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CTK Exchange
Bui Quang Tuan
Member since Jun-23-07
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Aug-11-09, 01:26 PM (EST) |
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"New Proof For Pythagorean Theorem?"
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Dear Alex and All My Friends,
I have found following proof for Pythagorean Theorem. Please check for me if it is a new one? In any case, it is very simple proof.ABC is a right triangle with right angle A, three sides a, b, c. L is a internal angle bisector of angle BAC. B', C' is orthogonal projections of B, C on L respectively. D is intersection of L with perpendicular bisector of BC. Suppose AC>=AB so B' is inside ABC and C' is outside ABC.
Area(ABB') = 1/4*c^2
Area(ACC') = 1/4*b^2
Area(DBC) = 1/4*a^2
b^2 + c^2 = a^2 <=> Area(DBC) = Area(ABB') + Area(ACC')
Two following right triangles are congruent DBB' = CDC' therefore
DC' = BB' = B'A
From this:
Area(BAB') = Area(BDC')
Area(CAB') = Area(CDC')
Area(CC'B') = Area(CC'B) (because BB'//CC')
So Area(DBC) = Area(ABB') + Area(ACC') and we are done. Thank you and best regards,
Bui Quang Tuan |
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alexb
Charter Member
2424 posts |
Aug-11-09, 01:35 PM (EST) |
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1. "RE: New Proof For Pythagorean Theorem?"
In response to message #0
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Bui Quang Tuan, at first sight, I do not remember seeing this proof elsewhere. I'll be on the road till the weekend and shall consult Loomis' when back. Thank you,
Alex |
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alexb
Charter Member
2424 posts |
Aug-15-09, 02:14 PM (EST) |
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2. "RE: New Proof For Pythagorean Theorem?"
In response to message #1
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Dear Bui Quang Tuan, I found in Loomis' book a proof based on about the same diagram and them realized that the idea behind one of Floor van Lamoen's proofs is more fundamental. I thus added the two to Floor's #64 without assigning either a special number. I hope you approve. Thank you,
Alex |
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Bui Quang Tuan
Member since Jun-23-07
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Aug-15-09, 04:51 PM (EST) |
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3. "RE: New Proof For Pythagorean Theorem?"
In response to message #2
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Dear Alex, Thank you very much for your interesting references. I think it is good idea to put my proof with Floor van Lamoen and Loomis #67 proofs. I would like to change some explain words to clearly meet my idea and diagram.
Please change first sentence as following: From the above, Area(BA'D) = Area(BB'C) and Area(AA'D) = Area(AB'C). Thank you and best regards,
Bui Quang Tuan >Dear Bui Quang Tuan,
>
>I found in Loomis' book a proof based on about the same
>diagram and them realized that the idea behind one of Floor
>van Lamoen's proofs is more fundamental.
>
>I thus added the two to Floor's #64 without assigning either
>a special number.
>
>I hope you approve.
>
>Thank you,
>Alex
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alexb
Charter Member
2424 posts |
Aug-15-09, 04:53 PM (EST) |
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4. "RE: New Proof For Pythagorean Theorem?"
In response to message #3
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>Please change first sentence as following:
>
>From the above, Area(BA'D) = Area(BB'C) and Area(AA'D) =
>Area(AB'C). Done. Thank you for looking into that. |
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Jprice2
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Aug-16-09, 01:26 PM (EST) |
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5. "RE: New Proof For Pythagorean Theorem?"
In response to message #0
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Geometry is not my strong suit. I am having trouble seeing why angle BDC is a right angle. Any hints or help would be appreciated.
Thanks |
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jprice2
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Aug-16-09, 02:33 PM (EST) |
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8. "RE: New Proof For Pythagorean Theorem?"
In response to message #7
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Ok, Thanks.
Could you direct me to a proof that shows that the bisector of the right angle formed by a point on a circle and the two end points of its diameter intersects the midpoint of the opposite semicircle?
Thanks and by the way this is a great website! |
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jprice2
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Aug-16-09, 03:49 PM (EST) |
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10. "RE: New Proof For Pythagorean Theorem?"
In response to message #9
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I see this now. The central angle is twice the inscribed angle and when the inscribed angle is bisected then the central angle is bisected and therefore the intersection of the two bisectors meets at the midpoint of the subtending arc.
Thank you. |
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