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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
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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