# Napoleon's Theorem via Two Rotations

*Napoleon's theorem* claims that the centers A', B', C', of the equilateral triangles A''BC, AB''C, ABC'', erected on the sides (either all inwardly or all outwardly) of a given triangle ABC form an equilateral triangle. The applet below serves to illustrate a very simple proof of this result.

Use the scroll bar at the bottom of the applet to rotate B'C' 30° clockwise about A and A'C' 30° counterclockwise about B. B'C' is mapped onto SR, with S on AC and R on AC''. This is because

What if applet does not run? |

In an equilateral triangle the distance from a vertex to the center equals

AS/AC | = AB'/AC | = κ, | |

AR/AC'' | = AC'/AB | = κ, | |

BP/BC | = BA'/BC | = κ, | |

BQ/BC'' | = BC'/AB | = κ. |

It follows that triangles ACC'' and ASR are similar as are triangles BCC'' and BPQ, implying that PQ||CC''||RS and, in addition,

RS | = κCC" = PQ |

So that B'C' = A'C'. Similarly, A'B' = B'C' = A'C'.

### References

- M. R. F. Smyth,
__MacCool's Proof of Napoleon's Theorem__,*Irish Math. Soc. Bulletin*59 (2007), 71-77 (available online)

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