ------ -*
   A*--                                                --*C
     *                                                  *
      *                                                *
       **                                            **
         *                                          *
          *                                        *
           ***                                  ***
              **                              **
                ***                        ***
                   *******          *******
                          **********

Proof 1: Extend line BC, and construct a point D on the line (outside 
the circle) such that DB = AB. Then you want to prove that DF = FC. 
How can you do this? Hint: Construct a circle through D, A, and C. 
Where is its center?

Proof 2: Construct a point D on BC such that DC = AB. Then it suffices 
to prove that BF = FD. How can you do this? Hint: Is there a triangle 
congruent to CDM?

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Conic Sections/Circles
High School Geometry

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