Geometrical Proof for Euclidian Triangle Circle Property

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In summary, 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. 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. Parallel lines that meet a a point. Hubris!
  • #1
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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  • #2
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:
 
  • #3
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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  • #4
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!
 
  • #5
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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  • #6
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:
 
  • #7
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:
 

FAQ: Geometrical Proof for Euclidian Triangle Circle Property

What is the Euclidian Triangle Circle Property?

The Euclidian Triangle Circle Property is a geometric proof that states that the angles inscribed in a semicircle are right angles. This means that if a triangle's vertices lie on the circumference of a semicircle, the angle formed by connecting any two of the triangle's vertices to the semicircle's center will always be a right angle.

Who discovered the Euclidian Triangle Circle Property?

The Euclidian Triangle Circle Property was first discovered by the ancient Greek mathematician Euclid in his book "Elements". This book laid the foundation for modern geometry and is still studied today.

How is the Euclidian Triangle Circle Property proven?

The Euclidian Triangle Circle Property is proven using theorems and postulates from Euclidean geometry, such as the Pythagorean Theorem and the fact that angles inscribed in a circle subtend the same arc. By combining these principles, it can be shown that the angles inscribed in a semicircle must be right angles.

What are some real-world applications of the Euclidian Triangle Circle Property?

The Euclidian Triangle Circle Property has many real-world applications, including in architecture, engineering, and navigation. It is also used in the design of circular objects, such as wheels, and in calculating distances and angles on maps and globes.

Are there any other properties of circles that are related to the Euclidian Triangle Circle Property?

Yes, there are several other properties of circles that are related to the Euclidian Triangle Circle Property, including the Inscribed Angle Theorem, which states that an angle inscribed in a circle is half the measure of the central angle that subtends the same arc. Additionally, the Tangent-Secant Theorem and the Chord-Chord Theorem are also related to the Euclidian Triangle Circle Property.

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