C.a.R. > Applications > Triangle Geometry > Traversals

On each side of a triangle, we have equilateral triangles. The blue traversals are then of equal length and intersect in the Fermat point. The angels at the intersection are 60° each.

For a proof, write

APa = AB + d(BC),

where d is the mapping rotating the vector BC by 60°. Two more rotations applied to both sides of this equation yield

d(d(APa) = d(d(AB)) + d(d(d(BC) = d(BA) + CB = CPc.

Thus two traversals are equal in length and meet at 120°.