- Given an equilateral triangle ABC in the plane, and a point P in the plane of the triangle ABC, the lengths PA, PB, and PC form the sides of a (maybe, degenerate) triangle.
The proof is quick. Consider a rotation of 60° about the point C. Assume A maps to B, and P maps to P '. Then we have , and . Hence triangle PCP ' is equilateral and . It is obvious that . Thus, triangle PBP ' has sides equal to PA, PB, and PC and the proof by construction is complete.
Further investigations reveal that if P is not in the interior of the triangle, but rather on the circumcircle, then PA, PB, PC form a degenerate triangle, with the largest being equal to the sum of the others.