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'.
- 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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