Geometrical Proof for Euclidian Triangle Circle Property

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Discussion Overview

The discussion revolves around the geometrical proof related to the property of a triangle inscribed in a circle, specifically focusing on a triangle with a fixed edge length and a fixed internal angle at the opposite vertex. Participants explore the relationship between the triangle's vertices and the circumcircle, as well as the properties of angles related to the circumcentre.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants propose that for a triangle with a fixed edge and angle, the opposite vertex lies on a circle, but they struggle to construct a geometrical proof.
  • One participant mentions that the angle to the circumcentre is fixed and can be proven using isosceles triangles, suggesting a relationship between the circumcentre and the triangle's angles.
  • Another participant questions the definition of the circumcentre and the angles involved, indicating a need for clarification on these concepts.
  • A participant discusses the method of drawing perpendicular bisectors to find the circumcentre and suggests that the angle at the circumcentre remains constant, contingent on the fixed angle of the triangle.
  • One participant introduces a circle theorem stating that a fixed chord subtends the same angle at any point on the circumference, proposing to use this theorem to approach the proof from a different angle.
  • There is a mention of the differences in classical geometry education between regions, with some participants expressing surprise at the lack of exposure to certain concepts in modern curricula.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and familiarity with classical geometry concepts, leading to a mix of agreement on certain properties while also highlighting gaps in knowledge and differing educational backgrounds. The discussion remains unresolved regarding the construction of a definitive geometrical proof.

Contextual Notes

Limitations include assumptions about the participants' prior knowledge of geometry, the dependence on definitions of terms like circumcentre, and the unresolved nature of the proposed proofs and theorems discussed.

Phrak
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No, this isn't a homework problem.

Given any triangle with a fixed length for one edge, and a fixed internal angle for the opposite vertex, the vertex will lay on a circle, where the other edges are allowed to vary in length. But I can't seem to construct a geometrical proof. Any takers? It's deceptively simple.
 

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Phrak said:
No, this isn't a homework problem.

Given any triangle with a fixed length for one edge, and a fixed internal angle for the opposite vertex, the vertex will lay on a circle, where the other edges are allowed to vary in length. But I can't seem to construct a geometrical proof. Any takers? It's deceptively simple.

Hi Phrak! :smile:

The angle to the circumcentre is fixed (it's double the given angle, and you can prove it using isoceles triangles), and the circumcentre lies on a fixed line. :wink:
 
tiny-tim said:
Hi Phrak! :smile:

The angle to the circumcentre is fixed (it's double the given angle, and you can prove it using isoceles triangles), and the circumcentre lies on a fixed line. :wink:

And I was just bemoaning the fact that no one responded. :smile: But wait. What's a circumcentre, and what angle from whence and to where?
 
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Phrak said:
And I was just bemoaning the fact that no one responded. :smile:

I think most PF members don't like "real" geometry! :biggrin:
But wait. What's a circumcentre, and what angle from whence and to where?

he he :smile:

The circumcentre is the centre of the circumcircle, which is the circle which goes through all three vertices of a triangle, and the angle is the angle subtended by the original line at the circumcentre. :wink:

:cool: join CAMREG … the CAMpaign for REal Geometry!​
 
So true. Parallel lines that meet a a point. Hubris!

I can't imagine how you know all this. Way classical education? I had one Euclidean geometry class in High School, sans circumcentres.

If I understand you correctly, I draw the pependicular bisectors through the remaining two sides. These will meet at an angle alpha=180-beta at the circumcentre I should expect alpha to be constant as much as beta is expected to be constant. But I can prove alpha is constant using isosoles triangles, right?
 
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Phrak said:
I can't imagine how you know all this. Classical education? I had one Euclidean geometry class in High School, sans circumcentres.

ooh, I'm sorry, Phrak :redface: …

i hadn't realized so little classical geometry is taught nowadays …

ok, it's best to introduce this idea the other away round, by starting with a circle, and seeing what happens inside it …

one of the most useful circle theorems is that a fixed chord AB subtends the same angle ACB at any point C on the circumference on the smaller arc, and 180º minus that angle on the larger arc, and that the angle AOB subtended at the centre is twice the latter angle …

to prove that, let O be the centre of the circle, and let the line CO (with C on the larger arc) poke a little beyond O to Q …

then OAC and OBC are isoceles triangles ('cos it's a circle! :wink:)

and so you can prove that AOQ = 2ACO and QOB = 2OCB …

and then for your problem, simply do the same proof "backwards" :smile:​
 
tiny-tim said:
ooh, I'm sorry, Phrak :redface: …

i hadn't realized so little classical geometry is taught nowadays …

No problem at all. Though it's been a while since high school for me. The classical studies are more common in the UK than the US, that makes the difference.

And thanks for the proof. :smile:
 

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