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Subject: "Broken Chord And Pythagorean Theorems"     Previous Topic | Next Topic
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Conferences The CTK Exchange College math Topic #738
Reading Topic #738
Bui Quang Tuan
Member since Jun-23-07
Jul-31-10, 04:27 PM (EST)
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"Broken Chord And Pythagorean Theorems"
 
   Dear All My Friends,

The Broken Chord Theorem is one of very small interesting theorems which credited to Archimedes. There are some proofs for it. Please see:
https://www.cut-the-knot.org/Curriculum/Geometry/BrokenChord.shtml
We rarely see applications of Broken Chord Theorem.

The Pythagorean Theorem is, off course, very famous and there are a lot of proofs. Please see:
https://www.cut-the-knot.org/pythagoras/index.shtml

Now we would like to provide one another proof for Broken Chord Theorem and use Broken Chord Theorem to prove Pythagorean theorem.

Proof of Broken Chord Theorem

I use drawing in the applet also my two attached images. Denote O as circumcenter of ABC.
P' = symmetry of P in O. Because P is midpoint of arc ACB so PP' is perpendicular bisector of AB and P, C are on one side with respect to AB.
Perpendicular from P' to AC cuts AC at M' and cuts circumcircle (O) again at M''
Two angles CAB and M''P'P are equal because their side lines are respectively perpendicular. Hence their chords are equal:
CB = M''P = M'M
By symmetry of circumcircle CM=AM'
From these:
AM = AM' + M'M = MC + CB
The proof is completed.

Proof of Pythagorean Theorem using Broken Chord Theorem

ABC is right triangle at C, AB is diameter of circumcircle O. In this case APB is right isosceles triangle at P.
Line PM cuts AB at N
B' is orthogonal projection of B on PM

AB=c, BC=a, CA=b
BCMB' is rectangle so: MB'=CB=a (1)

By Broken Chord Theorem:
AM=(a+b)/2 (2)
MC=(a+b)/2-a=(b-a)/2 (3)
PMC is right isosceles triangle at M so MP=MC=(b-a)/2 (4)

Area(MNB) = Area(ANB') because AM//B'B (5)

Now we calculate area of APB which is c^2/4

c^2/4 = Area(APB)
= Area(APM)+Area(PMB)+Area(AMN)+Area(MNB)
= Area(APM)+Area(PMB)+Area(AMN)+Area(ANB') (by (5))
= Area(APM)+Area(PMB)+Area(AMB')

Using (1), (2), (3), (4) to calculate areas of triangles:
Area(APB) = MP*AM/2 + MP*MC/2 + AM*MB'/2
= ((b-a)/2)*((a+b)/2)/2 + ((b-a)/2)*((b-a)/2)/2 + ((a+b)/2)*a/2
= (a^2+b^2)/4
It means c^2/4=(a^2 + b^2)/4 or c^2 = a^2 + b^2
The proof is completed.

Best regards,
Bui Quang Tuan

Attachments
https://www.cut-the-knot.org/htdocs/dcforum/User_files/4c54530e25fb49c6.jpg
https://www.cut-the-knot.org/htdocs/dcforum/User_files/4c5453292622a529.jpg

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mpdlc
Member since Mar-12-07
Aug-21-10, 09:15 AM (EST)
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1. "RE: Broken Chord And Pythagorean Theorems"
In response to message #0
 
   Definitively Bui you are a hard... hard worker, an indefatigable Math Marathonian.
I attach here another proof of the broken chord theorem it might be given for somebody else before, but I do not know, so I post it just in case.


Here is the proof almost identical to the one in CTK, maybe a bit'shorter

By extending PM to its intersection with the circunference we get point S
By drawing a line from S thru B will intersect the prolongation of AC on F

Angle MSA equal MSF since P bisects the original arc AB. Then segment AM = MF

We are to proof than CF equals CB

Angle CBF is suplementary of CBS, now since ACBS is an inscribed quadrilateral therefore CAS and CFB are equal for being suplementary both of CBS, which proof CF equal CB

mpdlc

Attachments
https://www.cut-the-knot.org/htdocs/dcforum/User_files/4c6fc9c462963ced.html

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mpdlc
Member since Mar-12-07
Aug-21-10, 10:58 AM (EST)
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2. "RE: Broken Chord And Pythagorean Theorems"
In response to message #1
 
   Including the missing attachment in a zip to pdf format.

mpdlc

Attachments
https://www.cut-the-knot.org/htdocs/dcforum/User_files/4c6feb03399ba175.zip

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alexbadmin
Charter Member
2574 posts
Aug-26-10, 10:31 AM (EST)
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3. "RE: Broken Chord And Pythagorean Theorems"
In response to message #2
 
   Thank you for that.

Managed finally to plug my computer into the internet. Here's your page:

https://www.cut-the-knot.org/triangle/BrokenChordmpdlc.shtml


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