Proving Segment Measure in Triangle ABC: BM+CM < AB+AC

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Homework Help Overview

The discussion revolves around proving the inequality involving segment measures in triangle ABC, specifically that BM + CM < AB + AC, where M is a point inside the triangle. The participants are exploring the application of the Triangle Inequality Theorem and considering various approaches to the proof.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using the Triangle Inequality Theorem and consider proving by contradiction or demonstrating that BM < AB and CM < AC. Some suggest drawing additional segments to create more inequalities, while others question the validity of certain assumptions regarding the position of point M.

Discussion Status

The discussion is ongoing, with participants offering various ideas and approaches. Some have suggested examining degenerate cases or considering the geometric properties of ellipses related to points B and C. There is no explicit consensus yet, but multiple lines of reasoning are being explored.

Contextual Notes

Participants note that the position of point M can vary, which complicates the proof. There are also references to constraints regarding the assumptions that can be made about the triangle and the nature of the inequalities involved.

Cantor
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I have a question involving the measure of segments

If m is a point inside a triangle ABC how could we prove that segment BM+CM < AB+AC. I am trying to use the Triangle Inequality Theorem (If A, B, C are three non-collinear points then AC < AB+BC) but I am stuck.

Should I prove by contradiction or show that BM < AB and CM <AC

Any help would be appreciated, Thanks:smile:

th_26657_triangle.jpg
 
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My first instinct is to draw some more lines. (e.g. maybe draw in the segment AM, or extend BM and CM to intersect the opposite sides of the triangle)

This would give me more inequalities to play with, and maybe I could derive something from there.


P.S. BM doesn't have to be less than AB. A counterexample isn't hard to find -- just draw AB and BM so that BM > AM, and then try and draw in the rest of the diagram.
 
Just make sure that whatever inequalities you end up working with, knowing they are true ensures that m is inside ABC.
 
Try thinking of points B and C as foci of an ellipse with A on the boundary.

Carl
 
I have tried all the possible inequalities but still nothing, the problem i am having is M could be anywhere so AB could be bigger or small them BM depending. Any other ideas by any chance.:smile:

Thanks

Cantor
 
Try thinking of points B and C as foci of an ellipse with A on the boundary.
Based on the title of the post, I think he's at a level where he's not allowed to simply assume that one ellipse lies inside the other, and he's not in a class where it's fair game to assume the Jordan curve theorem.

And the thought of studying conic sections in a (possibly) non-Euclidean setting makes me shudder!



Anyways, back to the original question -- I think looking at degenerate cases might help. They're often easier. What if M lies at one of the vertices? Or on one of the sides?

And, obviously, the theorem isn't true if M lies outside the triangle, so you have to use some fact that only works when M is inside.

(And, there's still the chance that inspired... or brute force... manipulation of the inequalities will provide the answer)
 
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