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 ∠B'AC = C'AC'' = 30 °. Similarly, A'C' maps onto PQ, with P on BC and Q on BC''.


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

In an equilateral triangle the distance from a vertex to the center equals κ = √3/3 of the side, meaning that

 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

  1. M. R. F. Smyth, MacCool's Proof of Napoleon's Theorem, Irish Math. Soc. Bulletin 59 (2007), 71-77 (available online)
[an error occurred while processing this directive]

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

[an error occurred while processing this directive]
[an error occurred while processing this directive]