First I want to thank an anonymous Person who Formulated this Post for me.

Napoleon’s Theorem:

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.

Proof:

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

(1) A+B+C=0

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

B+R(A-B)

The center of this equilateral triangle is then

(A+B+(B+R(A-B)))/3

Using (1), we may write this as

(3) ((B-C)+ R(A-B))/3

By symmetry the centers of the other equilateral triangles are

(4) ((C-A)+R(B-C))/3

(5) ((A-B)+R(C-A))/3

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

R^2(B-C)+R^3(A-B)

Applying (2) and the fact R^3=-1 turns this into

R(B-C)-(B-C)-(A-B)=(C-A)+R(B-C)

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.

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