T.R | Title | User | Personal Name | Date | Lines |
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850.1 | the easy case | ZFC::DERAMO | Take my advice, I'm not using it | Thu Mar 31 1988 21:11 | 8 |
| Well, if point a and point c are opposite ends of a diameter,
then as b moves along the circle angle abc is always a right
angle.
If points a and c are not opposite ends of a diameter, well,
this is just too much geometry for two days!
Dan
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850.2 | here's a sketch of proof | PULSAR::WALLY | Wally Neilsen-Steinhardt | Mon Apr 11 1988 13:24 | 20 |
| Yes, the theorem in .0 is true and provable. I don't think I could
give the full proof without a diagram, but here is a sketch:
Given points ABCD on a circle with center O.
Then there are 5 isosceles triangles OAD, OAB, OAC, OCD, OBC.
This leads to 5 equalities among angles.
And there are two equalities based on the fact that the sum
of angles in any two triangles, and therefore ABC and ADC, is
constant.
There are two more equalities based on angles contained within
angles (here's where the diagram is necessary).
Combine all these equalities in the right way and you get
<ABC = <ADC
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850.3 | almost got it | GORP::MARCOTTE | George Marcotte SWS Santa Clara | Mon Apr 11 1988 18:49 | 29 |
| > Given points ABCD on a circle with center O.
> Then there are 5 isosceles triangles OAD, OAB, OAC, OCD, OBC.
> This leads to 5 equalities among angles.
well <oab = <oba
<obc = <ocb
<OCD = <ODC
<OAD = <ODA
That is 4 whats the 5th one?
> And there are two equalities based on the fact that the sum
> of angles in any two triangles, and therefore ABC and ADC, is
> constant.
I can see that:
<ABC + <ADC + <bad + <dcb = 360
how is <ABC and <ADC constant?
> There are two more equalities based on angles contained within
> angles (here's where the diagram is necessary).
name an example
Its been tooooooo long...
George :-)
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850.4 | more hints | PULSAR::WALLY | Wally Neilsen-Steinhardt | Tue Apr 12 1988 14:05 | 24 |
| re: < Note 850.3 by GORP::MARCOTTE "George Marcotte SWS Santa Clara" >
-< almost got it >-
> That is 4 whats the 5th one?
<OAC = <OCA
> how is <ABC and <ADC constant?
What is constant is the sum of angles in a triangle:
<ABC + <ACB + <BAC = <ADC + <ACD + <DAC
>> There are two more equalities based on angles contained within
>> angles (here's where the diagram is necessary).
> name an example
<ABC = <ABO - <CBO
Note that the sign here depends on the way I drew the diagram, since
I follow the Euclidean convention that all angles are positive.
If you draw a different diagram, your sign may be different, but
the proof will still work out.
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