Geometrical Proof for Euclidian Triangle Circle Property

[Phrak]

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.

 

[tiny-tim]

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!

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.

 

[Phrak]

Hi Phrak!

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.

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

 

[tiny-tim]

And I was just bemoaning the fact that no one responded.

I think most PF members don't like "real" geometry!

But wait. What's a circumcentre, and what angle from whence and to where?

he he

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.

join CAMREG … the CAMpaign for REal Geometry!​

 

[Phrak]

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?

 

[tiny-tim]

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 …

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! )

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

and then for your problem, simply do the same proof "backwards" ​

 

[Phrak]

ooh, I'm sorry, Phrak …

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.