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