Theorem of the Broken ChordDate: 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/ |
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