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