First I want to thank an anonymous Person who Formulated this Post for me.
Let ABC be a triangle. Let equilateral triangles be constructed outward at each side of the triangle ABC. Then the midpoints of these equilateral triangles are the vertices of another equilateral triangle.
We use vector calculus. Let A, B, C denote the vectors pointing to the vertices. By translating the tirangle so that the barycenter of it becomes the origin, we may assume that
Let R be a the linear map which turns each vector by 60 degrees to the right. Note that R^3 = -1and for every vector X we have the identity
(2) R^2 X = RX – X
The tip of the equilateral triangle above the side from A to B is
The center of this equilateral triangle is then
Using (1), we may write this as
(3) ((B-C)+ R(A-B))/3
By symmetry the centers of the other equilateral triangles are
We need to prove that these three points are permuted when we rotate by 120 degrees aound the origin, that is R^2. For simplicity we may multiply the three vectors by 3. Applying R^2 to 3 times expression (3) gives
Applying (2) and the fact R^3=-1 turns this into
This evidently is three times (4). By symmetry, the three points (3),(4),(5) are permuted when rotated by R^2, which proves that they form an equilateral triangle.