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Bui Quang Tuan
Member since Jun2307

Aug1708, 01:53 PM (EST) 

"New Pythagorean Theorem Proof ?"

Dear All My Friends, Suppose ABC is right triangle with side lengths a, b, c and AB = c is hypotenuse. CH is altitude from C to AB. Denote (XYZ) = area of polygon XYZ. We translate ABC by direction HC to get triangle A'B'C' (not necessary exact how is CC'). A1 = intersection of AC, B'C' B1 = intersection of BC, A'C' D = intersection of BC, AA' D' = intersection of B'C', AA' By some parallelograms we can show: (ACB1) = (ACC') = (DCC') = (DCC'D')/2 (BCA1) = (BCC') = (BCC'B')/2 (ABB') = (ABB'A')/2 = (BDD'B')/2 therefore: (ABB') = (ACB1) + (BCA1) (1) By some right angles we can show that A1, B1, A, A', B, B' are on circumcircle of rectangle ABB'A' therefore: angle(BAB') = angle(BA1B') = angle(CBA1) = angle(CAB1) It means three right triangles ABB', BCA1, CAB1 are similar. Suppose k = BB'/AB = CA1/BC = CB1/CA then (1) can be written: k*c^2 = k*b^2 + k*a^2 or we have Pythagorean theorem: c^2 = b^2 + a^2May be it can be new proof for Pythagorean theorem? Thank you and best regards, Bui Quang Tuan 
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Bui Quang Tuan
Member since Jun2307

Aug1908, 07:17 AM (EST) 

6. "RE: New Pythagorean Theorem Proof ?"
In response to message #4

Dear Alex, Of course my proof is different because:  My area property of orthodiagonal inscribed quadrilateral is different  My proof is based on general fact when k = 1  It is very clear if we draw proof picture when k = 1 (please see attach file) But when draw this picture I see it is exactly proof 69 "Pythagoras' Theorem By Sheer Shearing": https://www.cuttheknot.org/pythagoras/PyhtI47PWW.shtml When look again generalization proof (with any k) I see that my proof method is the same Sheer Shearing method. Therefore, we can see all facts in this topic as:  Generalization of proof 69 or  Interesting area property of orthodiagonal inscribed quadrilateral with proof using Sheer Shearing method. Best regards, Bui Quang Tuan

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