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CTK Exchange
Bui Quang Tuan
Member since Jun-23-07
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Aug-17-08, 01:53 PM (EST) |
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"New Pythagorean Theorem Proof ?"
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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 Jun-23-07
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Aug-19-08, 07:17 AM (EST) |
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6. "RE: New Pythagorean Theorem Proof ?"
In response to message #4
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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.cut-the-knot.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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