# Napoleon's Theorem by Transformation

On each side of a triangle, erect an equilateral triangle, lying exterior to the original triangle. Then the segments connecting the centroids of the three equilateral triangles themselves form an equilateral triangle. |

We'll denote the Napoleonean triangles ABC", BCA", CAB" and there centers C', A', and B', respectively. The counterclockwise rotation through 120° around A' moves B into C; the rotation around B' similarly moves C into A and the rotation around C' moves A onto B. The product of the three rotation is a rotation through 360°, i.e., a translation with point B fixed. Which means that it is the identity transform.

(The applet below allows you to experiment with the 120° rotations around the centers of the Napoleonean triangles in turn. In addition to the three vertices A, B, C, one other point is moveable by the cursor. A sequence of rotations is selected by pressing the A', B', C' buttons, or by popping the most recent rotation. The sequence of rotations applies to the fourth point. You can verify the assertion that the sequence of rotations around A', B', C' leaves B fixed by placing the fourth point at B. Similarly, one can verify the next step of the proof that follows.)

What if applet does not run? |

Since the product of rotations in the sequence A', B', C' is the identity transform, it leaves C' fixed, but so does the rotation around C' alone. Thus the product of rotations around A' and B' also leaves C' fixed. Assume C' is rotated around A' into say P' which is then rotated around B' back into C'. In the quadrilateral C'A'P'B', angles C'A'P' and P'B'C' are 120° and the triangles C'A'P' and P'B'C' are isosceles. Their base angles are 30°, so that

I am grateful to Hubert Shutrick for bringing this proof to my attention. A proof based on similar consideration is found in the classic *Geometric Transformations I* by I. M. Yaglom as Problem 22(a). The product of rotations around A' and B' is found at the intersections of two lines inclined 60° to A'B' through the end points of that segment. This is a rotation through 240° that leaves C' fixed so that this is a rotation around C'. The two observations together imply that ΔA'B'C' is equilateral.

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