Minimum distance point proof for triangle with one angle >= 120

Hello,

I am having a little trouble developing a proof for the minimum distance point in the following scenario.

if i have a triangle with vertices A, B, and C.
measure of angle ABC is >=120 degrees
prove that the vertex B is the minimum distance point?

if there was another point call it P, then the distance from P to the vertices of the triangle which is
PA + PB + PC is greater than the distance from vertex B which is AB + BB+ BC = AB + BC.
That is saying that vertex B of triangle is the minimum distance point.

You might want to look at Fermat point.

Please help me construct a proof for this case.

Thank You

Quote:

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Originally Posted by hercules

Hello,

I am having a little trouble developing a proof for the minimum distance point in the following scenario.

if i have a triangle with vertices A, B, and C.
measure of angle ABC is >=120 degrees
prove that the vertex B is the minimum distance point?

if there was another point call it P, then the distance from P to the vertices of the triangle which is
PA + PB + PC is greater than the distance from vertex B which is AB + BB+ BC = AB + BC.
That is saying that vertex B of triangle is the minimum distance point.

You might want to look at Fermat point.

Please help me construct a proof for this case.

Thank You

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did the solution i suggested work out for this?

Quote:

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Originally Posted by Jhevon

did the solution i suggested work out for this?

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Yes, it did. We didn't even need to calculate the length using sine values
Since our triangle PBP' was isosceles by construction. We had only to realize that as the vertex angle decreased (which we found to be restricted between 0 and 60 degrees) the length of the base PP' also decreased.

Sorry, for the delayed response. I found math forum mail filtered.