Show Triangle ABC Similarity Type Independence of B,C

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Discussion Overview

The discussion revolves around the similarity type of triangle ABC formed by points A, B, and C, where points B and C lie on circles K and L, respectively. Participants explore whether the similarity type is independent of the positions of B and C on their respective circles, and they delve into the implications of the Inscribed Angle Theorem in this context.

Discussion Character

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

Main Points Raised

  • Some participants seek clarification on what is meant by the "similarity type" of a triangle.
  • There is a suggestion that the similarity type refers to the angles of the triangle being the same regardless of the positions of B and C.
  • One participant proposes that if angles at points B and C remain constant as they move along their respective circles, then triangles ABC and AB'C' are similar.
  • Another participant expresses doubt, suggesting that moving point B could result in a triangle that is not similar to the original.
  • Participants discuss the Inscribed Angle Theorem as a basis for why angles at points B and C remain constant.
  • A later reply raises a new question about demonstrating the independence of the length of segment BD from the positions of B and C.
  • There is a discussion about whether the angle at A being constant implies that the arc BD must also be the same, leading to a conclusion about the lengths of BD and the arc.

Areas of Agreement / Disagreement

Participants express differing views on whether the similarity type of triangle ABC is indeed independent of the positions of B and C. While some argue that the angles remain constant and thus the triangles are similar, others question this assumption. The discussion remains unresolved regarding the independence of the length of segment BD.

Contextual Notes

Participants mention the Inscribed Angle Theorem, but there are uncertainties about the implications of moving points B and C and how that affects similarity and lengths. The discussion does not resolve these uncertainties.

mathmari
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Hey! :o

Let K,L be circles with exactly two intersection points S and A. Let $C\in L$ and $B\in K$ with $C,B\neq S,A$. I want to show the similarity type of the triangle ABC is independent of the place of B and C.

Could you explain to me what I am supposed to show?
 
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Hey mathmari! (Smile)

Can you clarify what a similarity type of a triangle is?
Google doesn't seem to help.

I thought it might be the type of triangle, but whichever type we pick (equilateral. isoscales, scalene, acute, oblique, or right), we can draw circles through their corners.
And that would meet the conditions. (Thinking)
 
I like Serena said:
Can you clarify what a similarity type of a triangle is?
Google doesn't seem to help.

I thought it might be the type of triangle, but whichever type we pick (equilateral. isoscales, scalene, acute, oblique, or right), we can draw circles through their corners.
And that would meet the conditions. (Thinking)

I think that it is meant that a triangle ABC is always similar to an other triangle AB'C' no matter where B or B' are on the circle K and where C or C' are on the circle L.

So, do we have to consider two such triangles ans show that they are similar? (Wondering)
 
mathmari said:
I think that it is meant that a triangle ABC is always similar to an other triangle AB'C' no matter where B or B' are on the circle K and where C or C' are on the circle L.

So, do we have to consider two such triangles ans show that they are similar? (Wondering)

Well, if we make a drawing, I think we can immediately see that given a triangle ABC, if we pick B somewhere else, that it will not be similar. (Worried)
 
I like Serena said:
Well, if we make a drawing, I think we can immediately see that given a triangle ABC, if we pick B somewhere else, that it will not be similar. (Worried)
I thought now the following:

View attachment 6988

Given that AS is fixed we have that the angle of C will be the same no matter where it is on the circle, so Angle(C)=Angle(C').
The same holds also for B, i.e., Angle(B)=Angle(B').
Since Angle(C)=Angle(C') and Angle(B)=Angle(B'), we conclude that the two triangles ABC and AB'C' have all angles the same, and they are similar.
So, we conclude that all triangles are similar no matter where B and C are.

Is this correct? (Wondering)
 

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mathmari said:
I thought now the following:

Given that AS is fixed we have that the angle of C will be the same no matter where it is on the circle, so Angle(C)=Angle(C').
The same holds also for B, i.e., Angle(B)=Angle(B').
Since Angle(C)=Angle(C') and Angle(B)=Angle(B'), we conclude that the two triangles ABC and AB'C' have all angles the same, and they are similar.
So, we conclude that all triangles are similar no matter where B and C are.

Is this correct?

Aha! So we would have the extra restriction that S must be on BC.
Let me give that a try:
\begin{tikzpicture}[ultra thick]
\def\rK{3};
\def\rL{2};
\path
(0,0) coordinate (K)
+(30:\rK) coordinate (A) node[above] {A}
+(-30:\rK) coordinate (S) node[below] {S}
+(315:\rK) coordinate (C)
+(195:\rK) coordinate (C')
(4,0) coordinate (L)
+(60:\rL) coordinate (B)
+(-60:\rL) coordinate (B')
;
\draw (K) circle (\rK) (L) circle (\rL);
\draw[orange] (A) -- (B) node[above right] {B} -- (C) node
{C} -- cycle;
\draw[cyan] (A) -- (B') node[below right] {B'} -- (C') node
{C'} -- cycle;
\end{tikzpicture}

Yep. Looks as if that would work. (Happy)

Why would the angle at C be the same as the angle at C'? (Wondering)​
 
I like Serena said:
Why would the angle at C be the same as the angle at C'? (Wondering)

Isn't it because all the angles over the arc $AS$ on the circle are the same? (Wondering)
 
mathmari said:
Isn't it because all the angles over the arc $AS$ on the circle are the same?

Ah yes, the Inscribed angle theorem.

Then it's all good! (Nod)
 
I like Serena said:
Ah yes, the Inscribed angle theorem.

Then it's all good! (Nod)
(Nerd) (Yes)

If $CA$ intersects the circle $K$ at a second point $D$, how can we show that the length of $\overline{BD}$ is independent of the choice of $B$ and $C$ ? (Wondering)

Could you give me a hint? (Wondering)
 
Last edited by a moderator:
  • #10
We have the following drawing:

View attachment 6990

If we consider the arc $\overline{BD}$ we have that that angle $A$ is always the same (from the first question). Do we get from that the arc must also be the same? (Wondering)
 

Attachments

  • ge.png
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    5.1 KB · Views: 105
  • #11
mathmari said:
If $CA$ intersects the circle $K$ at a second point $D$, how can we show that the length of $\overline{BD}$ is independent of the choice of $B$ and $C$ ?

Could you give me a hint?

mathmari said:
We have the following drawing:

If we consider the arc $\overline{BD}$ we have that that angle $A$ is always the same (from the first question). Do we get from that the arc must also be the same?

Yes. I believe so.

Since we know that the angle $A$ is always the same, it follows by applying the Inscribed angle theorem in reverse, that the chord $BD$ always has the same length. Consequently the arc $BD$ will also always have the same length. (Thinking)
 
  • #12
I like Serena said:
Yes. I believe so.

Since we know that the angle $A$ is always the same, it follows by applying the Inscribed angle theorem in reverse, that the chord $BD$ always has the same length. Consequently the arc $BD$ will also always have the same length. (Thinking)

I see! Thank you very much! (Sun)
 

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