If P is a point in the triangle ABC, then a point A, a point B and a point C are taken as the perpendicular lines of PA, Pb and PC to form a new triangle. Then you can only prove that this new triangle is an equilateral triangle when the angle formed by every two straight lines of PA, Pb and PC is 120 degrees and the sum of PA+PB+PC is minimum?

(1) The opposite opening angle of fermat point is 120 degrees. ? △CC 1B and △AA 1B, BC = ba 1, Ba = bc 1, ∠ CBC 1 = ∠ b+60 degrees = ∠ ABA/kloc. △CC 1B and △AA 1B are congruent triangles, and ∠PCB=∠PA 1B? Similarly, ∠CBP=∠CA 1P? From ∠PA 1B+∠CA 1P=60 degrees, ∠PCB+∠CBP=60 degrees, then ∠CPB= 120 degrees? Similarly, ∠APB= 120 degrees, ∠APC= 120 degrees? (2)PA+PB+PC=AA 1？ Rotating △BPC 60 degrees around point B coincides with △BDA 1, and then connecting PD, △PDB is an equilateral triangle, so ∠BPD=60 degrees? BPA = 120 degrees, so A, P and D are on the same straight line. And ∠CPB=∠A 1DB= 120 degrees, ∠PDB=60 degrees, ∠PDA 1= 180 degrees, so a, p, d, a. ? (3)PA+PB+PC is the shortest? Take any point M (not coincident with point P) in △ABC, connect AM, BM and CM, rotate △BMC 60 degrees around point B to coincide with △BGA 1, connect AM, GM, A 1G (same as above), and then AA 1