# Can someone explain this proof to me?

I posted this in the geometry subforum, but I think that might have been the wrong place to post this.

Given two triangles with vertices A1, B1, C1 and A2, B2, C2 respectively. A1A2, B1B2, C1C2 are extended to meet at a point V say. Now, B1C1 and B2C2 are extended to meet at L, A1B1 and A2B2 meet at N and A1C1 and A2C2 meet at M. Prove that L, M and N are concurrent.

Proof (as given in text):

Let A1B1C1 be the reference triangle and V be the unit point (1,1,1). A2 is on the join of A1(1,0,0) and V(1,1,1), so it can be taken as (1+p,1,1). Similarly, the point B2 is given by (1,1+q,1) and C2 by (1,1,1+r).

Now, the line B2C2 is

$$\left|\stackrel{\stackrel{x}{1}}{1}\stackrel{\stackrel{y}{1+q}}{1}\stackrel{\stackrel{z}{1}}{1+r} \right|$$ = 0.

The point L is given by x = 0, y{1-(1+r)} + z{1 - (1+q)} = 0

i.e. x=0, $$\frac{y}{q} + \frac{z}{r}$$ = 0

Therefore, L lies on the line $$\frac{x}{p}+ \frac{y}{q}+ \frac{z}{r}$$ = 0. By symmetry, so do M and N.

Hence proved

From start to finish, I can't get it. Can someone please explain to me what all this means?

For the diagram that goes with the proof, please refer to my topic in the geometry section

tiny-tim
Homework Helper
Hi anirudh215!
Let A1B1C1 be the reference triangle and V be the unit point (1,1,1). A2 is on the join of A1(1,0,0) and V(1,1,1), so it can be taken as (1+p,1,1). Similarly, the point B2 is given by (1,1+q,1) and C2 by (1,1,1+r).

https://www.physicsforums.com/attachment.php?attachmentid=21094&d=1255422239"

Sorry, I don't understand this at all

how are A B and C (1,0,0) etc? and why is V (1,1,1)? and what's a "reference triangle"?

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Exactly what I want to know! I saw this in a geometry book entitled "Teaching of Higher Geometry in schools - A report for the Mathematical Association" Published by Bell and Sons. I've been blinking over many proofs in this text. What gets me down is it says it's for high school students!

tiny-tim
Homework Helper
mathworld and archive.org

hmm … that's at http://openlibrary.org/b/OL6159907M/teaching_of_higher_geometry_in_schools", but with no preview … why are you looking at a book published in 1953?

This proof seems to be using trilinear coordinates … see Eric W. Weisstein's http://mathworld.wolfram.com/ReferenceTriangle.html" [Broken]

You can find downloadable books on "trilinear coordinates" (or any subject!) by doing a search at archive.org … see http://www.archive.org/search.php?query=trilinear AND mediatype:texts"

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hmm … that's at http://openlibrary.org/b/OL6159907M/teaching_of_higher_geometry_in_schools", but with no preview … why are you looking at a book published in 1953?

Heh. I found it on my prof's shelf, and while he made me wait in his office (as all professors do), I picked up this one and began reading it. It had a few very cool proofs. This one was so intriguing, I just HAD TO find out about it.

So I did find out about it. It's not trilinear co ordinates but is something that is called barycentric co-ordinates. If you are also interested, you can find out more about it from H.M. Coxeter, Introduction to Geometry, page 216. This is damn interesting!!

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tiny-tim
Homework Helper
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Oops. Well, I don't know what trilinear coordinates are either, so I thought you'd made a mistake. Sorry. But that's basically it. Now how does one explain the problem?

tiny-tim