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Hello, that is my problem:
There is a triangle ABC. On the side AC there is a point D and on the side BC there is a point E. Besides you must notice that the line segment AD is as long as the line segment BE. So: AD=BE
Then one examines the circumscribed circles of the two triangles, which are a part of the triangle ABC, AEC and BDC. These circumscribed triangles subtend each other at the point C and at another point F. Finally it is to prove that the line segment with the starting point C and the ending point F is the bisector of the angle gamma (the angle at the point C).
Please help me, I am nearly desperate... I just don't know how tpó proof this...
Hello, that is my problem:
There is a triangle ABC. On the side AC there is a point D and on the side BC there is a point E. Besides you must notice that the line segment AD is as long as the line segment BE. So: AD=BE
Then one examines the circumscribed circles of the two triangles AEC and BDC, which are a part of the triangle ABC. These two circumscribed triangles subtend each other at the point C and at another point F. Finally it is to prove that the line segment with the starting point C and the ending point F is the bisector of the angle gamma (the angle at the point C).Please help me, I am nearly desperate... I just don't know how tpó proof this...
This is hard for me. I drew 3 scenarios, with odd lengths, and sure enough the sketches do tend to agree with the bisection of angle, but I have no idea how to prove this, sorry.
Last edited by John E. Franklin (2007-02-01 01:31:30)
igloo myrtilles fourmis
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This is hard for me. I drew 3 scenarios, with odd lengths, and sure enough the sketches do tend to agree with the bisection of angle, but I have no idea how to prove this, sorry.
Yeah, I've also alraedy drawn many scenarios, however, I was not able to prove it. darn, that difficult...
Can anyone draw a picture of it , I am confused since my English is poor
Numbers are the essence of the Universe
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Can anyone draw a picture of it , I am confused since my English is poor
I can.
What's your language?
Could I send it to you via e-mail?
Thomas11 wrote:Stanley_Marsh wrote:Can anyone draw a picture of it , I am confused since my English is poor
I can.
What's your language?
Could I send it to you via e-mail?could you please forward a copy to
it would be very helpfull,thanks; )
I've just sent you an e-mail.
Here's a picture of the problem
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Did the picture help you to understand my problem?
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Yeah , I am working on that now,don't know if I can solve it though
Numbers are the essence of the Universe
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What? how can the segment CF besect any angle? Name the angle gamma ,I don't really understand ,lol
Numbers are the essence of the Universe
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Stanley_Marsh wrote:What? how can the segment CF besect any angle? Name the angle gamma ,I don't really understand ,lol
Segment CF bissects angle gamma.
It's segment CD. Segment CF is a part of the triangle side a. Have a look at the picture.
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Hello, that is my problem:
There is a triangle ABC. On the side AC there is a point E and on the side BC there is a point F. Besides you must notice that the line segment AE is as long as the line segment BF. So: AE=BF
Then one examines the circumscribed circles of the two triangles, which are a part of the triangle ABC, AFC and BEC. These circumscribed triangles subtend each other at the point C and at another point D. Finally it is to prove that the line segment with the starting point C and the ending point D is the bisector of the angle gamma (the angle at the point C).Please help me, I am nearly desperate... I just don't know how tpó proof this...
Now the picture matches the exercise, I named s.th. wrong. Now it should be right.
Last edited by Thomas11 (2007-02-02 23:04:10)
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Has anybody found s.th. interesting or helpful yet? (me not -.-')
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Uh, please inform me, even if you only find s.th. that might be helpful, I'll appreciate your efforts.
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Help me...-.-
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